Question

The area enclosed by $y=x^{2}+cos\,x$ and its normal at $x=\frac{\pi}{2}$ in the first quadrant is

• A. $\frac{\pi^{5}}{32}-\frac{\pi^{4}}{64}+\frac{\pi^{3}}{32}+1$
• B. $\frac{\pi^{5}}{16}-\frac{\pi^{4}}{32}+\frac{\pi^{3}}{24}-1$
• C. $\frac{\pi^{5}}{32}-\frac{\pi^{4}}{32}+\frac{\pi^{3}}{16}$
• D. $\frac{\pi^{5}}{32}-\frac{\pi^{4}}{32}+\frac{\pi^{3}}{24}+1$

Answer 1

Answer: (D)

$f (x)=x^{2}+cos\,x$
$\Rightarrow f'(x)=2x-sin\,x$
$\Rightarrow f'\left(\frac{\pi}{2}\right)=\pi-1$
So, equation of normal at $x=\frac{\pi}{2}$ is
$\left(y-\frac{\pi^{2}}{4}\right)=\frac{1}{1-\pi}\left(x-\frac{\pi}{2}\right)$
It meets x-axis at $x=\frac{\left(\pi-1\right)\pi^{2}}{4}+\frac{\pi}{2}$
Also, $f'(x)=2x-sin\,x>\,0$ for $x>\,0$
and $f'(x)=2x-sin\,x<\,0$ for $x<\,0$
So, $f (x) $ increases for $x \in\left(0, \infty\right)$ and decreases for $x \in\left(-\infty, 0\right)$
Graph of the function is as shown in the following figure

Required area
= Area of $OABCO$ + Area of $\Delta BCD$
$=\int\limits_{0}^{\pi/ 2}\left(x^{2}+cos\,x\right)dx +\frac{1}{2}\left[\frac{\left(\pi-1\right)\pi^{2}}{4}+\frac{\pi}{2}-\frac{\pi}{2}\right]\times\frac{\pi^{2}}{4}$
$=\left[\frac{x^{3}}{3}+sin\,x\right]_{0}^{\pi /2} +\frac{\left(\pi-1\right)\pi^{4}}{32}$
$=\frac{\pi^{3}}{24}+1+\frac{\pi^{4}}{32}\left(\pi-1\right)$
$=\frac{\pi^{5}}{32}-\frac{\pi^{4}}{32}+\frac{\pi^{3}}{24}+1$