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

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

(A) $\frac{2}{3}(\sqrt{2}+1)$

(B) $\frac{4}{3}(\sqrt{2}-1)$

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Area Under Curves

The area bounded by the curves

y = x^{2} - 1

and

y = 1

Answer: (D)

The curve

y = ∣ x^{2} - 1∣

intersects

y = 1

when

∣ x^{2} - 1∣ = 1

, giving

x^{2} = 0

x^{2} = 2

, i.e.,

x = 0, \pm 2

. By symmetry, the area is

2 (\int_{0}^{1} (1 - (1 - x^{2})) d x + \int_{1}^{2} (1 - (x^{2} - 1)) d x) = 2 (\int_{0}^{1} x^{2} d x + \int_{1}^{2} (2 - x^{2}) d x) = 2 (\frac{1}{3} + [2 x - \frac{x ^{3}}{3}]_{1}^{2}) = 2 (\frac{1}{3} + 22 - \frac{2 2}{3} - 2 + \frac{1}{3}) = \frac{8}{3} (2 - 1)

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option