Examples (Revised) - Chapter 8 - Applications Of Integrals - Ncert Solutions class 12 - Maths

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Chapter 8 - Applications Of Integrals NCERT Solutions Class 12 Maths

Example 1

Find the area enclosed by the circle $x^2+y^2=a^2$.
Solution

From Fig 8.5, the whole area enclosed by the given circle

$=4($ area of the region $\mathrm{AOBA}$ bounded by the curve, $x$-axis and the ordinates $x=0$ and $x=a$ ) [as the circle is symmetrical about both $x$-axis and $y$-axis]
$
\begin{aligned}
& =4 \int_0^a y d x \text { (taking vertical strips) } \\
& =4 \int_0^a \sqrt{a^2-x^2} d x
\end{aligned}
$

Since $x^2+y^2=a^2$ gives $\quad y= \pm \sqrt{a^2-x^2}$

As the region AOBA lies in the first quadrant, $y$ is taken as positive. Integrating, we get the whole area enclosed by the given circle
$
\begin{aligned}
& =4\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a \\
& =4\left[\left(\frac{a}{2} \times 0+\frac{a^2}{2} \sin ^{-1} 1\right)-0\right]=4\left(\frac{a^2}{2}\right)\left(\frac{\pi}{2}\right)=\pi a^2
\end{aligned}
$

Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the region enclosed by circle

$
\begin{aligned}
& =4 \int_0^a x d y=4 \int_0^a \sqrt{a^2-y^2} d y \\
& =4\left[\frac{y}{2} \sqrt{a^2-y^2}+\frac{a^2}{2} \sin ^{-1} \frac{y}{a}\right]_0^a \\
& =4\left[\left(\frac{a}{2} \times 0+\frac{a^2}{2} \sin ^{-1} 1\right)-0\right] \\
& =4 \frac{a^2}{2} \frac{\pi}{2}=\pi a^2
\end{aligned}
$
Example 2

Find the area enclosed by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Solution

From Fig 8.7, the area of the region ABA'B'A bounded by the ellipse
$
=4\binom{\text { area of the region } A O B A \text { in the first quadrant bounded }}{\text { by the curve, } x-\text { axis and the ordinates } x=0, x=a}
$
(as the ellipse is symmetrical about both $x$-axis and $y$-axis)
$=4 \int_0^a y d x$ (taking vertical strips)
Now $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ gives $y= \pm \frac{b}{a} \sqrt{a^2-x^2}$, but as the region AOBA lies in the first quadrant, $y$ is taken as positive. So, the required area is

$
\begin{aligned}
& =4 \int_0^a \frac{b}{a} \sqrt{a^2-x^2} d x \\
& =\frac{4 b}{a}\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a \\
& =\frac{4 b}{a}\left[\left(\frac{a}{2} \times 0+\frac{a^2}{2} \sin ^{-1} 1\right)-0\right] \\
& =\frac{4 b}{a} \frac{a^2}{2} \frac{\pi}{2}=\pi a b
\end{aligned}
$

Alternatively, considering horizontal strips as shown in the Fig 8.8, the area of the ellipse is
$
\begin{aligned}
& =4 \int_0^b x d y=4 \frac{a}{b} \int_0^b \sqrt{b^2-y^2} d y \text { (Why?) } \\
& =\frac{4 a}{b}\left[\frac{y}{2} \sqrt{b^2-y^2}+\frac{b^2}{2} \sin ^{-1} \frac{y}{b}\right]_0^b \\
& =\frac{4 a}{b}\left[\left(\frac{b}{2} \times 0+\frac{b^2}{2} \sin ^{-1} 1\right)-0\right] \\
& =\frac{4 a}{b} \frac{b^2}{2} \frac{\pi}{2}=\pi a b
\end{aligned}
$

Example 3

Find the area of the region bounded by the line $y=3 x+2$, the $x$-axis and the ordinates $x=-1$ and $x=1$.

Solution

As shown in the Fig 8.9, the line $y=3 x+2$ meets $x$-axis at $x=\frac{-2}{3}$ and its graph lies below $x$-axis for $x \in\left(-1, \frac{-2}{3}\right)$ and above $x$-axis for $x \in\left(\frac{-2}{3}, 1\right)$.

The required area $=$ Area of the region $\mathrm{ACBA}+$ Area of the region ADEA

$\begin{aligned}
& =\left|\int_{-1}^{\frac{-2}{3}}(3 x+2) d x\right|+\int_{\frac{-2}{3}}^1(3 x+2) d x \\
& =\left|\left[\frac{3 x^2}{2}+2 x\right]_{-1}^{\frac{-2}{3}}\right|+\left[\frac{3 x^2}{2}+2 x\right]_{\frac{-2}{3}}^1=\frac{1}{6}+\frac{25}{6}=\frac{13}{3}
\end{aligned}$

Example 4

Find the area bounded by the curve $y=\cos x$ between $x=0$ and $x=2 \pi$.

Solution

From the Fig 8.10, the required area $=$ area of the region $\mathrm{OABO}+$ area of the region BCDB + area of the region DEFD.

Thus, we have the required area

$\begin{aligned}
& =\int_0^{\frac{\pi}{2}} \cos x d x+\left|\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \cos x d x\right|+\int_{\frac{3 \pi}{2}}^{2 \pi} \cos x d x \\
& =[\sin x]_0^{\frac{\pi}{2}}+\left|[\sin x]_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}\right|+[\sin x]_{\frac{3 \pi}{2}}^{2 \pi} \\
& =1+2+1=4
\end{aligned}$