Question
If $${c_1} = y = \frac{1}{{1 + {x^2}}}$$ and $${c_2} = y = \frac{{{x^2}}}{2}$$ be two curves lying in $$XY$$ -plane, then :
A.
area bounded by curve $$y = \frac{1}{{1 + {x^2}}}$$ and $$y = 0$$ is $$\frac{\pi }{2}$$
B.
area bounded by $${c_1}$$ and $${c_2}$$ is $$\frac{\pi }{2} - 1$$
C.
area bounded by $${c_1}$$ and $${c_2}$$ is $$1 - \frac{\pi }{2}$$
D.
area bounded by curve $$y = \frac{1}{{1 + {x^2}}}$$ and $$x$$-axis is $$\frac{\pi }{2}$$
Answer :
area bounded by $${c_1}$$ and $${c_2}$$ is $$\frac{\pi }{2} - 1$$
Solution :
Area bounded by $$y = \frac{1}{{1 + {x^2}}}$$ and $$x$$-axis is $$\int_{ - \infty }^\infty {\frac{1}{{1 + {x^2}}}} dx = \pi $$
Area bounded by two curves is $$ = \int_{ - 1}^1 {\left( {\frac{1}{{1 + {x^2}}} - \frac{{{x^2}}}{2}} \right)} dx = \frac{\pi }{2} - 1$$
Area bounded by $$y = \frac{1}{{1 + {x^2}}}$$ and $$x$$-axis is $$\int_{ - \infty }^\infty {\frac{1}{{1 + {x^2}}}} dx = \pi $$
Area bounded by two curves is $$ = \int_{ - 1}^1 {\left( {\frac{1}{{1 + {x^2}}} - \frac{{{x^2}}}{2}} \right)} dx = \frac{\pi }{2} - 1$$