The area of the region bounded by the curves y - x 2 and y -2-1- x 2 i A- -2-3B- -2-3D- -3

The correct option is A π 2/3 y = x^2, y = 2/1 + x^2⇒ x^2 = 2/1 + x^2⇒ x^2 + x^4 = 2 ⇒ x^4 + x^2 2 = 0 ⇒ (x^2 + 2)(x^2 1) = 0 ⇒ x^2 = 1 ⇒ x^2 = ± 1 ...

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Byju's Answer

Standard XII

Mathematics

Area between Two Curves

The area of t...

Question

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

π - \frac{2}{3}

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π + \frac{2}{3}

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\frac{π}{3}

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\frac{2 π}{3}

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Solution

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

$y = x^{2}, y = \frac{2}{1 + x^{2}} \Rightarrow x^{2} = \frac{2}{1 + x^{2}} \Rightarrow x^{2} + x^{4} = 2 \Rightarrow x^{4} + x^{2} - 2 = 0$
$\Rightarrow (x^{2} + 2) (x^{2} - 1) = 0 \Rightarrow x^{2} = 1 \Rightarrow x^{2} = \pm 1$
Required area = $\int_{- 1}^{1} (\frac{2}{1 + x^{2}} - x^{2}) d x$
$= 2 t a n^{- 1} x - x^{\frac{3}{3}}]_{- 1}^{1} = π - \frac{2}{3}$ sq.unit

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