Miscellaneous Exercise (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

Miscellaneous Exercise Question 1.

Find the area under the given curves and given lines:
(i) $y=x^2, x=1, x=2$ and $x$-axis.
(ii) $y=x^4, x=1, x=5$ and $x$-axis.

Answer.

(i) Equation of the curve (parabola) is
$
y=x^2
$

Required area bounded by curve (i), vertical line $x=1, x=2$ and $x$-axis
$
\begin{aligned}
& =\left|\int_1^2 y d x\right| \\
& =\left|\int_1^2 x^2 d x\right| \\
& =\left(\frac{x^3}{3}\right)_1^2
\end{aligned}
$

$=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}$ sq. units
(ii)Equation of the curve
$
y=x^4 \ldots . . \text { (i) }
$

It is clear that curve (i) passes through the origin because $x=0$, from (i) $y=0$.
Table of values for curve $y=x^4$ for $x=1$ and $x=5$ (given)

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| \\
& =\left(\frac{x^5}{5}\right)_1^5=\frac{5^5}{5}-\frac{1^5}{5} \\
& =\frac{3125-1}{5}=\frac{3124}{5} \\
& =624.8 \text { sq. units }
\end{aligned}
$

Miscellaneous Exercise Question 2.

Sketch the graph of $y=|x+3|$ and evaluate $\int_{-6}^0|x+3| d x$.

Answer.

Equation of the given curve is $y=|x+3|$ ...(i)

$\because y=|x+3| \geq 0$ for all real $x$
$\therefore$ Graph of curve is only above the $x$-axis i.e., in first and second quadrant only.
$
\begin{aligned}
& \therefore y=|x+3| \\
& =x+3
\end{aligned}
$

If $x+3 \geq 0$
$
\Rightarrow x \geq-3
$

$
\begin{aligned}
& \text { And } y=|x+3| \\
& =-(x+3)
\end{aligned}
$

If $x+3 \leq 0$
$
\Rightarrow x \leq-3
$

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

$\text { Table of values for } y=x+3 \text { for } x \leq-3$

$\begin{aligned}
& \text { Now, } \int_{-6}^0|x+3| d x \\
& =\int_{-5}^{-3}|x+3| d x+\int_{-3}^0|x+3| d x \\
& =\int_{-6}^{-3}-(x+3) d x+\int_{-3}^0(x+3) d x \\
& =-\left(\frac{x^2}{2}+3 x\right)_{-6}^{-3}+\left(\frac{x^2}{2}+3 x\right)_{-3}^0 \\
& =-\left[\frac{9}{2}-9-(18-18)\right]+\left[0-\left(\frac{9}{2}-9\right)\right] \\
& =-\frac{9}{2}+9+0+0-\frac{9}{2}+9 \\
& =18-\frac{18}{2}=18-9=9 \text { sq. units }
\end{aligned}$

Miscellaneous Exercise Question 3.

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

Answer.

Equation of the curve is $y=\sin x$ $\qquad$ .(i)
$\because y=\sin x \geq 0$ for $0 \leq x \leq \pi$ i.e., graph is in first and second quadrant.
And $y=\sin x \leq 0$ for $\pi \leq x \leq 2 \pi$ i.e., graph is in third and fourth quadrant.

If tangent is parallel to $x$-axis, then $\frac{d y}{d x}=0$
$
\begin{aligned}
& \Rightarrow \cos x=0 \\
& \Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2}
\end{aligned}
$

Table of values for curve $y=\sin x$ between $x=0$ and $x=2 \pi$

Now Required shaded area $=$ Area $\mathrm{OAB}+$ Area $B C D$
$
\begin{aligned}
& =\left|\int_0^\pi y d x\right|+\left|\int_\pi^{2 \pi} y d x\right| \\
& =\left|\int_0^\pi \sin x d x\right|+\left|\int_\pi^{2 \pi} \sin x d x\right|
\end{aligned}
$

$
\begin{aligned}
& =\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+1)| \\
& =2+2 \text { = 4 sq. units }
\end{aligned}
$

Miscellaneous Exercise Question 4.

Choose the correct answer:

Area bounded by the curve $y=x^3$ the $x-$ axis and the ordinate $x=-2$ and $x=1$ is:
(A) -9
(B) $\frac{-15}{4}$
(C) $\frac{15}{4}$
(D) $\frac{17}{4}$

Answer.

Equation of the curve is $y=x^3$
To find: Area OBN $\left(y=x^3\right.$ for $\left.-2 \leq x \leq 0\right)$ and Area OAM ( $y=x^3$ for $\left.0 \leq x \leq 1\right)$

$\therefore$ Required area $=$ Area $\mathrm{OBN}+$ Area OAM
$
\begin{aligned}
& =\left|\int_{-2}^0 x^3 d x\right|+\left|\int_0^1 x^3 d x\right| \\
& \left.=\left|\left(\frac{x^4}{4}\right)_{-2}^0\right|+\left\lvert\, \frac{x^4}{4}\right.\right)_0^1 \mid \\
& =\left|0-\frac{16}{4}\right|+\left|\frac{1}{4}-0\right| \\
& =4+\frac{1}{4}=\frac{17}{4} \text { sq. units }
\end{aligned}
$

Therefore, option (D) is correct,

Miscellaneous Exercise Question 5.

Choose the correct answer:

The area bounded by the curve $y=x|x|, x-$ axis and the ordinates $x=-1$ and $x=1$ is given by:
(A) 0
(B) $\frac{1}{3}$

(C) $\frac{2}{3}$
(D) $\frac{4}{3}$

Answer.

Equation of the curve is

$
y=x|x|=x(x)=x^2 \text { if } x \geq 0
$

And $y=x|x|=x(-x)=-x^2$ if $x \leq 0$
Required area $=$ Area ONBO + Area OAMO
$
\begin{aligned}
& =\left|\int_{-1}^0-x^2 d x\right|+\left|\int_0^1 x^2 d x\right| \\
& \left.=\left\lvert\, \frac{-x^3}{3}\right.\right) \left._{-1}^0|+|\left(\frac{x^3}{3}\right)_0^1 \right\rvert\,
\end{aligned}
$
$=0-\left(\frac{-1}{3}\right)+\frac{1}{3}-0=\frac{2}{3}$ sq. units
Therefore, option (C) is correct.