Explanation
we have,
$$y={{x}^{2}}\,\,......\,\,\left( 1 \right)$$
$$y=\dfrac{2}{{{x}^{2}}+1}\,\,.......\,\,\left( 2 \right)$$
Taking limit then,
By equation (1) and (2) to, and get,
$${{x}^{2}}=\dfrac{2}{{{x}^{2}}+1}$$
Put x=1 and we get,
$$ {{\left( 1 \right)}^{2}}=\dfrac{2}{{{\left( 1 \right)}^{2}}+1} $$
$$ 1=\dfrac{2}{2} $$
$$ 1=1 $$
Now, taking’
$$x=-1$$ and we get,
$$ {{\left( -1 \right)}^{2}}=\dfrac{2}{{{\left( -1 \right)}^{2}}+1} $$
Then, the limit of this function of \[1\,\,to\,\,-1\].
The required area\[=\int_{-1}^{1}{\left( \dfrac{2}{{{x}^{2}}+1}-{{x}^{2}} \right)dx}\]
$$ =\int_{-1}^{1}{\left( \dfrac{2}{{{x}^{2}}+1}-{{x}^{2}} \right)dx}=2\int_{0}^{1}{\left( \dfrac{2}{{{x}^{2}}+1}-{{x}^{2}} \right)dx} $$
$$ =2{{\left[ 2{{\tan }^{-1}}x-\dfrac{{{x}^{3}}}{3} \right]}_{0}}^{1} $$
$$ =2\left[ 2{{\tan }^{-1}}\left( 1 \right)-2{{\tan }^{-1}}0-\dfrac{1-0}{3} \right] $$
$$ =2\left[ 2\dfrac{\pi }{4}-2\times 0-\dfrac{1}{3} \right] $$
$$ =\pi -\dfrac{2}{3} $$
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