Definite Integration Question 443
Question: The area enclosed between the curves $ y=ax^{2} $ and $ x=ay^{2} $ (where a > 0) is 1 sq. unit, then the value of a is
Options:
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 $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ f(x)=x^{2}+\cos x $
$ \Rightarrow f’(x)=2x-sinx $
$ \Rightarrow f’( \frac{\pi }{2} )=\pi -1 $
So, equation of normal at $ x=\frac{\pi }{2} $ is $ ( y-\frac{{{\pi }^{2}}}{4} )=\frac{1}{1-\pi }( x-\frac{\pi }{2} ) $ 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). $
Required area =Area of OABCO + Area of $ \Delta ABC $
$ =\int\limits_0^{\pi /2}{(x^{2}+\cos x)dx+\frac{1}{2}[ \frac{(\pi -1){{\pi }^{2}}}{4}+\frac{\pi }{2}-\frac{\pi }{2} ]\times \frac{{{\pi }^{2}}}{4}} $
$ =[ \frac{x^{3}}{3}+\sin x ]_0^{\pi /2}+\frac{(\pi -1){{\pi }^{4}}}{32} $
$ =\frac{{{\pi }^{3}}}{24}+1+\frac{{{\pi }^{4}}}{32}(\pi -2) $
$ =\frac{{{\pi }^{5}}}{32}-\frac{{{\pi }^{4}}}{32}+\frac{{{\pi }^{3}}}{24}+1 $
BETA
