CBSE Class 12 Maths Chapter 8 Application Of Integrals Important Questions
Question 1. Using integration, find the area bounded by the curve y ²= 4x, y-axis, and y = 3.
Or,
Using integration, find the region’s area bounded by the line 2y = – x + 8, x-axis. x = 2 and x = 4.
Solution:
Given curve is y² = 4x ….(1)
and given line is y = 3 …..(2)
From equations (1) and (2):
Point of intersection is B(9/4, 3)
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⇒ Required Area = \(\int_0^3\) x dy
= \(\frac{1}{4} \int_0^3 y^2, d y=\frac{1}{4}\left[\frac{y^3}{3}\right]_0^3=\frac{1}{4}\left[\frac{3^3}{3}-\frac{0}{3}\right]=\left(\frac{1}{4}, \frac{27}{3}\right)=\frac{9}{4} \text { sq. units }\)
Or,
Given lines are 2y + x = 8, x = 2 and x = 4
⇒ Required area = \(\int_2^4 y \mathrm{dx}=\int_2^4\left(\frac{8-\mathrm{x}}{2}\right) \mathrm{dx}\)
= \(\int_2^4\left(4-\frac{1}{2} \mathrm{x}\right) \mathrm{dx}=\left[4 \mathrm{x}-\frac{1}{2}\left(\frac{\mathrm{x}^2}{2}\right)\right]_2^4\)
= \(\left[4 \times 4-\frac{1}{4}(4)^2\right]-\left[4 \times 2-\frac{1}{4}(2)^2\right]\)
=12-7=5
∴Required Area = 5 sq. units
Question 2. Using integration, find the area bounded by the circle x² + y² = 9.
Solution:
The whole area enclosed by the given circle will be 4 times the area of the region AOBA bounded by the curve, x-axis, and the ordinates x = 0 and x = 3 [as the circle is symmetrical about both the x-axis and y-axis]
⇒ Required area = \(4 \int_0^3 y d x=4 \int_0^3 \sqrt{3^2-x^2} d x \quad\left[x^2+y^2=3^2 \text { gives } y= \pm \sqrt{3^2-x^2}\right]\)
As the region AOBA lies in the first quadrant, v is taken as positive. Integrating, we get the whole area enclosed by the given circle
⇒ Required area = \(4\left[\frac{x}{2} \sqrt{3^2-x^2}+\frac{3^2}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right]_0^3\)
= \(4\left[\left(\frac{3}{2} \times 0+\frac{3^2}{2} \sin ^{-1}(1)\right)-0\right]=4\left(\frac{3^2}{2}\right)\left(\frac{\pi}{2}\right)=9 \pi \text { sq. units }\)
Question 3. Find the area of the region bounded by curve 4x²=y and the line y = 8x + 12. using integration.
Solution:
Given curve is 4x²= y….(1)
and given line is y = 8x + 12…..(2)
From equation (1) and (2), we get:
4x² – 8x – 12 = 0
⇒ x² – 2x – 3 = 0
⇒ (x-3) (x + 1) = 0 =3 x = 3, -1
From equation (1); when x = 3, y = 36 and when x = – 1 ⇒ y = 4.
So, point of intersection of the curve and line tire (3, 36) and (-1,4).
⇒ Required Area = \(\int_{-1}^3\left\{(8 x+12)-4 x^2\right\} d x=\left[\frac{8 x^2}{2}+12 x-\frac{4 x^3}{3}\right]_{-1}^3\)
= \((36+36-36)-\left(4-12+\frac{4}{3}\right)=36+\frac{20}{3}=\frac{128}{3} \text { sq. units }\)
Question 4. Using integration, find the area of the region bounded by the curves x² + y² = 4, x =√3y, and the x-axis lying in the first quadrant.
Solution:
Given curve x² + y² = 4 is a circle with center (0, 0) and radius 2.
And line is x = √3y
Now, for the point of intersection of the line and circle, we have:
⇒ 4y² = 4 ⇒ y = ± 1
For y = 1; x = √3
So. point C is (3, 1)
Required area = area of OACO + Area of ABCA
= \(\int_0^{\sqrt{3}} y_{\text ({line })} d x+\int_{\sqrt{3}}^2 y_{\mid \text {circle } \mid} d x=\frac{1}{\sqrt{3}} \int_0^{\sqrt{3}} x d x+\int_{\sqrt{3}}^2 \sqrt{4-x^2} d x\)
= \(\frac{1}{2 \sqrt{3}}\left[x^2\right]_0^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_3^{-2}\)
= \(\frac{1}{2 \sqrt{3}}(3)+\left\{2 \sin ^{-1}(1)-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}=\frac{\sqrt{3}}{2}+\left(2 \times \frac{\pi}{2}\right)-\frac{\sqrt{3}}{2}-2 \times \frac{\pi}{3}=\frac{\pi}{3} \text { sq.units }\)
Question 5. Using the method of integration, find the area of the triangle ABC. coordinates of whose vertices are A(2. 0), B(4, 5) and C(6, 3).
Solution:
Vertices of ΔABC are A(2,0), B(4,5) and C (6,3)
Equation of line AB: y = 5/2(x – 2)
Equation of line BC: y = 9-x
Equation of line AC: y= 3/4(x-2)
⇒ Required Area of ΔABC = \(\int_2^4(\text { line } \mathrm{AB}) \mathrm{dx}+\int_4^6(\text { line } \mathrm{BC}) \mathrm{dx}-\int_2^6(\text { line } \mathrm{AC}) \mathrm{dx}\)
= \(\frac{5}{2} \int_2^4(x-2) d x+\int_4^6(9-x) d x-\frac{3}{4} \int_2^6(x-2) d x\)
= \(\frac{5}{2}\left[\frac{x^2}{2}-2 x\right]_2^4+\left[9 x-\frac{x^2}{2}\right]_4^6-\frac{3}{4}\left[\frac{x^2}{2}-2 x\right]_2^6\)
= \(\frac{5}{2}[(8-8)-(2-4)]+[(54-18)-(36-8)]-\frac{3}{4}[(18-12)-(2-4)]\)
= \(\left(\frac{5}{2} \times 2\right)+8-\left(\frac{3}{4} \times 8\right)=5+8-6=7 \text { sq. units }\)
Question 6. Using the method of integration, find the area of a triangle whose vertices arc (1, 0), (2, 2), and (3, 1).
Solution:
Equation of line AB is y – 0 = \(\frac{(2-0)}{(2-1)}(x-1) \Rightarrow y=2(x-1)\)
Equation of line BC is y – 2 = \(\frac{(1-2)}{(3-2)}(x-2) \Rightarrow y=(-x+1)\)
Equation of line AC is y – 0 = \(\frac{(1-0)}{(3-1)}(x-1) \Rightarrow y=1/2(x-1)\)
⇒ Required Area = \(\int_1^2(\text { line } A B) d x+\int_2^3(\text { line } B C) d x-\int_1^3(\text { line } A C) d x\)
= \(\int_1^2 2(\mathrm{x}-1) \mathrm{dx}+\int_2^3(-\mathrm{x}+4) \mathrm{dx}-\int_1^3 \frac{1}{2}(\mathrm{x}-1) \mathrm{dx}\)
= \(2\left(\frac{\mathrm{x}^2}{2}-\mathrm{x}\right)_1^2+\left(\frac{-\mathrm{x}^2}{2}+4 \mathrm{x}\right)_2^3-\frac{1}{2}\left(\frac{\mathrm{x}^2}{2}-\mathrm{x}\right)_1^3\)
= \(2\left[(2-2)-\left(\frac{1}{2}-1\right)\right]+\left[\left(\frac{-9}{2}+12\right)-(-2+8)\right]-\frac{1}{2}\left[\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)\right] \)
= \(\left(2 \times \frac{1}{2}\right)+\left(\frac{-9}{2}+6\right)-\frac{1}{2}(4-2)=1+\frac{3}{2}-1=\frac{3}{2} \text { sq. units }\)
Question 7. Using integration, find the region’s area in the first quadrant enclosed by the x-axis, the line y = x, and the circle x² + y² = 32.
Solution:
Given line is y = x ….(1)
and given circle is x² + y² = 32….(2)
From equations (1) and (2); we have
2x² = 32 ⇒ x = ± 4
∴ y = ± 4
Now; (4, 4) lies in 1st quadrant
⇒ Required Area = \(\int_0^4 x d x+\int_1^{4 \sqrt{2}} \sqrt{32-x^2} d x\)
= \(\left(\frac{x^2}{2}\right)_0^4+\left[\frac{x}{2} \sqrt{32-x^2}+\frac{32}{2} \sin ^{-1}\left(\frac{x}{4 \sqrt{2}}\right)\right]_{-4}^{4 \sqrt{2}}\)
= \((8-0)+\left[\left(0+16 \times \frac{\pi}{2}\right)-\left(8+16 \times \frac{\pi}{4}\right)\right]=4 \pi \text { sq. units }\)