Question

The area of the region bounded by the curves $y=x^2$ and $y=\frac{2}{1+x^2}$ is :

• A. $\pi -\frac{2}{3}$
• B. $\pi +\frac{2}{3}$
• C. $\frac{\pi }{3}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is A $\pi -\frac{2}{3}$

$y=x^2$ and $y=\frac{2}{1+x^2}$

meeting point

$x^2(1+x^2)=2$

$x=1$ and $x=-1$ are roots $\Rightarrow (x^2-1)$ is factor

$(x^2+2)x^2-1)\overline{x^4+x^2-2}{x}^4-x^2\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_2x^2-2x^2+2=02x^2-2x^2=-2notpositive\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_0$

$\therefore area={\int }_{-1}^1(y_2-y_1)dx$

$=-\left[{\int }_{-1}^1(x^2-\frac{2}{(1+x^2)})dx\right]$

$=-[{\int }_{-1}^1{x}^{2}dx-{\int }_{-1}^1\frac{2}{1+x^2}dx]$

$=-[{\frac{x^2}{3}]}_{-1}^1-{2{\tan }^{-1}x]}_{-1}^1]$

$=-[[\frac{1}{3}-(-1/3)]-2[\pi /4-(-\pi /4)]]$

$=-[\frac{2}{3}-2[\pi /4+\pi /4]]$

$=\pi -\frac{2}{3}$