question_answer

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) \[\frac{{{\pi }^{5}}}{32}-\frac{{{\pi }^{4}}}{32}+\frac{{{\pi }^{3}}}{24}+1\]

Correct Answer: D

Solution :

[d] \[f(x)={{x}^{2}}+\cos x\] \[\Rightarrow f'(x)=2x-sinx\] \[\Rightarrow f'\left( \frac{\pi }{2} \right)=\pi -1\] 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 \[x=\frac{(\pi -1){{\pi }^{2}}}{4}+\frac{\pi }{2}\] 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 \[x\in (0,\infty )\]and decreases for \[x\in (-\infty ,0).\] Graph of the function is as shown in the following figure. Required area =Area of OABCO + Area of \[\Delta ABCD\] \[=\int\limits_{0}^{\pi /2}{({{x}^{2}}+\cos x)dx+\frac{1}{2}\left[ \frac{(\pi -1){{\pi }^{2}}}{4}+\frac{\pi }{2}-\frac{\pi }{2} \right]\times \frac{{{\pi }^{2}}}{4}}\]\[=\left[ \frac{{{x}^{3}}}{3}+\sin x \right]_{0}^{\pi /2}+\frac{(\pi -1)\,{{\pi }^{4}}}{32}\] \[=\frac{{{\pi }^{3}}}{24}+1+\frac{{{\pi }^{4}}}{32}(\pi -1)\] \[=\frac{{{\pi }^{5}}}{32}-\frac{{{\pi }^{4}}}{32}+\frac{{{\pi }^{3}}}{24}+1\]