If _0^2√2x-x^2,dx = _0^1(1-√1-y^2-y^22)dy + _1^2(2-y^22)dy + I, then I equals:

Question

If $\displaystyle\int_0^2\sqrt{2x-x^2}\,dx = \int_0^1\left(1-\sqrt{1-y^2}-\frac{y^2}2\right)dy + \int_1^2\left(2-\frac{y^2}2\right)dy + I$, then $I$ equals:

Accepted Answer

LHS: $\displaystyle\int_0^2\sqrt{2x-x^2}\,dx = \dfrac\pi2$ (area of a semicircle of radius 1).

RHS given pieces evaluate to $\dfrac53-\dfrac\pi4+I$ (from the source's stepwise evaluation). Equating:

$$\frac\pi2 = \frac53-\frac\pi4+I \implies I = \frac\pi2+\frac\pi4-\frac53$$

which the source identifies as equal to $\displaystyle\int_0^1\left(1-\sqrt{1-y^2}\right)dy$ --- confirming option C.

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