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Integration and its Applications

CBSE · Class 12 · Applied Mathematics

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

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82 Questions Solved · 10 Sections

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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

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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

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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

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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

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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.)*

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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}

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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}

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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.

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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

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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

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(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}

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(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

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(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}

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(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}

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(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]

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(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

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(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

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(ix)Evaluate ∫04x2+9 dx\int_0^4\sqrt{x^2+9}\,dx
(x)Evaluate ∫013t2(1+t3)(2+t3) dt\int_0^1\dfrac{3t^2}{(1+t^3)(2+t^3)}\,dt

Exercise 3.5

Q1(i)Evaluate ∫03f(x) dx\int_0^3 f(x)\,dx where f(x)={x+1x<12xx≥1f(x)=\begin{cases}x+1 & x<1\\ 2x & x\geq 1\end{cases}
Q1(ii)Evaluate ∫01x1−x dx\int_0^1 x\sqrt{1-x}\,dx
Q1(iii)Evaluate ∫04∣x−2∣ dx\int_0^4|x-2|\,dx
Q1(iv)Evaluate ∫15xx+6−x dx\int_1^5\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{6-x}}\,dx
Q1(v)Evaluate ∫0ax2020x2020+(a−x)2020 dx\int_0^a\dfrac{x^{2020}}{x^{2020}+(a-x)^{2020}}\,dx
Q1(vi)Evaluate ∫04(∣x∣+∣x−2∣+∣x−4∣) dx\int_0^4(|x|+|x-2|+|x-4|)\,dx
Q1(vii)Evaluate ∫−2211+ex dx\int_{-2}^{2}\dfrac{1}{1+\sqrt{e^x}}\,dx
Q1(viii)Evaluate ∫−11x3+∣x∣+1x2+2∣x∣+1 dx\int_{-1}^{1}\dfrac{x^3+|x|+1}{x^2+2|x|+1}\,dx
Q1(ix)Evaluate ∫01x(1−x)n dx\int_0^1 x(1-x)^n\,dx
Q1(x)Evaluate ∫−22(x3+12)4+x2 dx\int_{-2}^{2}\left(x^3+\dfrac{1}{2}\right)\sqrt{4+x^2}\,dx
Q1(xi)Evaluate ∫−11log⁡1−x1+x dx\int_{-1}^{1}\log\dfrac{1-x}{1+x}\,dx
Q1(xii)Evaluate ∫−11x21+ex dx\int_{-1}^{1}\dfrac{x^2}{1+e^x}\,dx
Q2Evaluate ∫02[x] dx\int_0^2[x]\,dx where [⋅][\cdot] denotes the Greatest Integer Function.

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.
2If the demand function for a commodity is p=25−x2p=25-x^2, find the consumers' surplus for p0=9p_0=9.
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.
4The demand function for a commodity is p=80−3x−x2p=80-3x-x^2. Find the consumers' surplus for p=40p=40.
5If the supply function is p=3x2+10p=3x^2+10 and x0=4x_0=4, find the producers' surplus.
6If the supply function is p=4−5x+x2p=4-5x+x^2, find the producers' surplus when the price is 18.
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?

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}
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}
Q3The value of xx at equilibrium is:
a) 1400/3 b) 600 c) 15 d) 200/3
Q4The equilibrium price is:
a) 400 b) 20 c) 600 d) 15
Q5The consumers' surplus at equilibrium price is:
a) 18009 b) 13500 c) 9000 d) 4500

Miscellaneous Exercise

Q1(i)Integrate x3ex2x^3 e^{x^2}
Q1(ii)Evaluate ∫(x4−x)1/4x5 dx\int\dfrac{(x^4-x)^{1/4}}{x^5}\,dx
Q1(iii)Evaluate ∫x4+xx4−9 dx\int\dfrac{x^4+x}{x^4-9}\,dx
Q1(iv)Evaluate ∫2xx4−1 dx\int\dfrac{2^x}{\sqrt{x^4-1}}\,dx
Q1(v)Evaluate ∫1(ex+1)5 dx\int\dfrac{1}{(e^x+1)^5}\,dx
Q1(vi)Evaluate ∫(1+x)log⁡x dx\int(1+x)\log x\,dx
Q2(i)Evaluate ∫23x3+1x(x−1) dx\int_2^3\dfrac{x^3+1}{x(x-1)}\,dx
Q2(ii)Evaluate ∫1/31(x−x3)1/3x4 dx\int_{1/3}^{1}\dfrac{(x-x^3)^{1/3}}{x^4}\,dx
Q2(iii)Evaluate ∫01log⁡(1x−1) dx\int_0^1\log\left(\dfrac{1}{x}-1\right)\,dx
Q2(iv)Evaluate ∫02x22−x dx\int_0^2 x^2\sqrt{2-x}\,dx
Q2(v)Evaluate ∫−1111+ex2 dx\int_{-1}^{1}\dfrac{1}{1+e^{x^2}}\,dx
Q2(vi)Evaluate ∫−11∣x∣−x dx\int_{-1}^{1}\sqrt{|x|-x}\,dx
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+C
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.

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