Chapter 8: Application of Integrals

Solution: Equation of ellipse is
$
\frac{x^2}{16}+\frac{y^2}{9}=1
$

Here $\quad a^2(=16)>b^2(=9)$
From (i), $\quad \frac{y^2}{9}=1-\frac{x^2}{16}$
$
\begin{array}{ll}
=\frac{16-x^2}{16} & \\
\Rightarrow & y^2=\frac{9}{16}\left(16-x^2\right) \\
\Rightarrow & y=\frac{3}{4} \sqrt{16-x^2}
\end{array}
$
for the arc of the ellipse in the first quadrant.
Ellipse (i) is symmetrical about $x$-axis.
( ∵ On changing $y \rightarrow-y$ in (i), it remains unchanged).
Ellipse (i) is symmetrical about $y$-axis.
( ∵ On changing $x \rightarrow-x$ in (i), it remains unchanged)
Intersections of ellipse (i) with $x$-axis ( $y=0$ )
Substituting $y=0$ in (i), $\frac{x^2}{16}=1 \quad \Rightarrow \quad x^2=16 \quad \Rightarrow \quad x= \pm 4$
∴ Intersections of ellipse ( $i$ ) with $x$-axis are ( 4,0 ) and ( $-4,0$ ).
Intersections of ellipse (i) with $\boldsymbol{y}$-axis ( $\boldsymbol{x}=0$ )
Substituting $x=0$ in (i), $\frac{y^2}{9}=1 \Rightarrow y^2=9 \Rightarrow y= \pm 3$.
∴ Intersections of ellipse ( $i$ ) with $y$-axis are ( 0,3 ) and ( $0,-3$ ).







∴ Area of region bounded by ellipse (i)
= Total shaded area
$=\mathbf{4} \times$ Area OAB of ellipse in first quadrant
$=4\left|\int_0^4 y d x\right| \quad(\because$ At end B of arc AB of ellipse; $x=0$ and at end A of $\operatorname{arc} \mathrm{AB} ; x=4)$
$=4\left|\int_0^4 \frac{3}{4} \sqrt{16-x^2} d x\right|$
[By (ii)]
$=3\left|\int_0^4 \sqrt{4^2-x^2} d x\right|=3\left[\frac{x}{2} \sqrt{4^2-x^2}+\frac{4^2}{2} \sin ^{-1} \frac{x}{4}\right]_0^4$
$\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]$
$=3\left[\frac{4}{2} \sqrt{16-16}+8 \sin ^{-1} 1-\left(0+8 \sin ^{-1} 0\right)\right]=3\left[0+\frac{8 \pi}{2}\right]$
$\left(\because \sin \frac{\pi}{2}=1 \Rightarrow \sin ^{-1} 1=\frac{\pi}{2}\right.$ and $\left.\sin 0=0 \Rightarrow \sin ^{-1} 0=0\right)$
$=3(4 \pi)=12 \pi$ sq. units.
Remark. We can also find area of this ellipse as $4\left|\int_0^3 \boldsymbol{x} \boldsymbol{d} \boldsymbol{y}\right|$

Solution: Equation of the ellipse is
$
\frac{x^2}{4}+\frac{y^2}{9}=1
$

Here $\quad a^2(=4)From (i), $\frac{y^2}{9}=1-\frac{x^2}{4}=\frac{4-x^2}{4}$
$
\Rightarrow \quad y^2=\frac{9}{4}\left(4-x^2\right) \quad \Rightarrow \quad y=\frac{3}{2} \sqrt{4-x^2}
$
for the arc of the ellipse in the first quadrant. Evidently ellipse (i) is symmetrical about $x$-axis and $y$-axis both.
[ ∵ On changing $y$ to $-y$ in ( $i$ ) or $x$ to $-x$ in ( $i$ ) keep it unchanged]
Intersections of ellipse (i) with $\boldsymbol{x}$-axis ( $\boldsymbol{y}=\mathbf{0}$ )
Substituting $y=0$ in (i), $\frac{x^2}{4}=1 \Rightarrow x^2=4 \Rightarrow x= \pm 2$





∴ Intersections of ellipse ( $i$ ) with $x$-axis are ( 2,0 ) and ( $-2,0$ )
Intersections of ellipse (i) with $\boldsymbol{y}$-axis ( $\boldsymbol{x}=\mathbf{0}$ )
Substituting $x=0$ in (i), $\frac{y^2}{9}=1 \Rightarrow y^2=9 \Rightarrow y= \pm 3$
∴ Intersections of ellipse ( $i$ ) with $y$-axis are ( 0,3 ) and ( $0,-3$ ).
∴ Area of region bounded by ellipse (i)
= Total shaded area
$=4 \times$ area OAB of ellipse in first quadrant
$=4\left|\int_0^2 y d x\right| \quad \because$ At end B of arc AB of ellipse $x=0$ and at end A of $\operatorname{arc} \mathrm{AB}, x=2$ )
$=4\left|\int_0^2 \frac{3}{2} \sqrt{4-x^2} d x\right|$
(By (ii))
$=4 \cdot \frac{3}{2}\left|\int_0^2 \sqrt{2^2-x^2} d x\right|=6\left[\frac{x}{2} \sqrt{2^2-x^2}+\frac{2^2}{2} \sin ^{-1} \frac{x}{2}\right]_0^2$
$\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]$
$=6\left[\frac{2}{2} \sqrt{4-4}+2 \sin ^{-1} 1-0-2 \sin ^{-1} 0\right]$
$=6\left[0+2 \cdot \frac{\pi}{2}-0\right]=6 \pi$ sq. units.

Solution: Equation of the circle is
$
x^2+y^2=4=2^2
$

Keep in mind that equation (i) represents a circle whose centre is origin and radius is 2 .
$
\begin{array}{rlrl}
\therefore & & y^2 & =2^2-x^2 \\
\therefore & y & =\sqrt{2^2-x^2}
\end{array}
$
for $\operatorname{arc} \mathrm{AB}$ of the circle in first quadrant.
∴ Required area lying in the first quadrant bounded by the






circle $x^2+y^2=4$ and the (vertical) lines $x=0$ and (tangent line) $x=2$.
$
\begin{aligned}
& \left.=\left|\int_0^2 y d x\right|=\left|\int_0^2 \sqrt{2^2-x^2} d x\right| \quad \text { By }(i i)\right) \\
& =\left|\left(\frac{x}{2} \sqrt{2^2-x^2}+\frac{2^2}{2} \sin ^{-1} \frac{x}{2}\right)_0^2\right| \\
& \quad\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right] \\
& =\frac{2}{2} \sqrt{4-4}+2 \sin ^{-1} 1-\left(0+2 \sin ^{-1} 0\right) \\
& =0+2 \cdot \frac{\pi}{2}-0-0=\pi \text { sq. units. } \\
& \quad\left[\because \sin 0=0 \Rightarrow \sin ^{-1} 0=0\right]
\end{aligned}
$
∴ Option (A) is the correct answer.

Solution: Equation of the curve (rightward parabola) is
$
y^2=4 x
$
∴ Required area of the region bounded by parabola (i), $y$-axis and the (horizontal) line $y=3$
$
\begin{aligned}
& =\text { Area OAM } \\
& =\left|\int_0^3 x d y\right|
\end{aligned}
$
[ ∵ For arc OA of the parabola (i), at point O, $y=0$ and at point A, $y=3$ ]
Substituting $x=\frac{y^2}{4}$ from (i) in (ii), required area
$
\begin{aligned}
& =\left|\int_0^3 \frac{y^2}{4} d y\right| \\
& =\frac{1}{4}\left|\left(\frac{y^3}{3}\right)_0^3\right|=\frac{1}{4}\left|\frac{27}{3}-0\right|=\frac{9}{4} \text { sq. units }
\end{aligned}
$
∴ Option (B) is correct answer.

Solution: (i) Equation of the curve (parabola) is
$
y=x^2 \text { i.e., } x^2=y
$

It is an upward parabola symmetrical about $y$-axis.
$[\because \quad$ Changing $x$ to $-x$ in (i) keeps it unchanged]
Required area bounded by curve (i) $y=x^2$, vertical lines




$
\begin{aligned}
x=1, x & =2 \text { and } x \text {-axis } \\
& =\left|\int_1^2 y d x\right|=\left|\int_1^2 x^2 d x\right| \\
& =\left|\left(\frac{x^3}{3}\right)_1^2\right|=\left|\frac{8}{3}-\frac{1}{3}\right|=\frac{7}{3} \text { sq. units. }
\end{aligned}
$
(ii) Equation of the curve is $y=x^4$
$\Rightarrow \quad y \geq 0$ for all real $x \quad(\because$ Power of $x$ is Even (4)) Curve (i) is symmetrical about $y$-axis.
[ ∵ On changing $x$ to $-x$ in (i), eqn. (i) remains unchanged] Evidently, curve (i) passes through the origin because for $x=0$, from (i) $y=0$.
Table of values for curve $y=x^4$ for $x=1$ to $x=5$ (given)

\begin{tabular}{|c|c|c|c|c|c|}
\hline$x$ & 1 & 2 & 3 & 4 & 5 \\
\hline$y$ & 1 & $2^4=16$ & $3^4=81$ & $4^4=256$ & $5^4=625$ \\
\hline
\end{tabular}

Required shaded area between the curve $y=x^4$, vertical lines $x=1, x=5$ and $x$-axis
$
\begin{aligned}
& =\left|\int_1^5 y d x\right|=\left|\int_1^5 x^4 d x\right| \quad \quad(\text { By }(i)) \\
& =\left|\left(\frac{x^5}{5}\right)_1^5\right|=\frac{5^5}{5}-\frac{1^5}{5}=\frac{3125-1}{5}=\frac{3124}{5} \\
& =\frac{3124 \times 2}{10}=624.8 \text { sq. units. }
\end{aligned}
$

Solution: Equation of the given curve is
$
y=|x+3|
$

Recall that from (i),
$
y=|x+3| \geq 0 \text { for all real } x .
$
∴ Graph of curve is only above the $x$-axis i.e., in first and second quadrants only.
From (i), $y=|x+3|=x+3$ if $x+3 \geq 0$ i.e., if $x \geq-3$ and $y=|x+3|=-(x+3)$ if $x+3 \leq 0$ i.e., if $x \leq-3$

Table of values for $y=x+3$ for $x \geq-3$

\begin{tabular}{|c|c|c|c|c|}
\hline$x$ & -3 & -2 & -1 & 0 \\
\hline$y$ & 0 & 1 & 2 & 3 \\
\hline
\end{tabular}

\begin{table}
\captionsetup{labelformat=empty}
\caption{Table of values for $y=-(x+3)$ for $x \leq-3$}
\begin{tabular}{|c|c|c|c|c|}
\hline$x$ & -3 & -4 & -5 & -6 \\
\hline$y$ & 0 & 1 & 2 & 3 \\
\hline
\end{tabular}
\end{table}
∴ Graph of curve (i) is as shown in the following figure.
∴ Graph of $y=|x+3|$ is L-shaped consisting of two rays above the $x$-axis at right angles to each other.
Now, $\quad \int_{-6}^0|x+3| d x=\int_{-6}^{-3}|x+3| d x+\int_{-3}^0|x+3| d x$
[ ∵ On substituting expression $x+3$ within modulus equal to zero, we get $x=-3$ and $\left.\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x\right]$



$
=\int_{-6}^{-3}-(x+3) d x \quad+\quad \int_{-3}^0(x+3) d x
$
(By (iii) because on
$
(-6,-3), x<-3 \Rightarrow x+3<0)
$
(By (ii) because on
$
(-3,0), x>-3 \Rightarrow x+3>0)
$
$
=-\left(\frac{x^2}{2}+3 x\right)_{-6}^{-3}+\left(\frac{x^2}{2}+3 x\right)_{-3}^0
$
$
=-\left[\frac{9}{2}-9-(18-18)\right]+\left[0-\left(\frac{9}{2}-9\right)\right]
$
$
=-\frac{9}{2}+9+0+0-\frac{9}{2}+9=18-\frac{18}{2}=18-9=9 \text { sq. units. }
$

Solution: Equation of the curve is $y=\sin x$

We begin by drawing the graph of $y=\sin x$ from $x=0$ to $x=2 \pi$
Now we know that $y=\sin x \geq 0$ for $0 \leq x \leq \pi$ i.e., in first and second quadrants
and $y=\sin x \leq 0$ for $\pi \leq x \leq 2 \pi$ i.e., in third and fourth quadrants.

To find points where tangent is parallel to $x$-axis, put $\frac{d y}{d x}=0$.
$
\Rightarrow \cos x=0 \Rightarrow x=\frac{\pi}{2}, x=\frac{3 \pi}{2}
$

Table of values for curve $\boldsymbol{y}=\boldsymbol{\operatorname { s i n }} \boldsymbol{x}$ between $x=0$ and $x=2 \pi$

\begin{tabular}{|c|c|c|c|c|c|}
\hline$x$ & 0 & $\frac{\pi}{2}$ & $\pi$ & $\frac{3 \pi}{2}$ & $2 \pi$ \\
\hline$y$ & 0 & 1 & 0 & -1 & 0 \\
\hline
\end{tabular}
$[\because \quad \sin n \pi=0$ for every integer $n$
and $\left.\sin \frac{3 \pi}{2}=\sin 270^{\circ}=\sin \left(180^{\circ}+90^{\circ}\right)=-\sin 90^{\circ}=-1\right]$



Required shaded area $=$ Area OAB + Area BCD
$
=\left|\int_0^\pi y d x\right|+\left|\int_\pi^{2 \pi} y d x\right|
$
[Here we will have to find area OAB and Area BCD separately because $y=\sin x \geq 0$ for $0 \leq x \leq \pi$
and $\quad y=\sin x \leq 0$ for $\pi \leq x \leq 2 \pi$ ]
Substituting $y=\sin x$ from (i),
$
\begin{aligned}
& =\left|\int_0^\pi \sin x d x\right|+\left|\int_\pi^{2 \pi} \sin x d x\right| \\
& =\left|-(\cos x)_0^\pi\right|+\left|-(\cos x)_\pi^{2 \pi}\right| \\
& =|-(\cos \pi-\cos 0)|+|-(\cos 2 \pi-\cos \pi)| \\
& =|-(-1-1)|+|-(1+1)| \\
& \quad\left[\because \cos n \pi=(-1)^n \text { for every integer } n\right. \\
& \quad \text { substituting } n=1,2 ; \cos \pi=-1, \cos 2 \pi=1] \\
& =2+2=4 \text { sq. units. }
\end{aligned}
$

Solution: Equation of the curve is $y=x^3$

We begin by drawing the graph of curve (i) for values of $x$ from $x=-2$ to $x=1$.

Table of Values for $\boldsymbol{y}=\boldsymbol{x}^{\mathbf{3}}$

\begin{tabular}{|l|l|l|l|l|}
\hline$x$ & -2 & -1 & 0 & 1 \\
\hline$y$ & -8 & -1 & 0 & 1 \\
\hline
\end{tabular}

We need to find the area of the total shaded region.
We will have to find the two shaded areas OBN and OAM separately because from the table,
*Limits of integration for parabola are $x=0$ to $x$ of point of intersection and for circle are $x$ of point of intersection to $x=$ radius of circle.






$y=x^3 \leq 0$ for $-2 \leq x \leq 0$ for the region OBN and $y=x^3 \geq 0$ for $0 \leq x \leq 1$ for the region OAM
Now area of region OBN $=\left|\int_{-2}^0 y d x\right|$
$
\begin{aligned}
& =\left|\int_{-2}^0 x^3 d x\right| \quad(\mathrm{By}(i)) \\
& =\left|\left(\frac{x^4}{4}\right)_{-2}^0\right| \\
& =\left|0-\frac{16}{4}\right|=|-4|=4
\end{aligned}
$

Again area of region $\mathrm{OAM}=\left|\int_0^1 y d x\right|$
$
\begin{aligned}
& =\left|\int_0^1 x^3 d x\right| \\
& =\left|\left(\frac{x^4}{4}\right)_0^1\right|=\left|\frac{1}{4}-0\right|=\frac{1}{4}
\end{aligned}
$

Adding areas (ii) and (iii), the total required shaded area
$
=4+\frac{1}{4}=\frac{16+1}{4}=\frac{17}{4} \text { sq. units }
$
∴ Option (D) is the correct answer.

Solution: Equation of the curve is
and
$
\begin{aligned}
y & =x|x|=x(x)=x^2 \text { if } x \geq 0 \\
y & =x|x|=x(-x) \\
& =-x^2 \text { if } x \leq 0
\end{aligned}
$

Eqn. (i) namely $x^2=y(x \geq 0)$ represents the arc of the upward parabola in first quadrant and equation (ii) namely $x^2=-y(x \leq 0)$ represents the arc of the downward parabola in the third quadrant.
We need to find the area bounded by the given curve, $x$-axis and the ordinates
$
\begin{aligned}
& x=-1 \text { and } x=1 . \\
& \text { Required area }=\text { Area ONBO } \\
& + \text { Area OAMO }
\end{aligned}
$


$
\begin{aligned}
= & \left|\int_{-1}^0 y d x\right| \text { (for } y \text { given by }(i i) \text { ) } \\
& +\left|\int_0^1 y d x\right| \quad(\text { for } y \text { given by }(i)) \\
= & \left|\int_{-1}^0-x^2 d x\right|+\left|\int_0^1 x^2 d x\right| \\
= & \left|\left(\frac{-x^3}{3}\right)_{-1}^0\right|+\left|\left(\frac{x^3}{3}\right)_0^1\right|=0-\left(\frac{-1}{3}\right)+\frac{1}{3}-0=\frac{2}{3}
\end{aligned}
$
∴ Option (C) is the correct answer.