NCERT Solutions: Class 12 Maths Chapter 8 - Application of Integrals

The ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a=4$ and $b=3$. The area bounded by an ellipse is $\pi ab$. Hence the required area is $\pi(4)(3)=12\pi$ square units.

Here $a^2=4$ and $b^2=9$, so $a=2$ and $b=3$. The area of the ellipse is $\pi ab=\pi(2)(3)=6\pi$ square units.

1. i. $\pi$
2. ii. $\dfrac{\pi}{2}$
3. iii. $\dfrac{\pi}{3}$
4. iv. $\dfrac{\pi}{4}$

In the first quadrant, the circle $x^2+y^2=4$ has radius $2$. The region bounded by $x=0$ and $x=2$ is the first-quadrant quarter of the circle. Hence area $=\dfrac14\pi(2)^2=\pi$, so option (i) is correct.

1. i. $2$
2. ii. $\dfrac94$
3. iii. $\dfrac93$
4. iv. $\dfrac92$

The curve gives $x=\dfrac{y^2}{4}$. The required area is measured with respect to $y$: $A=\int_0^3 x\,dy=\int_0^3\dfrac{y^2}{4}\,dy=\left[\dfrac{y^3}{12}\right]_0^3=\dfrac{27}{12}=\dfrac94$.

(i) The required area is $\int_1^2 x^2\,dx=\left[\dfrac{x^3}{3}\right]_1^2=\dfrac{8-1}{3}=\dfrac73$. (ii) The required area is $\int_1^5 x^4\,dx=\left[\dfrac{x^5}{5}\right]_1^5=\dfrac{3125-1}{5}=\dfrac{3124}{5}$.

The graph of $y=|x+3|$ is V-shaped with vertex at $(-3,0)$. Split the integral at $x=-3$: $\int_{-6}^{0}|x+3|\,dx=\int_{-6}^{-3}-(x+3)\,dx+\int_{-3}^{0}(x+3)\,dx$. The two parts are congruent right triangles of base $3$ and height $3$, so the total area is $2\cdot\dfrac12\cdot3\cdot3=9$.

$\sin x$ is non-negative on $[0,\pi]$ and non-positive on $[\pi,2\pi]$. Hence the area is $\int_0^\pi\sin x\,dx-\int_\pi^{2\pi}\sin x\,dx=2+2=4$ square units.

1. i. $-9$
2. ii. $-\dfrac{15}{4}$
3. iii. $\dfrac{15}{4}$
4. iv. $\dfrac{17}{4}$

Since $x^3$ is negative on $[-2,0]$ and positive on $[0,1]$, the area is $-\int_{-2}^{0}x^3\,dx+\int_0^1x^3\,dx=-\left[\dfrac{x^4}{4}\right]_{-2}^{0}+\left[\dfrac{x^4}{4}\right]_0^1=4+\dfrac14=\dfrac{17}{4}$.

1. i. $0$
2. ii. $\dfrac13$
3. iii. $\dfrac23$
4. iv. $\dfrac43$

For $x\lt0$, $y=x|x|=-x^2$; for $x\gt0$, $y=x|x|=x^2$. Therefore the area is $-\int_{-1}^{0}(-x^2)\,dx+\int_0^1x^2\,dx=\int_{-1}^{0}x^2\,dx+\int_0^1x^2\,dx=\dfrac13+\dfrac13=\dfrac23$.