The area of the region,bounded by the curves | vedclass

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(C) The function $y = \sin^{-1} x$ is defined for $-1 \leq x \leq 1$. In the first quadrant,we have $0 \leq x \leq 1$. Since $x(1-x) \geq 0$ for $x \i

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$\frac{1}{3}$

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(C) The function $y = \sin^{-1} x$ is defined for $-1 \leq x \leq 1$. In the first quadrant,we have $0 \leq x \leq 1$. Since $x(1-x) \geq 0$ for $x \in [0, 1]$,the curve $y_1 = \sin^{-1} x + x(1-x)$ lies above the curve $y_2 = \sin^{-1} x - x(1-x)$. The intersection points are found by setting $y_1 = y_2$,which gives $2x(1-x) = 0$,implying $x = 0$ or $x = 1$. The required area is given by the integral: $A = \int_{0}^{1} (y_1 - y_2) dx = \int_{0}^{1} [(\sin^{-1} x + x - x^2) - (\sin^{-1} x - x + x^2)] dx$ $A = \int_{0}^{1} (2x - 2x^2) dx = 2 \int_{0}^{1} (x - x^2) dx$ $A = 2 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 2 \left( \frac{1}{2} - \frac{1}{3} \right) = 2 \left( \frac{1}{6} \right) = \frac{1}{3}$.

The area of the region,bounded by the curves $y=\sin ^{-1} x+x(1-x)$ and $y=\sin ^{-1} x-x(1-x)$ in the first quadrant,is

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