Question

The area enclosed by $y=x^2+cosx$ 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)

$f(x)=x^2+cosx$
$\Rightarrow f^{\prime }(x)=2x-sinx$
$\Rightarrow f^{\prime }\left(\frac{\pi }{2}\right)=\pi -1$
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, $f^{\prime }(x)=2x-sinx>0$ for $x>0$
and $f^{\prime }(x)=2x-sinx<0$ 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 $\Delta BCD$
$=\underset{0}{\overset{\pi /2}{\int }}\left(x^2+cosx\right)dx+\frac{1}{2}\left[\frac{\left(\pi -1\right){\pi }^2}{4}+\frac{\pi }{2}-\frac{\pi }{2}\right]\times \frac{{\pi }^2}{4}$
$={\left[\frac{x^3}{3}+sinx\right]}_0^{\pi /2}+\frac{\left(\pi -1\right){\pi }^4}{32}$
$=\frac{{\pi }^3}{24}+1+\frac{{\pi }^4}{32}\left(\pi -1\right)$
$=\frac{{\pi }^5}{32}-\frac{{\pi }^4}{32}+\frac{{\pi }^3}{24}+1$