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Application of Integrals

CBSE · Class 12 · Mathematics

NCERT Solutions for Application of Integrals — CBSE Class 12 Mathematics.

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An illustration showing a curve y=f(x) bounded by the x-axis and vertical lines x=a and x=b. The area is approximated by numerous thin vertical rectangular strips, with one representative strip highli
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10 Questions Solved · 2 Sections

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EXERCISE 8.1

1Find the area of the region bounded by the ellipse x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1.Show solution
For an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, the area enclosed is πab\pi ab. Here,
x216+y29=1  ⟹  a=4, b=3. \frac{x^2}{16}+\frac{y^2}{9}=1 \implies a=4,\ b=3.
So the area is
πab=π(4)(3)=12π. \pi ab=\pi(4)(3)=12\pi.
Wait: since a2=16⇒a=4a^2=16\Rightarrow a=4 and b2=9⇒b=3b^2=9\Rightarrow b=3, the correct area is 12π12\pi, not 48π48\pi.

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2Find the area of the region bounded by the ellipse x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 1.Show solution
For the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, area =πab=\pi ab. Here,
x24+y29=1  ⟹  a=2, b=3. \frac{x^2}{4}+\frac{y^2}{9}=1 \implies a=2,\ b=3.
Hence area
πab=π(2)(3)=6π. \pi ab=\pi(2)(3)=6\pi.

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3Area lying in the first quadrant and bounded by the circle x2+y2=4x^2 + y^2 = 4 and the lines x=0x = 0 and x=2x = 2 isShow solution
In the first quadrant, the circle x2+y2=4x^2+y^2=4 has radius r=2r=2. The required region is the area under the quarter-circle from x=0x=0 to x=2x=2, which is one-fourth of the circle:
Area=14πr2=14π(2)2=π. \text{Area}=\frac{1}{4}\pi r^2=\frac{1}{4}\pi(2)^2=\pi.
So the correct option is (A).

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4Area of the region bounded by the curve y2=4xy^2 = 4x, yy-axis and the line y=3y = 3 isShow solution
Given y2=4xy^2=4x, we have x=y24x=\frac{y^2}{4}. The region is bounded by the y-axis and the line y=3y=3, so we use horizontal strips:
A=∫03x dy=∫03y24 dy. A=\int_0^3 x\,dy=\int_0^3 \frac{y^2}{4}\,dy.
Now,
A=14∫03y2 dy=14[y33]03=14⋅273=2712=94. A=\frac14\int_0^3 y^2\,dy=\frac14\left[\frac{y^3}{3}\right]_0^3 =\frac14\cdot\frac{27}{3}=\frac{27}{12}=\frac{9}{4}.
So the correct option is (B).

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Miscellaneous Exercise on Chapter 8

1(i)y=x2y = x^2, x=1x = 1, x=2x = 2 and xx-axisShow solution
Area under y=x2y=x^2 from x=1x=1 to x=2x=2 is
A=∫12x2 dx=[x33]12=83−13=73. A=\int_1^2 x^2\,dx=\left[\frac{x^3}{3}\right]_1^2 =\frac{8}{3}-\frac{1}{3}=\frac{7}{3}.
So the area is 73\frac{7}{3}.

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1(ii)y=x4y = x^4, x=1x = 1, x=5x = 5 and xx-axis
2Sketch the graph of y=∣x+3∣y = |x + 3| and evaluate ∫−60∣x+3∣dx\int_{-6}^{0} |x + 3| dx.
3Find the area bounded by the curve y=sin⁡xy = \sin x between x=0x = 0 and x=2πx = 2\pi.
4Area bounded by the curve y=x3y = x^3, the xx-axis and the ordinates x=−2x = -2 and x=1x = 1 is
5The area bounded by the curve y=x∣x∣y = x|x|, xx-axis and the ordinates x=−1x = -1 and x=1x = 1 is given by

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Frequently Asked Questions

What are the important topics in Application of Integrals for CBSE Class 12 Mathematics?
Application of Integrals 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 Application of Integrals — CBSE Class 12 Mathematics?
Understand the core concepts first, then work through the 98 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 Application of Integrals Class 12 Mathematics?
This page has free step-by-step NCERT Solutions for every exercise question in Application of Integrals (CBSE Class 12 Mathematics) — written the way examiners award marks: given, formula, working, answer.

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