The area bounded by the curves y = |x^{2} - 1 | vedclass

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(D) The curves are $y = |x^{2} - 1|$ and $y = 1$. To find the intersection points,set $|x^{2} - 1| = 1$. This implies $x^{2} - 1 = 1$ or $x^{2} - 1 =

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$\frac{8}{3}(\sqrt{2} - 1)$

Vedclass · Answer

(D) The curves are $y = |x^{2} - 1|$ and $y = 1$. To find the intersection points,set $|x^{2} - 1| = 1$. This implies $x^{2} - 1 = 1$ or $x^{2} - 1 = -1$. $x^{2} = 2 \implies x = \pm \sqrt{2}$ and $x^{2} = 0 \implies x = 0$. Due to symmetry about the $y$-axis,the total area is $2 	imes$ (area in the first quadrant). In the first quadrant,the region is bounded by $x = 0$ to $x = \sqrt{2}$. For $0 \le x \le 1$,$y = |x^{2} - 1| = 1 - x^{2}$. The area is $\int_{0}^{1} (1 - (1 - x^{2})) dx = \int_{0}^{1} x^{2} dx = [\frac{x^{3}}{3}]_{0}^{1} = \frac{1}{3}$. For $1 \le x \le \sqrt{2}$,$y = |x^{2} - 1| = x^{2} - 1$. The area is $\int_{1}^{\sqrt{2}} (1 - (x^{2} - 1)) dx = \int_{1}^{\sqrt{2}} (2 - x^{2}) dx = [2x - \frac{x^{3}}{3}]_{1}^{\sqrt{2}} = (2\sqrt{2} - \frac{2\sqrt{2}}{3}) - (2 - \frac{1}{3}) = \frac{4\sqrt{2}}{3} - \frac{5}{3}$. Total Area $= 2 	imes (\frac{1}{3} + \frac{4\sqrt{2} - 5}{3}) = 2 	imes (\frac{4\sqrt{2} - 4}{3}) = \frac{8}{3}(\sqrt{2} - 1)$.

The area bounded by the curves $y = |x^{2} - 1|$ and $y = 1$ is:

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