question_answer

The area of the region bounded by the curve \[y=\frac{{{a}^{3}}}{{{x}^{2}}+{{a}^{2}}}\]and \[x-axis\] is

A) \[\pi {{a}^{2}}sq\,\,unit\]                            

B) \[2\pi {{a}^{2}}sq\,\,unit\]

C) \[3\pi {{a}^{2}}sq\,\,unit\]                          

D)  None of these

Correct Answer: A

Solution :

We have, the equation of the curve                 \[y=\frac{{{a}^{3}}}{{{x}^{2}}+{{a}^{2}}}=f(x)\]                   (say) Here,     \[f(-x)=\frac{{{a}^{3}}}{{{(-x)}^{2}}+{{a}^{2}}}=\frac{{{a}^{3}}}{{{x}^{2}}+{{a}^{2}}}=f(x)\] \[\therefore \]The function is symmetrical about y-axis. Now, the required area                 \[=\int_{-\infty }^{\infty }{y}\,\,dx=2\int_{0}^{\infty }{y\,\,dy}\]                 \[=2\int_{0}^{\infty }{\frac{{{a}^{3}}}{{{x}^{2}}+{{a}^{2}}}dx}\]                 \[=2{{a}^{3}}\left[ \frac{1}{a}{{\tan }^{-1}}\frac{x}{a} \right]_{0}^{\infty }\]                 \[=2{{a}^{2}}({{\tan }^{-1}}\infty -{{\tan }^{-1}}0)\]                 \[=2{{a}^{2}}\cdot \left( \frac{\pi }{2}-0 \right)\]                 \[=\pi {{a}^{2}}sq\,\,unit\]