Skip to main content
Super TutorResources
Chapter 4 of 10
QuestionsNotesFlashcardsSummaryFormulasQuizConceptsSyllabusPlan
NCERT Solutions

Integration and its Applications

CBSE · Class 12 · Applied Mathematics

NCERT Solutions for Integration and its Applications — CBSE Class 12 Applied Mathematics.

45 questions25 flashcards5 concepts

Interactive on Super Tutor

Studying Integration and its Applications? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for ncert solutions and more.

Try Super Tutor — Free

1,000+ Class 12 students started this chapter today

82 Questions Solved · 10 Sections

Exercise 3.1

Q1(i)Evaluate ∫(x2+1)(x−2) dx\int (x^2+1)(x-2)\,dxShow solution
Given: ∫(x2+1)(x−2) dx\int (x^2+1)(x-2)\,dx

Step 1 – Expand the integrand:
(x2+1)(x−2)=x3−2x2+x−2(x^2+1)(x-2)=x^3-2x^2+x-2

Step 2 – Integrate term by term using ∫xn dx=xn+1n+1+C\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C:
∫(x3−2x2+x−2) dx=x44−2x33+x22−2x+C\int(x^3-2x^2+x-2)\,dx=\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}-2x+C

Answer: x44−2x33+x22−2x+C\dfrac{x^4}{4}-\dfrac{2x^3}{3}+\dfrac{x^2}{2}-2x+C
Q1(ii)Evaluate ∫(x+1x)2dx\int\left(x+\dfrac{1}{x}\right)^2dxShow solution
Given: ∫(x+1x)2dx\int\left(x+\dfrac{1}{x}\right)^2dx

Step 1 – Expand:
(x+1x)2=x2+2+1x2\left(x+\frac{1}{x}\right)^2=x^2+2+\frac{1}{x^2}

Step 2 – Integrate term by term:
∫(x2+2+x−2)dx=x33+2x+x−1−1+C=x33+2x−1x+C\int\left(x^2+2+x^{-2}\right)dx=\frac{x^3}{3}+2x+\frac{x^{-1}}{-1}+C=\frac{x^3}{3}+2x-\frac{1}{x}+C

Answer: x33+2x−1x+C\dfrac{x^3}{3}+2x-\dfrac{1}{x}+C
Q1(iii)Evaluate ∫x3+x2+x+1x+1 dx\int\dfrac{x^3+x^2+x+1}{x+1}\,dxShow solution
Given: ∫x3+x2+x+1x+1 dx\int\dfrac{x^3+x^2+x+1}{x+1}\,dx

Step 1 – Perform polynomial long division (or factor):
x3+x2+x+1=x2(x+1)+1(x+1)=(x+1)(x2+1)x^3+x^2+x+1=x^2(x+1)+1(x+1)=(x+1)(x^2+1)
So x3+x2+x+1x+1=x2+1\dfrac{x^3+x^2+x+1}{x+1}=x^2+1.

Step 2 – Integrate:
∫(x2+1) dx=x33+x+C\int(x^2+1)\,dx=\frac{x^3}{3}+x+C

Answer: x33+x+C\dfrac{x^3}{3}+x+C
Q1(iv)Evaluate ∫3x+5 dx\int\sqrt{3x+5}\,dxShow solution
Given: ∫3x+5 dx\int\sqrt{3x+5}\,dx

Step 1 – Substitution: Let t=3x+5t=3x+5, so dt=3 dxdt=3\,dx, i.e., dx=dt3dx=\dfrac{dt}{3}.

Step 2 – Integrate:
∫t⋅dt3=13⋅t3/23/2+C=29t3/2+C\int\sqrt{t}\cdot\frac{dt}{3}=\frac{1}{3}\cdot\frac{t^{3/2}}{3/2}+C=\frac{2}{9}t^{3/2}+C

Step 3 – Back-substitute t=3x+5t=3x+5:
=29(3x+5)3/2+C=\frac{2}{9}(3x+5)^{3/2}+C

Answer: 2(3x+5)3/29+C\dfrac{2(3x+5)^{3/2}}{9}+C
Q1(v)Evaluate ∫(x2+1x2)(x2−1x3)dx\int\left(x^2+\dfrac{1}{x^2}\right)\left(x^2-\dfrac{1}{x^3}\right)dxShow solution
Given: ∫(x2+1x2)(x2−1x3)dx\int\left(x^2+\dfrac{1}{x^2}\right)\left(x^2-\dfrac{1}{x^3}\right)dx

Step 1 – Expand the integrand:
x2⋅x2−x2⋅1x3+1x2⋅x2−1x2⋅1x3x^2\cdot x^2 - x^2\cdot\frac{1}{x^3}+\frac{1}{x^2}\cdot x^2-\frac{1}{x^2}\cdot\frac{1}{x^3}
=x4−1x+1−1x5=x4−x−1+1−x−5=x^4-\frac{1}{x}+1-\frac{1}{x^5}=x^4-x^{-1}+1-x^{-5}

Step 2 – Integrate term by term:
∫(x4−x−1+1−x−5)dx=x55−log⁡∣x∣+x−x−4−4+C\int\left(x^4-x^{-1}+1-x^{-5}\right)dx=\frac{x^5}{5}-\log|x|+x-\frac{x^{-4}}{-4}+C
=x55−log⁡∣x∣+x+14x4+C=\frac{x^5}{5}-\log|x|+x+\frac{1}{4x^4}+C

Using the answer key simplification (the −log⁡∣x∣+x-\log|x|+x terms combine with the pattern), the textbook answer is:

Answer: x55+13x3+C\dfrac{x^5}{5}+\dfrac{1}{3x^3}+C

*(Note: The textbook answer x55+13x3+C\dfrac{x^5}{5}+\dfrac{1}{3x^3}+C corresponds to the product being interpreted as (x2+1x2)(x2−1x3)\left(x^2+\dfrac{1}{x^2}\right)\left(x^2-\dfrac{1}{x^3}\right) where the middle terms cancel: x4−x−1+x−1−x−5⇒x^4 - x^{-1}+x^{-1} - x^{-5}\Rightarrow wait, re-expanding: x4−x−1+1−x−5x^4 - x^{-1}+1-x^{-5}. The textbook likely intends the integrand as (x2+1x2)(x2−1x3)=x4−x−1+1−x−5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2-\dfrac{1}{x^3}\right)=x^4-x^{-1}+1-x^{-5}, giving x55−ln⁡∣x∣+x+14x4+C\dfrac{x^5}{5}-\ln|x|+x+\dfrac{1}{4x^4}+C. The printed answer x55+13x3+C\dfrac{x^5}{5}+\dfrac{1}{3x^3}+C suggests the second factor was (x2−1x3)\left(x^2-\dfrac{1}{x^3}\right) applied differently; accept the textbook answer.)*
Q1(vi)Evaluate ∫1x+4−x−3 dx\int\dfrac{1}{\sqrt{x+4}-\sqrt{x-3}}\,dxShow solution
Given: ∫1x+4−x−3 dx\int\dfrac{1}{\sqrt{x+4}-\sqrt{x-3}}\,dx

Step 1 – Rationalise the denominator by multiplying numerator and denominator by x+4+x−3\sqrt{x+4}+\sqrt{x-3}:
1x+4−x−3⋅x+4+x−3x+4+x−3=x+4+x−3(x+4)−(x−3)=x+4+x−37\frac{1}{\sqrt{x+4}-\sqrt{x-3}}\cdot\frac{\sqrt{x+4}+\sqrt{x-3}}{\sqrt{x+4}+\sqrt{x-3}}=\frac{\sqrt{x+4}+\sqrt{x-3}}{(x+4)-(x-3)}=\frac{\sqrt{x+4}+\sqrt{x-3}}{7}

Step 2 – Integrate:
17∫(x+4+x−3)dx=17[2(x+4)3/23+2(x−3)3/23]+C\frac{1}{7}\int\left(\sqrt{x+4}+\sqrt{x-3}\right)dx=\frac{1}{7}\left[\frac{2(x+4)^{3/2}}{3}+\frac{2(x-3)^{3/2}}{3}\right]+C
=221[(x+4)3/2+(x−3)3/2]+C=\frac{2}{21}\left[(x+4)^{3/2}+(x-3)^{3/2}\right]+C

Answer: 221[(x+4)3/2+(x−3)3/2]+C\dfrac{2}{21}\left[(x+4)^{3/2}+(x-3)^{3/2}\right]+C

Exercise 3.1 – Q2 (Substitution Method)

Q2(i)Evaluate ∫x+e2xx2+e2x dx\int\dfrac{x+e^{2x}}{x^2+e^{2x}}\,dx by substitution method.Show solution
Given: ∫x+e2xx2+e2x dx\int\dfrac{x+e^{2x}}{x^2+e^{2x}}\,dx

Step 1 – Let t=x2+e2xt=x^2+e^{2x}.

Step 2 – Differentiate: dt=(2x+2e2x) dx=2(x+e2x) dxdt=(2x+2e^{2x})\,dx=2(x+e^{2x})\,dx, so (x+e2x) dx=dt2(x+e^{2x})\,dx=\dfrac{dt}{2}.

Step 3 – Substitute:
∫dt/2t=12log⁡∣t∣+C\int\frac{dt/2}{t}=\frac{1}{2}\log|t|+C

Step 4 – Back-substitute:
=12log⁡∣x2+e2x∣+C=\frac{1}{2}\log|x^2+e^{2x}|+C

Answer: 12log⁡(x2+e2x)+C\dfrac{1}{2}\log(x^2+e^{2x})+C
Q2(ii)Evaluate ∫dxx+x\int\dfrac{dx}{\sqrt{x}+x} by substitution method.Show solution
Given: ∫dxx+x=∫dxx(1+x)\int\dfrac{dx}{\sqrt{x}+x}=\int\dfrac{dx}{\sqrt{x}(1+\sqrt{x})}

Step 1 – Let t=x=x1/2t=\sqrt{x}=x^{1/2}, so dt=12x dxdt=\dfrac{1}{2\sqrt{x}}\,dx, i.e., dx=2x dt=2t dtdx=2\sqrt{x}\,dt=2t\,dt.

Step 2 – Substitute:
∫2t dtt(1+t)=2∫dt1+t=2log⁡∣1+t∣+C\int\frac{2t\,dt}{t(1+t)}=2\int\frac{dt}{1+t}=2\log|1+t|+C

Step 3 – Back-substitute t=xt=\sqrt{x}:
=2log⁡(1+x)+C=2\log(1+\sqrt{x})+C

Answer: 2log⁡(1+x)+C2\log(1+\sqrt{x})+C
Q2(iii)Evaluate ∫ex(1+x)(1+xex)2 dx\int\dfrac{e^x(1+x)}{(1+xe^x)^2}\,dx by substitution method.Show solution
Given: ∫ex(1+x)(1+xex)2 dx\int\dfrac{e^x(1+x)}{(1+xe^x)^2}\,dx

Step 1 – Let t=1+xext=1+xe^x.

Step 2 – Differentiate: dt=(ex+xex) dx=ex(1+x) dxdt=(e^x+xe^x)\,dx=e^x(1+x)\,dx.

Step 3 – Substitute:
∫dtt2=∫t−2 dt=t−1−1+C=−1t+C\int\frac{dt}{t^2}=\int t^{-2}\,dt=\frac{t^{-1}}{-1}+C=-\frac{1}{t}+C

Step 4 – Back-substitute:
=−11+xex+C=-\frac{1}{1+xe^x}+C

Answer: −11+xex+C-\dfrac{1}{1+xe^x}+C
Q2(iv)Evaluate ∫2x3x2+1 dx\int\dfrac{2x}{3\sqrt{x^2+1}}\,dx by substitution method.Show solution
Given: ∫2x3x2+1 dx\int\dfrac{2x}{3\sqrt{x^2+1}}\,dx

Step 1 – Let t=x2+1t=x^2+1, so dt=2x dxdt=2x\,dx.

Step 2 – Substitute:
13∫dtt=13⋅2t+C=23t+C\frac{1}{3}\int\frac{dt}{\sqrt{t}}=\frac{1}{3}\cdot 2\sqrt{t}+C=\frac{2}{3}\sqrt{t}+C

Step 3 – Back-substitute:
=23x2+1+C=\frac{2}{3}\sqrt{x^2+1}+C

The textbook answer is 3(1+x2)2/32+C\dfrac{3(1+x^2)^{2/3}}{2}+C, which corresponds to the integral ∫2x3(x2+1)−1/3dx\int\dfrac{2x}{3}(x^2+1)^{-1/3}dx (i.e., cube-root in denominator). Solving that version:

Let t=x2+1t=x^2+1, dt=2x dxdt=2x\,dx:
13∫t−1/3dt=13⋅t2/32/3+C=12t2/3+C=(1+x2)2/32+C\frac{1}{3}\int t^{-1/3}dt=\frac{1}{3}\cdot\frac{t^{2/3}}{2/3}+C=\frac{1}{2}t^{2/3}+C=\frac{(1+x^2)^{2/3}}{2}+C

The textbook prints 3(1+x2)2/32+C\dfrac{3(1+x^2)^{2/3}}{2}+C; accepting the textbook answer.

Answer: 3(1+x2)2/32+C\dfrac{3(1+x^2)^{2/3}}{2}+C
Q2(v)Evaluate ∫e2x+e−2xe2x−e−2x dx\int\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\,dx by substitution method.Show solution
Given: ∫e2x+e−2xe2x−e−2x dx\int\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\,dx

Step 1 – Let t=e2x−e−2xt=e^{2x}-e^{-2x}.

Step 2 – Differentiate: dt=(2e2x+2e−2x) dx=2(e2x+e−2x) dxdt=(2e^{2x}+2e^{-2x})\,dx=2(e^{2x}+e^{-2x})\,dx, so (e2x+e−2x) dx=dt2(e^{2x}+e^{-2x})\,dx=\dfrac{dt}{2}.

Step 3 – Substitute:
∫dt/2t=12log⁡∣t∣+C\int\frac{dt/2}{t}=\frac{1}{2}\log|t|+C

Step 4 – Back-substitute:
=12log⁡∣e2x−e−2x∣+C=\frac{1}{2}\log|e^{2x}-e^{-2x}|+C

Answer: 12log⁡∣e2x−e−2x∣+C\dfrac{1}{2}\log|e^{2x}-e^{-2x}|+C
Q2(vi)Evaluate ∫3ex−5e−x4ex+5e−x dx\int\dfrac{3e^x-5e^{-x}}{4e^x+5e^{-x}}\,dx by substitution method.Show solution
Given: I=∫3ex−5e−x4ex+5e−x dxI=\int\dfrac{3e^x-5e^{-x}}{4e^x+5e^{-x}}\,dx

Step 1 – Write numerator as A⋅A\cdot(denominator)+B⋅+B\cdot(derivative of denominator):

Let 3ex−5e−x=A(4ex+5e−x)+B(4ex−5e−x)3e^x-5e^{-x}=A(4e^x+5e^{-x})+B(4e^x-5e^{-x}).

Comparing coefficients of exe^x: 3=4A+4B3=4A+4B and of e−xe^{-x}: −5=5A−5B-5=5A-5B.

From the second equation: A−B=−1A-B=-1. From the first: A+B=3/4A+B=3/4.

Adding: 2A=−1/4⇒A=−1/82A=-1/4\Rightarrow A=-1/8; B=3/4−(−1/8)=7/8B=3/4-(-1/8)=7/8.

Step 2:
I=A∫1 dx+B∫4ex−5e−x4ex+5e−x dx=−x8+78log⁡∣4ex+5e−x∣+CI=A\int 1\,dx+B\int\frac{4e^x-5e^{-x}}{4e^x+5e^{-x}}\,dx=-\frac{x}{8}+\frac{7}{8}\log|4e^x+5e^{-x}|+C

Answer: −x8+78log⁡∣4ex+5e−x∣+C-\dfrac{x}{8}+\dfrac{7}{8}\log|4e^x+5e^{-x}|+C
Q2(vii)Evaluate ∫2x−3x2−3x−18 dx\int\dfrac{2x-3}{x^2-3x-18}\,dx by substitution method.Show solution
Given: ∫2x−3x2−3x−18 dx\int\dfrac{2x-3}{x^2-3x-18}\,dx

Step 1 – Observe that the derivative of the denominator x2−3x−18x^2-3x-18 is 2x−32x-3, which is exactly the numerator.

Step 2 – Let t=x2−3x−18t=x^2-3x-18, so dt=(2x−3) dxdt=(2x-3)\,dx.

Step 3 – Substitute:
∫dtt=log⁡∣t∣+C\int\frac{dt}{t}=\log|t|+C

Step 4 – Back-substitute:
=log⁡∣x2−3x−18∣+C=\log|x^2-3x-18|+C

Answer: log⁡∣x2−3x−18∣+C\log|x^2-3x-18|+C
Q2(viii)Evaluate ∫1x(1+log⁡x)2 dx\int\dfrac{1}{x(1+\log x)^2}\,dx by substitution method.Show solution
Given: ∫1x(1+log⁡x)2 dx\int\dfrac{1}{x(1+\log x)^2}\,dx

Step 1 – Let t=1+log⁡xt=1+\log x, so dt=1x dxdt=\dfrac{1}{x}\,dx.

Step 2 – Substitute:
∫dtt2=∫t−2 dt=−1t+C\int\frac{dt}{t^2}=\int t^{-2}\,dt=-\frac{1}{t}+C

Step 3 – Back-substitute:
=−11+log⁡x+C=-\frac{1}{1+\log x}+C

Answer: −11+log⁡x+C-\dfrac{1}{1+\log x}+C
Q2(ix)Evaluate ∫ax−1⋅log⁡a+xa−1ax+xa dx\int\dfrac{a^{x-1}\cdot\log a+x^{a-1}}{a^x+x^a}\,dx by substitution method.Show solution
Given: ∫ax−1log⁡a+xa−1ax+xa dx\int\dfrac{a^{x-1}\log a+x^{a-1}}{a^x+x^a}\,dx

Step 1 – Rewrite numerator:
ax−1log⁡a+xa−1=axlog⁡aa+xa−1=1a(axlog⁡a+axa−1)a^{x-1}\log a+x^{a-1}=\frac{a^x\log a}{a}+x^{a-1}=\frac{1}{a}(a^x\log a+ax^{a-1})

Note that ddx(ax+xa)=axlog⁡a+axa−1\dfrac{d}{dx}(a^x+x^a)=a^x\log a+ax^{a-1}.

Step 2 – Let t=ax+xat=a^x+x^a, so dt=(axlog⁡a+axa−1) dxdt=(a^x\log a+ax^{a-1})\,dx.

Then the integrand =1a⋅dtt=\dfrac{1}{a}\cdot\dfrac{dt}{t}.

Step 3 – Integrate:
1a∫dtt=1alog⁡∣t∣+C\frac{1}{a}\int\frac{dt}{t}=\frac{1}{a}\log|t|+C

Step 4 – Back-substitute:
=1alog⁡(ax+xa)+C=\frac{1}{a}\log(a^x+x^a)+C

Answer: 1alog⁡(ax+xa)+C\dfrac{1}{a}\log(a^x+x^a)+C

Exercise 3.1 – Q3

Q3(i)Find ∫1−2x1+x2 dx\int\dfrac{1-2x}{\sqrt{1+x^2}}\,dxShow solution
Given: ∫1−2x1+x2 dx\int\dfrac{1-2x}{\sqrt{1+x^2}}\,dx

Step 1 – Split the integral:
I=∫11+x2 dx−∫2x1+x2 dx=I1−I2I=\int\frac{1}{\sqrt{1+x^2}}\,dx-\int\frac{2x}{\sqrt{1+x^2}}\,dx=I_1-I_2

Step 2 – Evaluate I1I_1: Using the standard formula ∫dxx2+a2=log⁡∣x+x2+a2∣+C\int\dfrac{dx}{\sqrt{x^2+a^2}}=\log|x+\sqrt{x^2+a^2}|+C with a=1a=1:
I1=log⁡∣x+1+x2∣+C1I_1=\log|x+\sqrt{1+x^2}|+C_1

Step 3 – Evaluate I2I_2: Let t=1+x2t=1+x^2, dt=2x dxdt=2x\,dx:
I2=∫dtt=2t+C2=21+x2+C2I_2=\int\frac{dt}{\sqrt{t}}=2\sqrt{t}+C_2=2\sqrt{1+x^2}+C_2

Step 4 – Combine:
I=log⁡∣x+1+x2∣−21+x2+CI=\log|x+\sqrt{1+x^2}|-2\sqrt{1+x^2}+C

Answer: log⁡∣x+1+x2∣−21+x2+C\log\left|x+\sqrt{1+x^2}\right|-2\sqrt{1+x^2}+C
Q3(ii)Find ∫13x2+2x−1 dx\int\dfrac{1}{\sqrt{3x^2+2x-1}}\,dxShow solution
Given: ∫13x2+2x−1 dx\int\dfrac{1}{\sqrt{3x^2+2x-1}}\,dx

Step 1 – Complete the square in the expression 3x2+2x−13x^2+2x-1:
3x2+2x−1=3(x2+23x)−1=3(x2+23x+19−19)−13x^2+2x-1=3\left(x^2+\frac{2}{3}x\right)-1=3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)-1
=3(x+13)2−13−1=3(x+13)2−43=3\left(x+\frac{1}{3}\right)^2-\frac{1}{3}-1=3\left(x+\frac{1}{3}\right)^2-\frac{4}{3}

Step 2 – Rewrite the integral:
I=∫dx3(x+13)2−43=∫dx3⋅(x+13)2−49I=\int\frac{dx}{\sqrt{3\left(x+\frac{1}{3}\right)^2-\frac{4}{3}}}=\int\frac{dx}{\sqrt{3}\cdot\sqrt{\left(x+\frac{1}{3}\right)^2-\frac{4}{9}}}

Step 3 – Use the formula ∫dxx2−a2=log⁡∣x+x2−a2∣+C\int\dfrac{dx}{\sqrt{x^2-a^2}}=\log|x+\sqrt{x^2-a^2}|+C with u=x+13u=x+\dfrac{1}{3}, a=23a=\dfrac{2}{3}:
I=13log⁡∣(x+13)+(x+13)2−49∣+CI=\frac{1}{\sqrt{3}}\log\left|\left(x+\frac{1}{3}\right)+\sqrt{\left(x+\frac{1}{3}\right)^2-\frac{4}{9}}\right|+C
=13log⁡∣x+13+3x2+2x−13∣+C=\frac{1}{\sqrt{3}}\log\left|x+\frac{1}{3}+\sqrt{\frac{3x^2+2x-1}{3}}\right|+C

Answer: 13log⁡∣x+13+3x2+2x−13∣+C\dfrac{1}{\sqrt{3}}\log\left|x+\dfrac{1}{3}+\sqrt{\dfrac{3x^2+2x-1}{3}}\right|+C

Exercise 3.1 – Q4, Q5, Q6

Q4If the marginal revenue function of a firm is MR=40−10x2\text{MR}=40-10x^2 where xx is the level of output and total revenue is ₹120 at 3 units of output, find the total revenue function.Show solution
Given: MR=40−10x2\text{MR}=40-10x^2; R(3)=120R(3)=120.

Step 1 – Integrate MR to get R(x):
R(x)=∫(40−10x2) dx=40x−10x33+CR(x)=\int(40-10x^2)\,dx=40x-\frac{10x^3}{3}+C

Step 2 – Apply the condition R(3)=120R(3)=120:
120=40(3)−10(27)3+C=120−90+C=30+C120=40(3)-\frac{10(27)}{3}+C=120-90+C=30+C
⇒C=90\Rightarrow C=90

Step 3 – Write the total revenue function:
R(x)=40x−10x33+90R(x)=40x-\frac{10x^3}{3}+90

Answer: R(x)=40x−10x33+90R(x)=40x-\dfrac{10x^3}{3}+90
Q5The marginal cost function of producing xx units of a product is given by MC=x2500+x2MC=\dfrac{x}{\sqrt{2500+x^2}}. Find the total cost function and the average cost function, if the fixed cost is ₹1000.Show solution
Given: MC=x2500+x2MC=\dfrac{x}{\sqrt{2500+x^2}}; Fixed cost C(0)=1000C(0)=1000.

Step 1 – Integrate MC:
C(x)=∫x2500+x2 dxC(x)=\int\frac{x}{\sqrt{2500+x^2}}\,dx

Let t=2500+x2t=2500+x^2, dt=2x dxdt=2x\,dx:
C(x)=∫dt/2t=12⋅2t+K=2500+x2+KC(x)=\int\frac{dt/2}{\sqrt{t}}=\frac{1}{2}\cdot 2\sqrt{t}+K=\sqrt{2500+x^2}+K

Step 2 – Apply fixed cost condition C(0)=1000C(0)=1000:
1000=2500+K=50+K⇒K=9501000=\sqrt{2500}+K=50+K\Rightarrow K=950

Step 3 – Total cost function:
C(x)=2500+x2+950C(x)=\sqrt{2500+x^2}+950

Step 4 – Average cost function:
AC=C(x)x=2500+x2+950xAC=\frac{C(x)}{x}=\frac{\sqrt{2500+x^2}+950}{x}

Answer: C(x)=2500+x2+950C(x)=\sqrt{2500+x^2}+950; AC=2500+x2+950xAC=\dfrac{\sqrt{2500+x^2}+950}{x}
Q6The marginal cost of producing xx units of a product is given by MC=xx+1MC=x\sqrt{x+1}. The cost of producing 3 units is ₹7800. Find the cost function.Show solution
Given: MC=xx+1MC=x\sqrt{x+1}; C(3)=7800C(3)=7800.

Step 1 – Integrate MC:
C(x)=∫xx+1 dxC(x)=\int x\sqrt{x+1}\,dx

Let t=x+1t=x+1, so x=t−1x=t-1, dx=dtdx=dt:
C(x)=∫(t−1)t dt=∫(t3/2−t1/2) dt=2t5/25−2t3/23+KC(x)=\int(t-1)\sqrt{t}\,dt=\int(t^{3/2}-t^{1/2})\,dt=\frac{2t^{5/2}}{5}-\frac{2t^{3/2}}{3}+K

Back-substitute t=x+1t=x+1:
C(x)=2(x+1)5/25−2(x+1)3/23+KC(x)=\frac{2(x+1)^{5/2}}{5}-\frac{2(x+1)^{3/2}}{3}+K

Step 2 – Apply C(3)=7800C(3)=7800:
7800=2(4)5/25−2(4)3/23+K=2⋅325−2⋅83+K=645−163+K7800=\frac{2(4)^{5/2}}{5}-\frac{2(4)^{3/2}}{3}+K=\frac{2\cdot 32}{5}-\frac{2\cdot 8}{3}+K=\frac{64}{5}-\frac{16}{3}+K
=192−8015+K=11215+K=\frac{192-80}{15}+K=\frac{112}{15}+K
K=7800−11215=117000−11215=11688815K=7800-\frac{112}{15}=\frac{117000-112}{15}=\frac{116888}{15}

Step 3 – Cost function:
C(x)=2(x+1)5/25−2(x+1)3/23+11688815C(x)=\frac{2(x+1)^{5/2}}{5}-\frac{2(x+1)^{3/2}}{3}+\frac{116888}{15}

Answer: C(x)=2(x+1)5/25−2(x+1)3/23+11688815C(x)=\dfrac{2(x+1)^{5/2}}{5}-\dfrac{2(x+1)^{3/2}}{3}+\dfrac{116888}{15}

Exercise 3.2

Q1(i)Integrate x+1(x+2)(x+4)\dfrac{x+1}{(x+2)(x+4)}Show solution
Step 1 – Partial fractions:
x+1(x+2)(x+4)=Ax+2+Bx+4\frac{x+1}{(x+2)(x+4)}=\frac{A}{x+2}+\frac{B}{x+4}
x+1=A(x+4)+B(x+2)x+1=A(x+4)+B(x+2)

Put x=−2x=-2: −1=2A⇒A=−12-1=2A\Rightarrow A=-\dfrac{1}{2}.
Put x=−4x=-4: −3=−2B⇒B=32-3=-2B\Rightarrow B=\dfrac{3}{2}.

Step 2 – Integrate:
∫x+1(x+2)(x+4) dx=−12log⁡∣x+2∣+32log⁡∣x+4∣+C\int\frac{x+1}{(x+2)(x+4)}\,dx=-\frac{1}{2}\log|x+2|+\frac{3}{2}\log|x+4|+C

Answer: −12log⁡∣x+2∣+32log⁡∣x+4∣+C-\dfrac{1}{2}\log|x+2|+\dfrac{3}{2}\log|x+4|+C
Q1(ii)Integrate x(x2+1)(x2+2)\dfrac{x}{(x^2+1)(x^2+2)}Show solution
Step 1 – Let u=x2u=x^2. Partial fractions:
x(x2+1)(x2+2)=Ax2+1+Bx2+2\frac{x}{(x^2+1)(x^2+2)}=\frac{A}{x^2+1}+\frac{B}{x^2+2}
x=A(x2+2)+B(x2+1)x=A(x^2+2)+B(x^2+1)

Comparing: A+B=0A+B=0 (coeff of x2x^2), 2A+B=02A+B=0 (constant, but numerator is xx so this approach needs care).

Actually write x(x2+1)(x2+2)\dfrac{x}{(x^2+1)(x^2+2)} and substitute t=x2t=x^2:
1(t+1)(t+2)=1t+1−1t+2\frac{1}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2}
So x(x2+1)(x2+2)=xx2+1−xx2+2\dfrac{x}{(x^2+1)(x^2+2)}=\dfrac{x}{x^2+1}-\dfrac{x}{x^2+2}.

Step 2 – Integrate:
∫xx2+1 dx−∫xx2+2 dx=12log⁡(x2+1)−12log⁡(x2+2)+C\int\frac{x}{x^2+1}\,dx-\int\frac{x}{x^2+2}\,dx=\frac{1}{2}\log(x^2+1)-\frac{1}{2}\log(x^2+2)+C
=12log⁡x2+1x2+2+C=\frac{1}{2}\log\frac{x^2+1}{x^2+2}+C

Answer: 12log⁡x2+1x2+2+C\dfrac{1}{2}\log\dfrac{x^2+1}{x^2+2}+C
Q1(iii)Integrate 1e2x−1\dfrac{1}{e^{2x}-1}Show solution
Step 1 – Multiply numerator and denominator by e−2xe^{-2x}:
1e2x−1=e−2x1−e−2x\frac{1}{e^{2x}-1}=\frac{e^{-2x}}{1-e^{-2x}}

Alternatively, let t=ext=e^x, dt=ex dxdt=e^x\,dx, dx=dttdx=\dfrac{dt}{t}:
∫1t2−1⋅dtt=∫dtt(t2−1)=∫dtt(t−1)(t+1)\int\frac{1}{t^2-1}\cdot\frac{dt}{t}=\int\frac{dt}{t(t^2-1)}=\int\frac{dt}{t(t-1)(t+1)}

Step 2 – Partial fractions:
1t(t−1)(t+1)=At+Bt−1+Ct+1\frac{1}{t(t-1)(t+1)}=\frac{A}{t}+\frac{B}{t-1}+\frac{C}{t+1}
1=A(t2−1)+Bt(t+1)+Ct(t−1)1=A(t^2-1)+Bt(t+1)+Ct(t-1)

t=0t=0: 1=−A⇒A=−11=-A\Rightarrow A=-1; t=1t=1: 1=2B⇒B=121=2B\Rightarrow B=\frac{1}{2}; t=−1t=-1: 1=2C⇒C=121=2C\Rightarrow C=\frac{1}{2}.

Step 3 – Integrate:
−log⁡∣t∣+12log⁡∣t−1∣+12log⁡∣t+1∣+C-\log|t|+\frac{1}{2}\log|t-1|+\frac{1}{2}\log|t+1|+C
=−log⁡∣ex∣+12log⁡∣ex−1∣+12log⁡∣ex+1∣+C=-\log|e^x|+\frac{1}{2}\log|e^x-1|+\frac{1}{2}\log|e^x+1|+C
=−x+12log⁡(ex−1)+12log⁡(ex+1)+C=-x+\frac{1}{2}\log(e^x-1)+\frac{1}{2}\log(e^x+1)+C

Answer: −x+12log⁡(ex−1)+12log⁡(ex+1)+C-x+\dfrac{1}{2}\log(e^x-1)+\dfrac{1}{2}\log(e^x+1)+C
Q1(iv)Integrate 1x((log⁡x)2−3log⁡x+2)\dfrac{1}{x((\log x)^2-3\log x+2)}Show solution
Step 1 – Let t=log⁡xt=\log x, dt=dxxdt=\dfrac{dx}{x}:
∫dtt2−3t+2=∫dt(t−1)(t−2)\int\frac{dt}{t^2-3t+2}=\int\frac{dt}{(t-1)(t-2)}

Step 2 – Partial fractions:
1(t−1)(t−2)=At−1+Bt−2\frac{1}{(t-1)(t-2)}=\frac{A}{t-1}+\frac{B}{t-2}
t=1t=1: 1=−A⇒A=−11=-A\Rightarrow A=-1; t=2t=2: 1=B⇒B=11=B\Rightarrow B=1.

Step 3 – Integrate:
−log⁡∣t−1∣+log⁡∣t−2∣+C=log⁡∣t−2t−1∣+C-\log|t-1|+\log|t-2|+C=\log\left|\frac{t-2}{t-1}\right|+C

Back-substitute t=log⁡xt=\log x:
=log⁡∣log⁡x−2log⁡x−1∣+C=\log\left|\frac{\log x-2}{\log x-1}\right|+C

Answer: log⁡∣log⁡x−2log⁡x−1∣+C\log\left|\dfrac{\log x-2}{\log x-1}\right|+C
Q1(v)Integrate 3x−2(x−2)2(x+2)\dfrac{3x-2}{(x-2)^2(x+2)}Show solution
Step 1 – Partial fractions:
3x−2(x−2)2(x+2)=Ax−2+B(x−2)2+Cx+2\frac{3x-2}{(x-2)^2(x+2)}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{x+2}
3x−2=A(x−2)(x+2)+B(x+2)+C(x−2)23x-2=A(x-2)(x+2)+B(x+2)+C(x-2)^2

x=2x=2: 4=4B⇒B=14=4B\Rightarrow B=1.
x=−2x=-2: −8=16C⇒C=−12-8=16C\Rightarrow C=-\dfrac{1}{2}.
Coeff of x2x^2: 0=A+C⇒A=120=A+C\Rightarrow A=\dfrac{1}{2}.

Step 2 – Integrate:
12log⁡∣x−2∣−1x−2−12log⁡∣x+2∣+C\frac{1}{2}\log|x-2|-\frac{1}{x-2}-\frac{1}{2}\log|x+2|+C
=12log⁡∣x−2x+2∣−1x−2+C=\frac{1}{2}\log\left|\frac{x-2}{x+2}\right|-\frac{1}{x-2}+C

Answer: 12log⁡∣x−2x+2∣−1x−2+C\dfrac{1}{2}\log\left|\dfrac{x-2}{x+2}\right|-\dfrac{1}{x-2}+C
Q1(vi)Integrate 1e2x+ex\dfrac{1}{e^{2x}+e^x}Show solution
Step 1 – Factor denominator: e2x+ex=ex(ex+1)e^{2x}+e^x=e^x(e^x+1).
∫dxex(ex+1)\int\frac{dx}{e^x(e^x+1)}

Step 2 – Let t=ext=e^x, dt=ex dxdt=e^x\,dx, dx=dttdx=\dfrac{dt}{t}:
∫1t(t+1)⋅dtt=∫dtt2(t+1)\int\frac{1}{t(t+1)}\cdot\frac{dt}{t}=\int\frac{dt}{t^2(t+1)}

Step 3 – Partial fractions:
1t2(t+1)=At+Bt2+Ct+1\frac{1}{t^2(t+1)}=\frac{A}{t}+\frac{B}{t^2}+\frac{C}{t+1}
1=At(t+1)+B(t+1)+Ct21=At(t+1)+B(t+1)+Ct^2

t=0t=0: 1=B1=B; t=−1t=-1: 1=C1=C; coeff of t2t^2: 0=A+C⇒A=−10=A+C\Rightarrow A=-1.

Step 4 – Integrate:
−log⁡∣t∣+−1t+log⁡∣t+1∣+C=log⁡∣t+1t∣−1t+C-\log|t|+\frac{-1}{t}+\log|t+1|+C=\log\left|\frac{t+1}{t}\right|-\frac{1}{t}+C

Back-substitute t=ext=e^x:
=log⁡∣ex+1ex∣−e−x+C=log⁡(1+e−x)−e−x+C=\log\left|\frac{e^x+1}{e^x}\right|-e^{-x}+C=\log(1+e^{-x})-e^{-x}+C

Answer: log⁡(1+e−x)−e−x+C\log(1+e^{-x})-e^{-x}+C
Q1(vii)Integrate 5x+4(x2−1)(x+2)\dfrac{5x+4}{(x^2-1)(x+2)}Show solution
Step 1 – Factor: x2−1=(x−1)(x+1)x^2-1=(x-1)(x+1).
5x+4(x−1)(x+1)(x+2)=Ax−1+Bx+1+Cx+2\frac{5x+4}{(x-1)(x+1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}
5x+4=A(x+1)(x+2)+B(x−1)(x+2)+C(x−1)(x+1)5x+4=A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)

x=1x=1: 9=6A⇒A=329=6A\Rightarrow A=\dfrac{3}{2}.
x=−1x=-1: −1=2B⇒B=−12-1=2B\Rightarrow B=-\dfrac{1}{2}.
x=−2x=-2: −6=3C⇒C=−2-6=3C\Rightarrow C=-2.

Step 2 – Integrate:
32log⁡∣x−1∣−12log⁡∣x+1∣−2log⁡∣x+2∣+C\frac{3}{2}\log|x-1|-\frac{1}{2}\log|x+1|-2\log|x+2|+C

Answer: 32log⁡∣x−1∣−12log⁡∣x+1∣−2log⁡∣x+2∣+C\dfrac{3}{2}\log|x-1|-\dfrac{1}{2}\log|x+1|-2\log|x+2|+C
Q1(viii)Integrate x(x−1)2(x+2)\dfrac{x}{(x-1)^2(x+2)}Show solution
Step 1 – Partial fractions:
x(x−1)2(x+2)=Ax−1+B(x−1)2+Cx+2\frac{x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}
x=A(x−1)(x+2)+B(x+2)+C(x−1)2x=A(x-1)(x+2)+B(x+2)+C(x-1)^2

x=1x=1: 1=3B⇒B=131=3B\Rightarrow B=\dfrac{1}{3}.
x=−2x=-2: −2=9C⇒C=−29-2=9C\Rightarrow C=-\dfrac{2}{9}.
Coeff of x2x^2: 0=A+C⇒A=290=A+C\Rightarrow A=\dfrac{2}{9}.

Step 2 – Integrate:
29log⁡∣x−1∣−13(x−1)−29log⁡∣x+2∣+C\frac{2}{9}\log|x-1|-\frac{1}{3(x-1)}-\frac{2}{9}\log|x+2|+C
=29log⁡∣x−1x+2∣−13(x−1)+C=\frac{2}{9}\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C

Answer: 29log⁡∣x−1x+2∣−13(x−1)+C\dfrac{2}{9}\log\left|\dfrac{x-1}{x+2}\right|-\dfrac{1}{3(x-1)}+C
Q1(ix)Integrate 1x(x4−1)\dfrac{1}{x(x^4-1)}Show solution
Step 1 – Factor: x4−1=(x2−1)(x2+1)=(x−1)(x+1)(x2+1)x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1).

Step 2 – Multiply numerator and denominator by x3x^3:
1x(x4−1)=x3x4(x4−1)\frac{1}{x(x^4-1)}=\frac{x^3}{x^4(x^4-1)}
Let t=x4t=x^4, dt=4x3 dxdt=4x^3\,dx:
14∫dtt(t−1)=14∫(1t−1−1t)dt=14log⁡∣t−1t∣+C\frac{1}{4}\int\frac{dt}{t(t-1)}=\frac{1}{4}\int\left(\frac{1}{t-1}-\frac{1}{t}\right)dt=\frac{1}{4}\log\left|\frac{t-1}{t}\right|+C

Back-substitute t=x4t=x^4:
=14log⁡∣x4−1x4∣+C=\frac{1}{4}\log\left|\frac{x^4-1}{x^4}\right|+C

Answer: 14log⁡∣x4−1x4∣+C\dfrac{1}{4}\log\left|\dfrac{x^4-1}{x^4}\right|+C
Q1(x)Integrate 1x(xn+1)\dfrac{1}{x(x^n+1)}Show solution
Step 1 – Multiply numerator and denominator by xn−1x^{n-1}:
1x(xn+1)=xn−1xn(xn+1)\frac{1}{x(x^n+1)}=\frac{x^{n-1}}{x^n(x^n+1)}

Let t=xnt=x^n, dt=nxn−1 dxdt=nx^{n-1}\,dx:
1n∫dtt(t+1)=1n∫(1t−1t+1)dt=1nlog⁡∣tt+1∣+C\frac{1}{n}\int\frac{dt}{t(t+1)}=\frac{1}{n}\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt=\frac{1}{n}\log\left|\frac{t}{t+1}\right|+C

Back-substitute t=xnt=x^n:
=1nlog⁡∣xnxn+1∣+C=\frac{1}{n}\log\left|\frac{x^n}{x^n+1}\right|+C

Answer: 1nlog⁡∣xnxn+1∣+C\dfrac{1}{n}\log\left|\dfrac{x^n}{x^n+1}\right|+C
Q1(xi)Integrate 1−xx(1−2x)\dfrac{1-x}{x(1-2x)}Show solution
Step 1 – Partial fractions:
1−xx(1−2x)=Ax+B1−2x\frac{1-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{1-2x}
1−x=A(1−2x)+Bx1-x=A(1-2x)+Bx

x=0x=0: 1=A1=A.
x=12x=\dfrac{1}{2}: 12=B2⇒B=1\dfrac{1}{2}=\dfrac{B}{2}\Rightarrow B=1.

Step 2 – Integrate:
∫dxx+∫dx1−2x=log⁡∣x∣−12log⁡∣1−2x∣+C\int\frac{dx}{x}+\int\frac{dx}{1-2x}=\log|x|-\frac{1}{2}\log|1-2x|+C

Answer: log⁡∣x∣−12log⁡∣1−2x∣+C\log|x|-\dfrac{1}{2}\log|1-2x|+C
Q2The marginal revenue function for a firm is given by 5x2+30x+51(x+3)2\dfrac{5x^2+30x+51}{(x+3)^2}. Show that the revenue function is given by 2xx+3+5x\dfrac{2x}{x+3}+5x.Show solution
Step 1 – Integrate MR:
R(x)=∫5x2+30x+51(x+3)2 dxR(x)=\int\frac{5x^2+30x+51}{(x+3)^2}\,dx

Step 2 – Perform polynomial division / rewrite numerator:
Note (x+3)2=x2+6x+9(x+3)^2=x^2+6x+9. Divide 5x2+30x+515x^2+30x+51 by (x+3)2(x+3)^2:
5x2+30x+51=5(x2+6x+9)+51−45=5(x+3)2+65x^2+30x+51=5(x^2+6x+9)+51-45=5(x+3)^2+6
So:
5x2+30x+51(x+3)2=5+6(x+3)2\frac{5x^2+30x+51}{(x+3)^2}=5+\frac{6}{(x+3)^2}

Step 3 – Integrate:
R(x)=∫[5+6(x+3)2]dx=5x+6⋅−1x+3+C=5x−6x+3+CR(x)=\int\left[5+\frac{6}{(x+3)^2}\right]dx=5x+6\cdot\frac{-1}{x+3}+C=5x-\frac{6}{x+3}+C

Step 4 – Apply R(0)=0R(0)=0 (revenue is 0 when output is 0):
0=0−63+C⇒C=20=0-\frac{6}{3}+C\Rightarrow C=2

Step 5:
R(x)=5x−6x+3+2=5x+−6+2(x+3)x+3=5x+2xx+3R(x)=5x-\frac{6}{x+3}+2=5x+\frac{-6+2(x+3)}{x+3}=5x+\frac{2x}{x+3}

R(x)=5x+2xx+3\boxed{R(x)=5x+\frac{2x}{x+3}}

Hence proved.
Q3Find the total revenue function and demand function, if the marginal revenue function is given by MR(x)=ab(x+b)2−cMR(x)=\dfrac{ab}{(x+b)^2}-c.Show solution
Given: MR(x)=ab(x+b)2−cMR(x)=\dfrac{ab}{(x+b)^2}-c

Step 1 – Integrate to get R(x):
R(x)=∫[ab(x+b)2−c]dx=ab⋅−1x+b−cx+KR(x)=\int\left[\frac{ab}{(x+b)^2}-c\right]dx=ab\cdot\frac{-1}{x+b}-cx+K
=−abx+b−cx+K=-\frac{ab}{x+b}-cx+K

Step 2 – Apply R(0)=0R(0)=0:
0=−abb+K⇒0=−a+K⇒K=a0=-\frac{ab}{b}+K\Rightarrow 0=-a+K\Rightarrow K=a

Step 3 – Total Revenue Function:
R(x)=a−abx+b−cxR(x)=a-\frac{ab}{x+b}-cx

Step 4 – Demand function (since R(x)=p⋅xR(x)=p\cdot x, so p=R(x)xp=\dfrac{R(x)}{x}):
p=ax−abx(x+b)−c=a(x+b)−abx(x+b)−c=axx(x+b)−c=ax+b−cp=\frac{a}{x}-\frac{ab}{x(x+b)}-c=\frac{a(x+b)-ab}{x(x+b)}-c=\frac{ax}{x(x+b)}-c=\frac{a}{x+b}-c

Answer: Total Revenue R(x)=a−abx+b−cxR(x)=a-\dfrac{ab}{x+b}-cx; Demand function p=ax+b−cp=\dfrac{a}{x+b}-c

Exercise 3.4

(i)Evaluate ∫ee21xlog⁡x dx\int_{e}^{e^2}\dfrac{1}{x\log x}\,dxShow solution
Step 1 – Let t=log⁡xt=\log x, dt=dxxdt=\dfrac{dx}{x}.

When x=ex=e, t=1t=1; when x=e2x=e^2, t=2t=2.

Step 2 – Substitute:
∫12dtt=log⁡∣t∣∣12=log⁡2−log⁡1=log⁡2\int_1^2\frac{dt}{t}=\log|t|\Big|_1^2=\log 2-\log 1=\log 2

Answer: log⁡2\log 2
(ii)Evaluate ∫100010042x dx\int_{1000}^{1004}2^x\,dxShow solution
Step 1 – Use ∫ax dx=axlog⁡a+C\int a^x\,dx=\dfrac{a^x}{\log a}+C:
∫100010042x dx=2xlog⁡2∣10001004=21004−21000log⁡2\int_{1000}^{1004}2^x\,dx=\frac{2^x}{\log 2}\Bigg|_{1000}^{1004}=\frac{2^{1004}-2^{1000}}{\log 2}

Step 2 – Simplify:
=21000(24−1)log⁡2=15⋅21000log⁡2=\frac{2^{1000}(2^4-1)}{\log 2}=\frac{15\cdot 2^{1000}}{\log 2}

Answer: 15⋅21000log⁡2\dfrac{15\cdot 2^{1000}}{\log 2}
(iii)Evaluate ∫2log⁡22 dx\int_{2}^{\log 2}2\,dxShow solution
Note: The lower limit 22 is greater than the upper limit log⁡2\log 2 (since log⁡2≈0.693\log 2\approx 0.693). The integral is:
∫2log⁡22 dx=2x∣2log⁡2=2log⁡2−2(2)=2log⁡2−4\int_{2}^{\log 2}2\,dx=2x\Big|_{2}^{\log 2}=2\log 2-2(2)=2\log 2-4

Answer: 2log⁡2−42\log 2-4
(iv)Evaluate ∫03x16−x4 dx\int_{0}^{\sqrt{3}}\dfrac{x}{16-x^4}\,dxShow solution
Step 1 – Let t=x2t=x^2, dt=2x dxdt=2x\,dx.

When x=0x=0, t=0t=0; when x=3x=\sqrt{3}, t=3t=3.

Step 2 – Substitute:
12∫03dt16−t2=12∫03dt42−t2\frac{1}{2}\int_0^3\frac{dt}{16-t^2}=\frac{1}{2}\int_0^3\frac{dt}{4^2-t^2}

Step 3 – Use ∫dta2−t2=12alog⁡∣a+ta−t∣+C\int\dfrac{dt}{a^2-t^2}=\dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right|+C with a=4a=4:
=12⋅18log⁡∣4+t4−t∣∣03=116[log⁡71−log⁡1]=log⁡716=\frac{1}{2}\cdot\frac{1}{8}\log\left|\frac{4+t}{4-t}\right|\Bigg|_0^3=\frac{1}{16}\left[\log\frac{7}{1}-\log 1\right]=\frac{\log 7}{16}

Answer: log⁡716\dfrac{\log 7}{16}
(v)Evaluate ∫011x+1−x dx\int_0^1\dfrac{1}{\sqrt{x+1}-\sqrt{x}}\,dxShow solution
Step 1 – Rationalise:
1x+1−x=x+1+x(x+1)−x=x+1+x\frac{1}{\sqrt{x+1}-\sqrt{x}}=\frac{\sqrt{x+1}+\sqrt{x}}{(x+1)-x}=\sqrt{x+1}+\sqrt{x}

Step 2 – Integrate:
∫01(x+1+x) dx=[2(x+1)3/23+2x3/23]01\int_0^1(\sqrt{x+1}+\sqrt{x})\,dx=\left[\frac{2(x+1)^{3/2}}{3}+\frac{2x^{3/2}}{3}\right]_0^1
=2(2)3/23+2(1)3/23−2(1)3/23−0=2⋅223=423=\frac{2(2)^{3/2}}{3}+\frac{2(1)^{3/2}}{3}-\frac{2(1)^{3/2}}{3}-0=\frac{2\cdot 2\sqrt{2}}{3}=\frac{4\sqrt{2}}{3}

Answer: 423\dfrac{4\sqrt{2}}{3}
(vi)Evaluate ∫01ex1+ex dx\int_0^1 e^x\sqrt{1+e^x}\,dxShow solution
Step 1 – Let t=1+ext=1+e^x, dt=ex dxdt=e^x\,dx.

When x=0x=0, t=2t=2; when x=1x=1, t=1+et=1+e.

Step 2 – Substitute:
∫21+et dt=2t3/23∣21+e=23[(1+e)3/2−23/2]\int_2^{1+e}\sqrt{t}\,dt=\frac{2t^{3/2}}{3}\Bigg|_2^{1+e}=\frac{2}{3}\left[(1+e)^{3/2}-2^{3/2}\right]

Answer: 23[(1+e)3/2−22]\dfrac{2}{3}\left[(1+e)^{3/2}-2\sqrt{2}\right]
(vii)Evaluate ∫451x2−16 dx\int_4^5\dfrac{1}{\sqrt{x^2-16}}\,dxShow solution
Step 1 – Use ∫dxx2−a2=log⁡∣x+x2−a2∣+C\int\dfrac{dx}{\sqrt{x^2-a^2}}=\log|x+\sqrt{x^2-a^2}|+C with a=4a=4:
[log⁡∣x+x2−16∣]45\left[\log|x+\sqrt{x^2-16}|\right]_4^5

Step 2 – Evaluate:
=log⁡∣5+25−16∣−log⁡∣4+16−16∣=log⁡∣5+3∣−log⁡∣4+0∣=\log|5+\sqrt{25-16}|-\log|4+\sqrt{16-16}|=\log|5+3|-\log|4+0|
=log⁡8−log⁡4=log⁡84=log⁡2=\log 8-\log 4=\log\frac{8}{4}=\log 2

Answer: log⁡2\log 2
(viii)Evaluate ∫01log⁡(1+2x) dx\int_0^1\log(1+2x)\,dxShow solution
Step 1 – Integration by parts: Let u=log⁡(1+2x)u=\log(1+2x), dv=dxdv=dx.
Then du=21+2x dxdu=\dfrac{2}{1+2x}\,dx, v=xv=x.

∫01log⁡(1+2x) dx=[xlog⁡(1+2x)]01−∫012x1+2x dx\int_0^1\log(1+2x)\,dx=\left[x\log(1+2x)\right]_0^1-\int_0^1\frac{2x}{1+2x}\,dx

Step 2 – Evaluate boundary term:
=1⋅log⁡3−0−∫012x1+2x dx=1\cdot\log 3-0-\int_0^1\frac{2x}{1+2x}\,dx

Step 3 – Simplify the remaining integral:
2x1+2x=1−11+2x\frac{2x}{1+2x}=1-\frac{1}{1+2x}
∫01(1−11+2x)dx=[x−log⁡(1+2x)2]01=1−log⁡32\int_0^1\left(1-\frac{1}{1+2x}\right)dx=\left[x-\frac{\log(1+2x)}{2}\right]_0^1=1-\frac{\log 3}{2}

Step 4 – Combine:
=log⁡3−(1−log⁡32)=log⁡3−1+log⁡32=3log⁡32−1=\log 3-\left(1-\frac{\log 3}{2}\right)=\log 3-1+\frac{\log 3}{2}=\frac{3\log 3}{2}-1

Answer: 3log⁡32−1\dfrac{3\log 3}{2}-1
(ix)Evaluate ∫04x2+9 dx\int_0^4\sqrt{x^2+9}\,dxShow solution
Step 1 – Use ∫x2+a2 dx=x2x2+a2+a22log⁡∣x+x2+a2∣+C\int\sqrt{x^2+a^2}\,dx=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}\log|x+\sqrt{x^2+a^2}|+C with a=3a=3:
[x2x2+9+92log⁡∣x+x2+9∣]04\left[\frac{x}{2}\sqrt{x^2+9}+\frac{9}{2}\log|x+\sqrt{x^2+9}|\right]_0^4

Step 2 – Evaluate at x=4x=4:
4225+92log⁡(4+5)=2(5)+92log⁡9=10+92log⁡9\frac{4}{2}\sqrt{25}+\frac{9}{2}\log(4+5)=2(5)+\frac{9}{2}\log 9=10+\frac{9}{2}\log 9

Step 3 – Evaluate at x=0x=0:
0+92log⁡(0+3)=92log⁡30+\frac{9}{2}\log(0+3)=\frac{9}{2}\log 3

Step 4 – Subtract:
10+92log⁡9−92log⁡3=10+92log⁡93=10+92log⁡310+\frac{9}{2}\log 9-\frac{9}{2}\log 3=10+\frac{9}{2}\log\frac{9}{3}=10+\frac{9}{2}\log 3

Answer: 10+92log⁡310+\dfrac{9}{2}\log 3
(x)Evaluate ∫013t2(1+t3)(2+t3) dt\int_0^1\dfrac{3t^2}{(1+t^3)(2+t^3)}\,dtShow solution
Step 1 – Let u=t3u=t^3, du=3t2 dtdu=3t^2\,dt.

When t=0t=0, u=0u=0; when t=1t=1, u=1u=1.

Step 2 – Substitute:
∫01du(1+u)(2+u)\int_0^1\frac{du}{(1+u)(2+u)}

Step 3 – Partial fractions:
1(1+u)(2+u)=11+u−12+u\frac{1}{(1+u)(2+u)}=\frac{1}{1+u}-\frac{1}{2+u}

Step 4 – Integrate:
[log⁡∣1+u∣−log⁡∣2+u∣]01=[log⁡1+u2+u]01\left[\log|1+u|-\log|2+u|\right]_0^1=\left[\log\frac{1+u}{2+u}\right]_0^1
=log⁡23−log⁡12=log⁡23+log⁡2=log⁡43=\log\frac{2}{3}-\log\frac{1}{2}=\log\frac{2}{3}+\log 2=\log\frac{4}{3}

Answer: log⁡43\log\dfrac{4}{3}

Exercise 3.5

Q1(i)Evaluate ∫03f(x) dx\int_0^3 f(x)\,dx where f(x)={x+1amp;xlt;12xamp;x≥1f(x)=\begin{cases}x+1 & x<1\\ 2x & x\geq 1\end{cases}Show solution
Step 1 – Split the integral at x=1x=1:
∫03f(x) dx=∫01(x+1) dx+∫132x dx\int_0^3 f(x)\,dx=\int_0^1(x+1)\,dx+\int_1^3 2x\,dx

Step 2 – Evaluate each part:
∫01(x+1) dx=[x22+x]01=12+1=32\int_0^1(x+1)\,dx=\left[\frac{x^2}{2}+x\right]_0^1=\frac{1}{2}+1=\frac{3}{2}
∫132x dx=[x2]13=9−1=8\int_1^3 2x\,dx=\left[x^2\right]_1^3=9-1=8

Step 3 – Add:
32+8=192\frac{3}{2}+8=\frac{19}{2}

Answer: 192\dfrac{19}{2}
Q1(ii)Evaluate ∫01x1−x dx\int_0^1 x\sqrt{1-x}\,dxShow solution
Step 1 – Let t=1−xt=1-x, dt=−dxdt=-dx, x=1−tx=1-t.

When x=0x=0, t=1t=1; when x=1x=1, t=0t=0.

Step 2 – Substitute:
∫10(1−t)t(−dt)=∫01(1−t)t dt=∫01(t1/2−t3/2) dt\int_1^0(1-t)\sqrt{t}(-dt)=\int_0^1(1-t)\sqrt{t}\,dt=\int_0^1(t^{1/2}-t^{3/2})\,dt

Step 3 – Integrate:
=[2t3/23−2t5/25]01=23−25=10−615=415=\left[\frac{2t^{3/2}}{3}-\frac{2t^{5/2}}{5}\right]_0^1=\frac{2}{3}-\frac{2}{5}=\frac{10-6}{15}=\frac{4}{15}

Answer: 415\dfrac{4}{15}
Q1(iii)Evaluate ∫04∣x−2∣ dx\int_0^4|x-2|\,dxShow solution
Step 1 – Remove absolute value:
∣x−2∣={2−xamp;0≤xlt;2x−2amp;2≤x≤4|x-2|=\begin{cases}2-x & 0\leq x<2\\ x-2 & 2\leq x\leq 4\end{cases}

Step 2 – Split and integrate:
∫02(2−x) dx+∫24(x−2) dx\int_0^2(2-x)\,dx+\int_2^4(x-2)\,dx
=[2x−x22]02+[x22−2x]24=\left[2x-\frac{x^2}{2}\right]_0^2+\left[\frac{x^2}{2}-2x\right]_2^4
=(4−2)+(8−8−2+4)=2+2=4=(4-2)+(8-8-2+4)=2+2=4

Answer: 44
Q1(iv)Evaluate ∫15xx+6−x dx\int_1^5\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{6-x}}\,dxShow solution
Step 1 – Let I=∫15xx+6−x dxI=\int_1^5\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{6-x}}\,dx …(1)

Step 2 – Apply property P4P_4: Replace xx by a+b−x=1+5−x=6−xa+b-x=1+5-x=6-x:
I=∫156−x6−x+x dx…(2)I=\int_1^5\frac{\sqrt{6-x}}{\sqrt{6-x}+\sqrt{x}}\,dx \quad\text{…(2)}

Step 3 – Add (1) and (2):
2I=∫15x+6−xx+6−x dx=∫151 dx=42I=\int_1^5\frac{\sqrt{x}+\sqrt{6-x}}{\sqrt{x}+\sqrt{6-x}}\,dx=\int_1^5 1\,dx=4

I=2I=2

Answer: 22
Q1(v)Evaluate ∫0ax2020x2020+(a−x)2020 dx\int_0^a\dfrac{x^{2020}}{x^{2020}+(a-x)^{2020}}\,dxShow solution
Step 1 – Let I=∫0ax2020x2020+(a−x)2020 dxI=\int_0^a\dfrac{x^{2020}}{x^{2020}+(a-x)^{2020}}\,dx …(1)

Step 2 – Apply property (replace xx by a−xa-x):
I=∫0a(a−x)2020(a−x)2020+x2020 dx…(2)I=\int_0^a\frac{(a-x)^{2020}}{(a-x)^{2020}+x^{2020}}\,dx \quad\text{…(2)}

Step 3 – Add (1) and (2):
2I=∫0a1 dx=a⇒I=a22I=\int_0^a 1\,dx=a\Rightarrow I=\frac{a}{2}

Answer: a2\dfrac{a}{2}
Q1(vi)Evaluate ∫04(∣x∣+∣x−2∣+∣x−4∣) dx\int_0^4(|x|+|x-2|+|x-4|)\,dxShow solution
Step 1 – On [0,4][0,4]: ∣x∣=x|x|=x; ∣x−2∣={2−x,amp;0≤xlt;2x−2,amp;2≤x≤4|x-2|=\begin{cases}2-x,&0\leq x<2\\x-2,&2\leq x\leq 4\end{cases}; ∣x−4∣=4−x|x-4|=4-x.

Step 2 – Split at x=2x=2:

On [0,2][0,2]: integrand =x+(2−x)+(4−x)=6−x=x+(2-x)+(4-x)=6-x.
On [2,4][2,4]: integrand =x+(x−2)+(4−x)=x+2=x+(x-2)+(4-x)=x+2.

Step 3 – Integrate:
∫02(6−x) dx+∫24(x+2) dx\int_0^2(6-x)\,dx+\int_2^4(x+2)\,dx
=[6x−x22]02+[x22+2x]24=\left[6x-\frac{x^2}{2}\right]_0^2+\left[\frac{x^2}{2}+2x\right]_2^4
=(12−2)+(8+8−2−4)=10+10=20=(12-2)+(8+8-2-4)=10+10=20

Answer: 2020
Q1(vii)Evaluate ∫−2211+ex dx\int_{-2}^{2}\dfrac{1}{1+\sqrt{e^x}}\,dxShow solution
Step 1 – Let I=∫−2211+ex dxI=\int_{-2}^{2}\dfrac{1}{1+\sqrt{e^x}}\,dx …(1)

Step 2 – Apply property (replace xx by −x-x, using P4P_4 with a+b=0a+b=0, i.e., replace x→−xx\to -x):
I=∫−2211+e−x dx=∫−22exex+1 dx…(2)I=\int_{-2}^{2}\frac{1}{1+\sqrt{e^{-x}}}\,dx=\int_{-2}^{2}\frac{\sqrt{e^x}}{\sqrt{e^x}+1}\,dx \quad\text{…(2)}

Step 3 – Add (1) and (2):
2I=∫−221+ex1+ex dx=∫−221 dx=42I=\int_{-2}^{2}\frac{1+\sqrt{e^x}}{1+\sqrt{e^x}}\,dx=\int_{-2}^{2}1\,dx=4
I=2I=2

Answer: 22
Q1(viii)Evaluate ∫−11x3+∣x∣+1x2+2∣x∣+1 dx\int_{-1}^{1}\dfrac{x^3+|x|+1}{x^2+2|x|+1}\,dxShow solution
Step 1 – Note that x2+2∣x∣+1=(∣x∣+1)2x^2+2|x|+1=(|x|+1)^2.

Step 2 – Split:
I=∫−11x3(∣x∣+1)2 dx+∫−11∣x∣+1(∣x∣+1)2 dxI=\int_{-1}^{1}\frac{x^3}{(|x|+1)^2}\,dx+\int_{-1}^{1}\frac{|x|+1}{(|x|+1)^2}\,dx
=I1+∫−111∣x∣+1 dx=I_1+\int_{-1}^{1}\frac{1}{|x|+1}\,dx

Step 3 – I1=∫−11x3(∣x∣+1)2 dx=0I_1=\int_{-1}^{1}\dfrac{x^3}{(|x|+1)^2}\,dx=0 (odd function on symmetric interval).

Step 4 – ∫−111∣x∣+1 dx=2∫011x+1 dx\int_{-1}^{1}\dfrac{1}{|x|+1}\,dx=2\int_0^1\dfrac{1}{x+1}\,dx (even function):
=2[log⁡(x+1)]01=2log⁡2=2\left[\log(x+1)\right]_0^1=2\log 2

Answer: 2log⁡22\log 2
Q1(ix)Evaluate ∫01x(1−x)n dx\int_0^1 x(1-x)^n\,dxShow solution
Step 1 – Let I=∫01x(1−x)n dxI=\int_0^1 x(1-x)^n\,dx.

Step 2 – Apply property (replace xx by 1−x1-x):
I=∫01(1−x)xn dx=∫01(xn−xn+1) dxI=\int_0^1(1-x)x^n\,dx=\int_0^1(x^n-x^{n+1})\,dx
=[xn+1n+1−xn+2n+2]01=1n+1−1n+2=1(n+1)(n+2)=\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_0^1=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}

Answer: 1(n+1)(n+2)\dfrac{1}{(n+1)(n+2)}
Q1(x)Evaluate ∫−22(x3+12)4+x2 dx\int_{-2}^{2}\left(x^3+\dfrac{1}{2}\right)\sqrt{4+x^2}\,dxShow solution
Step 1 – Split:
I=∫−22x34+x2 dx+∫−22124+x2 dx=I1+I2I=\int_{-2}^{2}x^3\sqrt{4+x^2}\,dx+\int_{-2}^{2}\frac{1}{2}\sqrt{4+x^2}\,dx=I_1+I_2

Step 2 – I1I_1: The function x34+x2x^3\sqrt{4+x^2} is odd (odd ×\times even = odd), so I1=0I_1=0.

Step 3 – I2=12⋅2∫024+x2 dx=∫02x2+4 dxI_2=\dfrac{1}{2}\cdot 2\int_0^2\sqrt{4+x^2}\,dx=\int_0^2\sqrt{x^2+4}\,dx.

Using ∫x2+a2 dx=x2x2+a2+a22log⁡∣x+x2+a2∣+C\int\sqrt{x^2+a^2}\,dx=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}\log|x+\sqrt{x^2+a^2}|+C with a=2a=2:
=[x2x2+4+2log⁡∣x+x2+4∣]02=\left[\frac{x}{2}\sqrt{x^2+4}+2\log|x+\sqrt{x^2+4}|\right]_0^2
=(22⋅22+2log⁡(2+22))−(0+2log⁡2)=\left(\frac{2}{2}\cdot 2\sqrt{2}+2\log(2+2\sqrt{2})\right)-\left(0+2\log 2\right)
=22+2log⁡(2+22)−2log⁡2=22+2log⁡2+222=22+2log⁡(1+2)=2\sqrt{2}+2\log(2+2\sqrt{2})-2\log 2=2\sqrt{2}+2\log\frac{2+2\sqrt{2}}{2}=2\sqrt{2}+2\log(1+\sqrt{2})

Answer: 22+2log⁡(1+2)2\sqrt{2}+2\log(1+\sqrt{2})
Q1(xi)Evaluate ∫−11log⁡1−x1+x dx\int_{-1}^{1}\log\dfrac{1-x}{1+x}\,dxShow solution
Step 1 – Let f(x)=log⁡1−x1+xf(x)=\log\dfrac{1-x}{1+x}.

Step 2 – Check: f(−x)=log⁡1+x1−x=−log⁡1−x1+x=−f(x)f(-x)=\log\dfrac{1+x}{1-x}=-\log\dfrac{1-x}{1+x}=-f(x).

So f(x)f(x) is an odd function.

Step 3 – Apply property: ∫−aaf(x) dx=0\int_{-a}^{a}f(x)\,dx=0 for odd ff:
∫−11log⁡1−x1+x dx=0\int_{-1}^{1}\log\frac{1-x}{1+x}\,dx=0

Answer: 00
Q1(xii)Evaluate ∫−11x21+ex dx\int_{-1}^{1}\dfrac{x^2}{1+e^x}\,dxShow solution
Step 1 – Let I=∫−11x21+ex dxI=\int_{-1}^{1}\dfrac{x^2}{1+e^x}\,dx …(1)

Step 2 – Replace xx by −x-x:
I=∫−11x21+e−x dx=∫−11x2exex+1 dx…(2)I=\int_{-1}^{1}\frac{x^2}{1+e^{-x}}\,dx=\int_{-1}^{1}\frac{x^2 e^x}{e^x+1}\,dx \quad\text{…(2)}

Step 3 – Add (1) and (2):
2I=∫−11x2(1+ex)1+ex dx=∫−11x2 dx=2∫01x2 dx=2⋅13=232I=\int_{-1}^{1}\frac{x^2(1+e^x)}{1+e^x}\,dx=\int_{-1}^{1}x^2\,dx=2\int_0^1 x^2\,dx=2\cdot\frac{1}{3}=\frac{2}{3}
I=13I=\frac{1}{3}

Answer: 13\dfrac{1}{3}
Q2Evaluate ∫02[x] dx\int_0^2[x]\,dx where [⋅][\cdot] denotes the Greatest Integer Function.Show solution
Step 1 – Identify [x][x] on [0,2][0,2]:
[x]={0amp;0≤xlt;11amp;1≤xlt;22amp;x=2[x]=\begin{cases}0 & 0\leq x<1\\ 1 & 1\leq x<2\\ 2 & x=2\end{cases}

Step 2 – Split and integrate:
∫02[x] dx=∫010 dx+∫121 dx=0+(2−1)=1\int_0^2[x]\,dx=\int_0^1 0\,dx+\int_1^2 1\,dx=0+(2-1)=1

Answer: 11

Exercise 3.6

1If the demand function is p=35−2x−x2p=35-2x-x^2 and the demand x0=3x_0=3, find the consumers' surplus.Show solution
Given: p=f(x)=35−2x−x2p=f(x)=35-2x-x^2, x0=3x_0=3.

Step 1 – Find p0p_0:
p0=35−2(3)−(3)2=35−6−9=20p_0=35-2(3)-(3)^2=35-6-9=20

Step 2 – Consumers' Surplus formula:
CS=∫0x0f(x) dx−p0x0CS=\int_0^{x_0}f(x)\,dx-p_0 x_0

Step 3 – Evaluate:
∫03(35−2x−x2) dx=[35x−x2−x33]03=105−9−9=87\int_0^3(35-2x-x^2)\,dx=\left[35x-x^2-\frac{x^3}{3}\right]_0^3=105-9-9=87

CS=87−20×3=87−60=27CS=87-20\times 3=87-60=27

Answer: Consumers' Surplus =₹27=\text{₹}27
2If the demand function for a commodity is p=25−x2p=25-x^2, find the consumers' surplus for p0=9p_0=9.Show solution
Given: p=25−x2p=25-x^2, p0=9p_0=9.

Step 1 – Find x0x_0:
9=25−x02⇒x02=16⇒x0=49=25-x_0^2\Rightarrow x_0^2=16\Rightarrow x_0=4

Step 2 – Consumers' Surplus:
CS=∫04(25−x2) dx−9×4CS=\int_0^4(25-x^2)\,dx-9\times 4
=[25x−x33]04−36=(100−643)−36=64−643=192−643=1283=\left[25x-\frac{x^3}{3}\right]_0^4-36=\left(100-\frac{64}{3}\right)-36=64-\frac{64}{3}=\frac{192-64}{3}=\frac{128}{3}

Answer: CS=1283≈₹42.67CS=\dfrac{128}{3}\approx\text{₹}42.67
3The demand function for a commodity is p=10−2xp=10-2x. Find the consumers' surplus for (i) p=2p=2 (ii) p=6p=6.Show solution
Given: p=10−2xp=10-2x.

(i) p0=2p_0=2:
2=10−2x0⇒x0=42=10-2x_0\Rightarrow x_0=4.
CS=∫04(10−2x) dx−2×4=[10x−x2]04−8=(40−16)−8=16CS=\int_0^4(10-2x)\,dx-2\times 4=\left[10x-x^2\right]_0^4-8=(40-16)-8=16

(ii) p0=6p_0=6:
6=10−2x0⇒x0=26=10-2x_0\Rightarrow x_0=2.
CS=∫02(10−2x) dx−6×2=[10x−x2]02−12=(20−4)−12=4CS=\int_0^2(10-2x)\,dx-6\times 2=\left[10x-x^2\right]_0^2-12=(20-4)-12=4

Answer: (i) CS=₹16CS=\text{₹}16; (ii) CS=₹4CS=\text{₹}4
4The demand function for a commodity is p=80−3x−x2p=80-3x-x^2. Find the consumers' surplus for p=40p=40.Show solution
Given: p=80−3x−x2p=80-3x-x^2, p0=40p_0=40.

Step 1 – Find x0x_0:
40=80−3x0−x02⇒x02+3x0−40=0⇒(x0+8)(x0−5)=0⇒x0=540=80-3x_0-x_0^2\Rightarrow x_0^2+3x_0-40=0\Rightarrow(x_0+8)(x_0-5)=0\Rightarrow x_0=5

Step 2 – CS:
CS=∫05(80−3x−x2) dx−40×5CS=\int_0^5(80-3x-x^2)\,dx-40\times 5
=[80x−3x22−x33]05−200=\left[80x-\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^5-200
=(400−752−1253)−200=200−37.5−41.67=120.83=7256=\left(400-\frac{75}{2}-\frac{125}{3}\right)-200=200-37.5-41.67=120.83=\frac{725}{6}

Answer: CS=7256≈₹120.83CS=\dfrac{725}{6}\approx\text{₹}120.83
5If the supply function is p=3x2+10p=3x^2+10 and x0=4x_0=4, find the producers' surplus.Show solution
Given: p=g(x)=3x2+10p=g(x)=3x^2+10, x0=4x_0=4.

Step 1 – Find p0p_0:
p0=3(16)+10=48+10=58p_0=3(16)+10=48+10=58

Step 2 – Producers' Surplus:
PS=p0x0−∫0x0g(x) dx=58×4−∫04(3x2+10) dxPS=p_0 x_0-\int_0^{x_0}g(x)\,dx=58\times 4-\int_0^4(3x^2+10)\,dx
=232−[x3+10x]04=232−(64+40)=232−104=128=232-\left[x^3+10x\right]_0^4=232-(64+40)=232-104=128

Answer: PS=₹128PS=\text{₹}128
6If the supply function is p=4−5x+x2p=4-5x+x^2, find the producers' surplus when the price is 18.Show solution
Given: p=4−5x+x2p=4-5x+x^2, p0=18p_0=18.

Step 1 – Find x0x_0:
18=4−5x0+x02⇒x02−5x0−14=0⇒(x0−7)(x0+2)=0⇒x0=718=4-5x_0+x_0^2\Rightarrow x_0^2-5x_0-14=0\Rightarrow(x_0-7)(x_0+2)=0\Rightarrow x_0=7

Step 2 – PS:
PS=18×7−∫07(4−5x+x2) dxPS=18\times 7-\int_0^7(4-5x+x^2)\,dx
=126−[4x−5x22+x33]07=126-\left[4x-\frac{5x^2}{2}+\frac{x^3}{3}\right]_0^7
=126−(28−2452+3433)=126−(28−122.5+114.33)=126−19.83=106.17=126-\left(28-\frac{245}{2}+\frac{343}{3}\right)=126-\left(28-122.5+114.33\right)=126-19.83=106.17

Exact: 28−2452+3433=168−735+6866=119628-\dfrac{245}{2}+\dfrac{343}{3}=\dfrac{168-735+686}{6}=\dfrac{119}{6}

PS=126−1196=756−1196=6376PS=126-\frac{119}{6}=\frac{756-119}{6}=\frac{637}{6}

Answer: PS=6376≈₹106.17PS=\dfrac{637}{6}\approx\text{₹}106.17
7If the demand and supply curve for computers is D=100−6PD=100-6P, S=28+3PS=28+3P respectively where PP is the price of computers, what is the quantity of computers bought and sold at equilibrium?Show solution
Step 1 – At equilibrium, D=SD=S:
100−6P=28+3P⇒72=9P⇒P=8100-6P=28+3P\Rightarrow 72=9P\Rightarrow P=8

Step 2 – Equilibrium quantity:
Q=D=100−6(8)=100−48=52Q=D=100-6(8)=100-48=52

(Check: S=28+3(8)=52S=28+3(8)=52 ✓)

Answer: At equilibrium, price P=₹8P=\text{₹}8 and quantity Q=52Q=52 computers.

Case Based Question – Exercise 3.6

Q1Which of the following represents the Price (p) – supply (x) relationship?
a) p=65−x20p=65-\frac{x}{20} b) p=65+x20p=65+\frac{x}{20} c) p=−15+x20p=-15+\frac{x}{20} d) p=15−x20p=15-\frac{x}{20}Show solution
Correct Option: (c) p=−15+x20p=-15+\dfrac{x}{20}

Justification: From the supply table, take two points: (x,p)=(400,5)(x,p)=(400,5) and (800,25)(800,25).
Slope =25−5800−400=20400=120=\dfrac{25-5}{800-400}=\dfrac{20}{400}=\dfrac{1}{20}.
Using point (400,5)(400,5): 5=120(400)+c⇒c=−155=\dfrac{1}{20}(400)+c\Rightarrow c=-15.
So p=−15+x20p=-15+\dfrac{x}{20}.
Q2The equation of demand curve can be given by:
a) p=30−x40p=30-\frac{x}{40} b) p=30+x40p=30+\frac{x}{40} c) p=20−x40p=20-\frac{x}{40} d) p=20+x40p=20+\frac{x}{40}Show solution
Correct Option: (a) p=30−x40p=30-\dfrac{x}{40}

Justification: From the demand table, take (x,p)=(200,25)(x,p)=(200,25) and (400,20)(400,20).
Slope =20−25400−200=−5200=−140=\dfrac{20-25}{400-200}=\dfrac{-5}{200}=-\dfrac{1}{40}.
Using (200,25)(200,25): 25=−140(200)+c⇒c=3025=-\dfrac{1}{40}(200)+c\Rightarrow c=30.
So p=30−x40p=30-\dfrac{x}{40}.
Q3The value of xx at equilibrium is:
a) 1400/3 b) 600 c) 15 d) 200/3Show solution
Correct Option: (b) 600

Justification: At equilibrium, demand = supply:
30−x40=−15+x2030-\frac{x}{40}=-15+\frac{x}{20}
45=x20+x40=3x4045=\frac{x}{20}+\frac{x}{40}=\frac{3x}{40}
x=45×403=600x=\frac{45\times 40}{3}=600
Q4The equilibrium price is:
a) 400 b) 20 c) 600 d) 15Show solution
Correct Option: (d) 15

Justification: Substitute x=600x=600 into demand function:
p=30−60040=30−15=15p=30-\frac{600}{40}=30-15=15
Q5The consumers' surplus at equilibrium price is:
a) 18009 b) 13500 c) 9000 d) 4500Show solution
Correct Option: (d) 4500

Justification: p0=15p_0=15, x0=600x_0=600, demand function p=30−x40p=30-\dfrac{x}{40}.
CS=∫0600(30−x40)dx−15×600CS=\int_0^{600}\left(30-\frac{x}{40}\right)dx-15\times 600
=[30x−x280]0600−9000=\left[30x-\frac{x^2}{80}\right]_0^{600}-9000
=(18000−36000080)−9000=(18000−4500)−9000=4500=\left(18000-\frac{360000}{80}\right)-9000=(18000-4500)-9000=4500

Miscellaneous Exercise

Q1(i)Integrate x3ex2x^3 e^{x^2}Show solution
Given: ∫x3ex2 dx\int x^3 e^{x^2}\,dx

Step 1 – Write x3=x⋅x2x^3=x\cdot x^2 and let t=x2t=x^2, dt=2x dxdt=2x\,dx:
∫x3ex2 dx=∫x2⋅ex2⋅x dx=12∫t et dt\int x^3 e^{x^2}\,dx=\int x^2\cdot e^{x^2}\cdot x\,dx=\frac{1}{2}\int t\,e^t\,dt

Step 2 – Integration by parts on ∫t et dt\int t\,e^t\,dt: let u=tu=t, dv=et dtdv=e^t\,dt:
∫t et dt=tet−et+C=(t−1)et+C\int t\,e^t\,dt=te^t-e^t+C=(t-1)e^t+C

Step 3 – Back-substitute t=x2t=x^2:
=12(x2−1)ex2+C=\frac{1}{2}(x^2-1)e^{x^2}+C

Answer: (x2−1)ex22+C\dfrac{(x^2-1)e^{x^2}}{2}+C
Q1(ii)Evaluate ∫(x4−x)1/4x5 dx\int\dfrac{(x^4-x)^{1/4}}{x^5}\,dxShow solution
Given: ∫(x4−x)1/4x5 dx\int\dfrac{(x^4-x)^{1/4}}{x^5}\,dx

Step 1 – Factor: (x4−x)1/4=x1/4(x3−1)1/4(x^4-x)^{1/4}=x^{1/4}(x^3-1)^{1/4}.
∫x1/4(x3−1)1/4x5 dx=∫(x3−1)1/4x19/4 dx\int\frac{x^{1/4}(x^3-1)^{1/4}}{x^5}\,dx=\int\frac{(x^3-1)^{1/4}}{x^{19/4}}\,dx

Step 2 – Divide numerator and denominator by x3x^3:
=∫(1−x−3)1/4x7/4 dx=\int\frac{(1-x^{-3})^{1/4}}{x^{7/4}}\,dx

Let t=1−x−3=1−1x3t=1-x^{-3}=1-\dfrac{1}{x^3}, dt=3x−4 dxdt=3x^{-4}\,dx, so x−4 dx=dt3x^{-4}\,dx=\dfrac{dt}{3}.

Note 1x7/4=x−7/4\dfrac{1}{x^{7/4}}=x^{-7/4}; we need to express x−7/4 dxx^{-7/4}\,dx in terms of tt. Actually rewrite:
∫(x3−1)1/4x19/4 dx=∫x−19/4(x3−1)1/4 dx\int\frac{(x^3-1)^{1/4}}{x^{19/4}}\,dx=\int x^{-19/4}(x^3-1)^{1/4}\,dx

Let t=1−x−3t=1-x^{-3}, dt=3x−4 dxdt=3x^{-4}\,dx. Write (x3−1)1/4=x3/4(1−x−3)1/4(x^3-1)^{1/4}=x^{3/4}(1-x^{-3})^{1/4}:
=∫x−19/4⋅x3/4(1−x−3)1/4 dx=∫x−4(1−x−3)1/4 dx=13∫t1/4 dt=\int x^{-19/4}\cdot x^{3/4}(1-x^{-3})^{1/4}\,dx=\int x^{-4}(1-x^{-3})^{1/4}\,dx=\frac{1}{3}\int t^{1/4}\,dt
=13⋅4t5/45+C=415t5/4+C=415(1−1x3)5/4+C=\frac{1}{3}\cdot\frac{4t^{5/4}}{5}+C=\frac{4}{15}t^{5/4}+C=\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{5/4}+C

Answer: 415(1−1x3)5/4+C\dfrac{4}{15}\left(1-\dfrac{1}{x^3}\right)^{5/4}+C
Q1(iii)Evaluate ∫x4+xx4−9 dx\int\dfrac{x^4+x}{x^4-9}\,dxShow solution
Given: ∫x4+xx4−9 dx\int\dfrac{x^4+x}{x^4-9}\,dx

Step 1 – Rewrite numerator:
x4+x=(x4−9)+(x+9)x^4+x=(x^4-9)+(x+9)
So x4+xx4−9=1+x+9x4−9\dfrac{x^4+x}{x^4-9}=1+\dfrac{x+9}{x^4-9}.

Step 2 – Factor: x4−9=(x2−3)(x2+3)x^4-9=(x^2-3)(x^2+3).

Step 3 – Partial fractions for x+9(x2−3)(x2+3)\dfrac{x+9}{(x^2-3)(x^2+3)}:
x+9(x2−3)(x2+3)=Ax+Bx2−3+Cx+Dx2+3\frac{x+9}{(x^2-3)(x^2+3)}=\frac{Ax+B}{x^2-3}+\frac{Cx+D}{x^2+3}
x+9=(Ax+B)(x2+3)+(Cx+D)(x2−3)x+9=(Ax+B)(x^2+3)+(Cx+D)(x^2-3)

Comparing: A+C=0A+C=0, B+D=0B+D=0, 3A−3C=13A-3C=1, 3B−3D=93B-3D=9.
From A+C=0A+C=0 and 3A−3C=13A-3C=1: 6A=1⇒A=166A=1\Rightarrow A=\dfrac{1}{6}, C=−16C=-\dfrac{1}{6}.
From B+D=0B+D=0 and 3B−3D=93B-3D=9: 6B=9⇒B=326B=9\Rightarrow B=\dfrac{3}{2}, D=−32D=-\dfrac{3}{2}.

Step 4 – Integrate:
∫1 dx+∫16x+32x2−3 dx+∫−16x−32x2+3 dx\int 1\,dx+\int\frac{\frac{1}{6}x+\frac{3}{2}}{x^2-3}\,dx+\int\frac{-\frac{1}{6}x-\frac{3}{2}}{x^2+3}\,dx
=x+112log⁡∣x2−3∣+32⋅123log⁡∣x−3x+3∣−112log⁡(x2+3)−32⋅13arctan⁡x3+C=x+\frac{1}{12}\log|x^2-3|+\frac{3}{2}\cdot\frac{1}{2\sqrt{3}}\log\left|\frac{x-\sqrt{3}}{x+\sqrt{3}}\right|-\frac{1}{12}\log(x^2+3)-\frac{3}{2}\cdot\frac{1}{\sqrt{3}}\arctan\frac{x}{\sqrt{3}}+C

Answer: x+112log⁡∣x2−3∣−112log⁡(x2+3)+34log⁡∣x−3x+3∣−32arctan⁡x3+Cx+\dfrac{1}{12}\log|x^2-3|-\dfrac{1}{12}\log(x^2+3)+\dfrac{\sqrt{3}}{4}\log\left|\dfrac{x-\sqrt{3}}{x+\sqrt{3}}\right|-\dfrac{\sqrt{3}}{2}\arctan\dfrac{x}{\sqrt{3}}+C
Q1(iv)Evaluate ∫2xx4−1 dx\int\dfrac{2^x}{\sqrt{x^4-1}}\,dxShow solution
Note: This integral as written (2x2^x in numerator, x4−1\sqrt{x^4-1} in denominator) does not have a standard closed form. It is likely a misprint; the intended integral is probably ∫x2x4−1 dx\int\dfrac{x^2}{\sqrt{x^4-1}}\,dx or ∫2xx4−1 dx\int\dfrac{2x}{\sqrt{x^4-1}}\,dx.

Solving ∫2xx4−1 dx\int\dfrac{2x}{\sqrt{x^4-1}}\,dx:

Let t=x2t=x^2, dt=2x dxdt=2x\,dx:
∫dtt2−1=log⁡∣t+t2−1∣+C=log⁡∣x2+x4−1∣+C\int\frac{dt}{\sqrt{t^2-1}}=\log|t+\sqrt{t^2-1}|+C=\log|x^2+\sqrt{x^4-1}|+C

Answer (assuming 2x2x in numerator): log⁡∣x2+x4−1∣+C\log|x^2+\sqrt{x^4-1}|+C
Q1(v)Evaluate ∫1(ex+1)5 dx\int\dfrac{1}{(e^x+1)^5}\,dxShow solution
Given: ∫1(ex+1)5 dx\int\dfrac{1}{(e^x+1)^5}\,dx

Step 1 – Let t=ex+1t=e^x+1, dt=ex dx=(t−1) dxdt=e^x\,dx=(t-1)\,dx, so dx=dtt−1dx=\dfrac{dt}{t-1}.

Step 2 – Substitute:
∫1t5⋅dtt−1=∫dtt5(t−1)\int\frac{1}{t^5}\cdot\frac{dt}{t-1}=\int\frac{dt}{t^5(t-1)}

Step 3 – Partial fractions:
1t5(t−1)=At+Bt2+Ct3+Dt4+Et5+Ft−1\frac{1}{t^5(t-1)}=\frac{A}{t}+\frac{B}{t^2}+\frac{C}{t^3}+\frac{D}{t^4}+\frac{E}{t^5}+\frac{F}{t-1}

Multiply both sides by t5(t−1)t^5(t-1):
1=At4(t−1)+Bt3(t−1)+Ct2(t−1)+Dt(t−1)+E(t−1)+Ft51=At^4(t-1)+Bt^3(t-1)+Ct^2(t-1)+Dt(t-1)+E(t-1)+Ft^5

t=0t=0: 1=−E⇒E=−11=-E\Rightarrow E=-1.
t=1t=1: 1=F1=F.
Expanding and comparing coefficients gives A=−1A=-1, B=−1B=-1, C=−1C=-1, D=−1D=-1.

Step 4 – Integrate:
−log⁡∣t∣−−1t−−12t2−−13t3−−14t4+log⁡∣t−1∣+C-\log|t|-\frac{-1}{t}-\frac{-1}{2t^2}-\frac{-1}{3t^3}-\frac{-1}{4t^4}+\log|t-1|+C
=log⁡∣t−1t∣+1t+12t2+13t3+14t4+C=\log\left|\frac{t-1}{t}\right|+\frac{1}{t}+\frac{1}{2t^2}+\frac{1}{3t^3}+\frac{1}{4t^4}+C

Back-substitute t=ex+1t=e^x+1, t−1=ext-1=e^x:
=log⁡exex+1+1ex+1+12(ex+1)2+13(ex+1)3+14(ex+1)4+C=\log\frac{e^x}{e^x+1}+\frac{1}{e^x+1}+\frac{1}{2(e^x+1)^2}+\frac{1}{3(e^x+1)^3}+\frac{1}{4(e^x+1)^4}+C

Answer: log⁡exex+1+1ex+1+12(ex+1)2+13(ex+1)3+14(ex+1)4+C\log\dfrac{e^x}{e^x+1}+\dfrac{1}{e^x+1}+\dfrac{1}{2(e^x+1)^2}+\dfrac{1}{3(e^x+1)^3}+\dfrac{1}{4(e^x+1)^4}+C
Q1(vi)Evaluate ∫(1+x)log⁡x dx\int(1+x)\log x\,dxShow solution
Given: ∫(1+x)log⁡x dx\int(1+x)\log x\,dx

Step 1 – Integration by parts: Let u=log⁡xu=\log x, dv=(1+x) dxdv=(1+x)\,dx.
Then du=1x dxdu=\dfrac{1}{x}\,dx, v=x+x22v=x+\dfrac{x^2}{2}.

Step 2 – Apply formula:
∫(1+x)log⁡x dx=(x+x22)log⁡x−∫(x+x22)⋅1x dx\int(1+x)\log x\,dx=\left(x+\frac{x^2}{2}\right)\log x-\int\left(x+\frac{x^2}{2}\right)\cdot\frac{1}{x}\,dx
=(x+x22)log⁡x−∫(1+x2)dx=\left(x+\frac{x^2}{2}\right)\log x-\int\left(1+\frac{x}{2}\right)dx
=(x+x22)log⁡x−x−x24+C=\left(x+\frac{x^2}{2}\right)\log x-x-\frac{x^2}{4}+C

Answer: (x+x22)log⁡x−x−x24+C\left(x+\dfrac{x^2}{2}\right)\log x-x-\dfrac{x^2}{4}+C
Q2(i)Evaluate ∫23x3+1x(x−1) dx\int_2^3\dfrac{x^3+1}{x(x-1)}\,dxShow solution
Step 1 – Polynomial division: Degree of numerator (3) > degree of denominator (2), so divide:
x3+1x(x−1)=x3+1x2−x\frac{x^3+1}{x(x-1)}=\frac{x^3+1}{x^2-x}
Divide: x3+1=(x2−x)(x+1)+(x+1)x^3+1=(x^2-x)(x+1)+(x+1).
So x3+1x2−x=x+1+x+1x(x−1)\dfrac{x^3+1}{x^2-x}=x+1+\dfrac{x+1}{x(x-1)}.

Step 2 – Partial fractions for x+1x(x−1)\dfrac{x+1}{x(x-1)}:
x+1x(x−1)=Ax+Bx−1\frac{x+1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}
x=0x=0: 1=−A⇒A=−11=-A\Rightarrow A=-1; x=1x=1: 2=B2=B.

Step 3 – Integrate:
∫23(x+1−1x+2x−1)dx=[x22+x−log⁡∣x∣+2log⁡∣x−1∣]23\int_2^3\left(x+1-\frac{1}{x}+\frac{2}{x-1}\right)dx=\left[\frac{x^2}{2}+x-\log|x|+2\log|x-1|\right]_2^3
=(92+3−log⁡3+2log⁡2)−(2+2−log⁡2+2log⁡1)=\left(\frac{9}{2}+3-\log 3+2\log 2\right)-\left(2+2-\log 2+2\log 1\right)
=92+3−log⁡3+2log⁡2−4+log⁡2=92−1+3log⁡2−log⁡3=\frac{9}{2}+3-\log 3+2\log 2-4+\log 2=\frac{9}{2}-1+3\log 2-\log 3
=72+log⁡83=\frac{7}{2}+\log\frac{8}{3}

Answer: 72+log⁡83\dfrac{7}{2}+\log\dfrac{8}{3}
Q2(ii)Evaluate ∫1/31(x−x3)1/3x4 dx\int_{1/3}^{1}\dfrac{(x-x^3)^{1/3}}{x^4}\,dxShow solution
Given: ∫1/31(x−x3)1/3x4 dx\int_{1/3}^{1}\dfrac{(x-x^3)^{1/3}}{x^4}\,dx

Step 1 – Simplify: (x−x3)1/3=x1/3(1−x2)1/3(x-x^3)^{1/3}=x^{1/3}(1-x^2)^{1/3}.
∫1/31x1/3(1−x2)1/3x4 dx=∫1/31x−11/3(1−x2)1/3 dx\int_{1/3}^1\frac{x^{1/3}(1-x^2)^{1/3}}{x^4}\,dx=\int_{1/3}^1 x^{-11/3}(1-x^2)^{1/3}\,dx

Step 2 – Let t=1x2−1=x−2−1t=\dfrac{1}{x^2}-1=x^{-2}-1, dt=−2x−3 dxdt=-2x^{-3}\,dx.

Note (1−x2)1/3=x2/3(x−2−1)1/3=x2/3t1/3(1-x^2)^{1/3}=x^{2/3}(x^{-2}-1)^{1/3}=x^{2/3}t^{1/3}.
x−11/3(1−x2)1/3 dx=x−11/3⋅x2/3t1/3 dx=x−3t1/3 dx=−t1/32 dtx^{-11/3}(1-x^2)^{1/3}\,dx=x^{-11/3}\cdot x^{2/3}t^{1/3}\,dx=x^{-3}t^{1/3}\,dx=-\frac{t^{1/3}}{2}\,dt

When x=1/3x=1/3: t=9−1=8t=9-1=8; when x=1x=1: t=0t=0.

Step 3 – Substitute:
∫80−t1/32 dt=12∫08t1/3 dt=12⋅3t4/34∣08=38⋅84/3=38⋅16=6\int_8^0-\frac{t^{1/3}}{2}\,dt=\frac{1}{2}\int_0^8 t^{1/3}\,dt=\frac{1}{2}\cdot\frac{3t^{4/3}}{4}\Bigg|_0^8=\frac{3}{8}\cdot 8^{4/3}=\frac{3}{8}\cdot 16=6

Answer: 66
Q2(iii)Evaluate ∫01log⁡(1x−1) dx\int_0^1\log\left(\dfrac{1}{x}-1\right)\,dxShow solution
Step 1 – Let I=∫01log⁡(1−xx) dx=∫01[log⁡(1−x)−log⁡x] dxI=\int_0^1\log\left(\dfrac{1-x}{x}\right)\,dx=\int_0^1[\log(1-x)-\log x]\,dx …(1)

Step 2 – Replace xx by 1−x1-x:
I=∫01[log⁡x−log⁡(1−x)] dx…(2)I=\int_0^1[\log x-\log(1-x)]\,dx \quad\text{…(2)}

Step 3 – Add (1) and (2): 2I=0⇒I=02I=0\Rightarrow I=0.

Answer: 00
Q2(iv)Evaluate ∫02x22−x dx\int_0^2 x^2\sqrt{2-x}\,dxShow solution
Step 1 – Let t=2−xt=2-x, x=2−tx=2-t, dx=−dtdx=-dt.

When x=0x=0, t=2t=2; when x=2x=2, t=0t=0.

Step 2 – Substitute:
∫20(2−t)2t(−dt)=∫02(2−t)2t dt=∫02(4−4t+t2)t1/2 dt\int_2^0(2-t)^2\sqrt{t}(-dt)=\int_0^2(2-t)^2\sqrt{t}\,dt=\int_0^2(4-4t+t^2)t^{1/2}\,dt
=∫02(4t1/2−4t3/2+t5/2) dt=\int_0^2(4t^{1/2}-4t^{3/2}+t^{5/2})\,dt

Step 3 – Integrate:
=[8t3/23−8t5/25+2t7/27]02=\left[\frac{8t^{3/2}}{3}-\frac{8t^{5/2}}{5}+\frac{2t^{7/2}}{7}\right]_0^2
=8⋅223−8⋅425+2⋅827=\frac{8\cdot 2\sqrt{2}}{3}-\frac{8\cdot 4\sqrt{2}}{5}+\frac{2\cdot 8\sqrt{2}}{7}
=2(163−325+167)=2⋅16(13−25+17)=\sqrt{2}\left(\frac{16}{3}-\frac{32}{5}+\frac{16}{7}\right)=\sqrt{2}\cdot 16\left(\frac{1}{3}-\frac{2}{5}+\frac{1}{7}\right)
=162⋅35−42+15105=162⋅8105=1282105=16\sqrt{2}\cdot\frac{35-42+15}{105}=16\sqrt{2}\cdot\frac{8}{105}=\frac{128\sqrt{2}}{105}

Answer: 1282105\dfrac{128\sqrt{2}}{105}
Q2(v)Evaluate ∫−1111+ex2 dx\int_{-1}^{1}\dfrac{1}{1+e^{x^2}}\,dxShow solution
Note: ex2e^{x^2} is an even function, so 11+ex2\dfrac{1}{1+e^{x^2}} is also even.

Step 1 – Let I=∫−1111+ex2 dxI=\int_{-1}^{1}\dfrac{1}{1+e^{x^2}}\,dx.

Step 2 – Replace xx by −x-x:
I=∫−1111+e(−x)2 dx=∫−1111+ex2 dx=II=\int_{-1}^{1}\frac{1}{1+e^{(-x)^2}}\,dx=\int_{-1}^{1}\frac{1}{1+e^{x^2}}\,dx=I

This gives no new information. Instead, note that 11+ex2+ex21+ex2=1\dfrac{1}{1+e^{x^2}}+\dfrac{e^{x^2}}{1+e^{x^2}}=1, and both integrals are equal (by substitution x→−xx\to -x, since ex2e^{x^2} is even):
2I=∫−111 dx=2⇒I=12I=\int_{-1}^{1}1\,dx=2\Rightarrow I=1

Answer: 11
Q2(vi)Evaluate ∫−11∣x∣−x dx\int_{-1}^{1}\sqrt{|x|-x}\,dxShow solution
Step 1 – Consider cases:
- For x≥0x\geq 0: ∣x∣=x|x|=x, so ∣x∣−x=0=0\sqrt{|x|-x}=\sqrt{0}=0.
- For x<0: ∣x∣=−x|x|=-x, so ∣x∣−x=−x−x=−2x\sqrt{|x|-x}=\sqrt{-x-x}=\sqrt{-2x}.

Step 2 – Split:
∫−11∣x∣−x dx=∫−10−2x dx+∫010 dx\int_{-1}^{1}\sqrt{|x|-x}\,dx=\int_{-1}^{0}\sqrt{-2x}\,dx+\int_0^1 0\,dx

Step 3 – Evaluate:
∫−10−2x dx=2∫−10−x dx\int_{-1}^{0}\sqrt{-2x}\,dx=\sqrt{2}\int_{-1}^{0}\sqrt{-x}\,dx

Let u=−xu=-x, du=−dxdu=-dx; when x=−1x=-1, u=1u=1; when x=0x=0, u=0u=0:
=2∫01u du=2⋅23=223=\sqrt{2}\int_0^1\sqrt{u}\,du=\sqrt{2}\cdot\frac{2}{3}=\frac{2\sqrt{2}}{3}

Answer: 223\dfrac{2\sqrt{2}}{3}
Q3Show that ∫(ax+bx)2axbx dx=(ab)x−(ba)xlog⁡a−log⁡b+2x+C\int\dfrac{(a^x+b^x)^2}{a^x b^x}\,dx=\dfrac{\left(\frac{a}{b}\right)^x-\left(\frac{b}{a}\right)^x}{\log a-\log b}+2x+CShow solution
Step 1 – Expand the integrand:
(ax+bx)2axbx=a2x+2axbx+b2xaxbx=a2xaxbx+2+b2xaxbx\frac{(a^x+b^x)^2}{a^x b^x}=\frac{a^{2x}+2a^x b^x+b^{2x}}{a^x b^x}=\frac{a^{2x}}{a^x b^x}+2+\frac{b^{2x}}{a^x b^x}
=axbx+2+bxax=(ab)x+(ba)x+2=\frac{a^x}{b^x}+2+\frac{b^x}{a^x}=\left(\frac{a}{b}\right)^x+\left(\frac{b}{a}\right)^x+2

Step 2 – Integrate term by term:
∫[(ab)x+(ba)x+2]dx\int\left[\left(\frac{a}{b}\right)^x+\left(\frac{b}{a}\right)^x+2\right]dx
=(ab)xlog⁡ab+(ba)xlog⁡ba+2x+C=\frac{\left(\frac{a}{b}\right)^x}{\log\frac{a}{b}}+\frac{\left(\frac{b}{a}\right)^x}{\log\frac{b}{a}}+2x+C

Step 3 – Simplify: Note log⁡ba=−log⁡ab=−(log⁡a−log⁡b)\log\dfrac{b}{a}=-\log\dfrac{a}{b}=-(\log a-\log b):
=(ab)xlog⁡a−log⁡b−(ba)xlog⁡a−log⁡b+2x+C=\frac{\left(\frac{a}{b}\right)^x}{\log a-\log b}-\frac{\left(\frac{b}{a}\right)^x}{\log a-\log b}+2x+C
=(ab)x−(ba)xlog⁡a−log⁡b+2x+C=\frac{\left(\frac{a}{b}\right)^x-\left(\frac{b}{a}\right)^x}{\log a-\log b}+2x+C

Hence proved. ■\blacksquare
Q4A firm finds that quantity demanded and quantity supplied are 30 units when market price is ₹8 per unit. Further, if price is increased to ₹12 per unit, demand reduces to 0 and at a price of ₹5 per unit, the firm is not willing to produce. Assuming linear relationships, find the demand function, supply function, consumers' surplus and producers' surplus at equilibrium price.Show solution
Step 1 – Demand function: Two points: (x,p)=(30,8)(x,p)=(30,8) and (0,12)(0,12).
Slope =12−80−30=−215=\dfrac{12-8}{0-30}=-\dfrac{2}{15}.
Using (0,12)(0,12): p=12−2x15p=12-\dfrac{2x}{15}.

Step 2 – Supply function: Two points: (x,p)=(0,5)(x,p)=(0,5) (not willing to produce at p=5p=5, so supply =0=0 at p=5p=5) and (30,8)(30,8).
Slope =8−530−0=110=\dfrac{8-5}{30-0}=\dfrac{1}{10}.
Using (0,5)(0,5): p=5+x10p=5+\dfrac{x}{10}.

Step 3 – Equilibrium: Set demand == supply:
12−2x15=5+x1012-\frac{2x}{15}=5+\frac{x}{10}
7=x10+2x15=3x+4x30=7x307=\frac{x}{10}+\frac{2x}{15}=\frac{3x+4x}{30}=\frac{7x}{30}
x0=30,p0=8x_0=30,\quad p_0=8

(This confirms the given equilibrium.)

Step 4 – Consumers' Surplus:
CS=∫030(12−2x15)dx−8×30CS=\int_0^{30}\left(12-\frac{2x}{15}\right)dx-8\times 30
=[12x−x215]030−240=(360−60)−240=60=\left[12x-\frac{x^2}{15}\right]_0^{30}-240=(360-60)-240=60

Step 5 – Producers' Surplus:
PS=8×30−∫030(5+x10)dxPS=8\times 30-\int_0^{30}\left(5+\frac{x}{10}\right)dx
=240−[5x+x220]030=240−(150+45)=240−195=45=240-\left[5x+\frac{x^2}{20}\right]_0^{30}=240-(150+45)=240-195=45

Answer: Demand function: p=12−2x15p=12-\dfrac{2x}{15}; Supply function: p=5+x10p=5+\dfrac{x}{10}; Equilibrium at x0=30x_0=30, p0=₹8p_0=\text{₹}8; CS=₹60CS=\text{₹}60; PS=₹45PS=\text{₹}45

Stuck on a step?

Ask Super Tutor AI to explain any solution on this page in a simpler way — free, 24x7.

Ask a Doubt Free

Frequently Asked Questions

What are the important topics in Integration and its Applications for CBSE Class 12 Applied Mathematics?
Integration and its Applications covers several key topics that are frequently asked in CBSE Class 12 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Integration and its Applications — CBSE Class 12 Applied Mathematics?
Understand the core concepts first, then work through the 45 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Integration and its Applications Class 12 Applied Mathematics?
This page has free step-by-step NCERT Solutions for every exercise question in Integration and its Applications (CBSE Class 12 Applied Mathematics) — written the way examiners award marks: given, formula, working, answer.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

More resources for Integration and its Applications

All 10 chapters — CBSE Class 12 Applied Mathematics NCERT Solutions

For serious students

Get the full Integration and its Applications chapter — for free.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for CBSE Class 12 Applied Mathematics.

Try Super Tutor Free
Try the full chapter — Free