If $\int_{0}^{2}\left(\sqrt{2 x}-\sqrt{2 x-x^{2}}\right) dx = \int_{0}^{1}\left(1-\sqrt{1-y^{2}}-\frac{y^{2}}{2}\right) dy+\int_{1}^{2}\left(2-\frac{y^{2}}{2}\right) dy+I$, then $I$ equals:

(C) $\int_{0}^{1}\left(1-\sqrt{1-y^{2}}\right) dy$

(A) $\int_{0}^{1}\left(1+\sqrt{1-y^{2}}\right) dy$

(B) $\int_{0}^{1}\left(\frac{y^{2}}{2}-\sqrt{1-y^{2}}+1\right) dy$

(D) $\int_{0}^{1}\left(\frac{y^{2}}{2}+\sqrt{1-y^{2}}+1\right) dy$

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Definite Integrals

\int_{0}^{2} (2 x - 2 x - x^{2}) d x = \int_{0}^{1} (1 - 1 - y^{2} - \frac{y ^{2}}{2}) d y + \int_{1}^{2} (2 - \frac{y ^{2}}{2}) d y + I

, then

I

equals:

Answer: (C)

Compute LHS:

\int_{0}^{2} 2 x d x = \frac{2}{3} (2 x)^{3/2}_{0}^{2} = \frac{8}{3}

. For

\int_{0}^{2} 2 x - x^{2} d x

, complete the square:

2 x - x^{2} = 1 - (x - 1)^{2}

. This is a semicircle of radius 1, so the integral equals

\frac{π}{2}

. Thus LHS

= \frac{8}{3} - \frac{π}{2}

. For RHS, compute the two definite integrals: they evaluate to

\frac{5}{3} - \frac{π}{4}

. Therefore

I = \frac{8}{3} - \frac{π}{2} - \frac{5}{3} + \frac{π}{4} = 1 - \frac{π}{4} = \int_{0}^{1} (1 - 1 - y^{2}) d y

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option