MCQ
$\text { If } \int\limits_{0}^{2}\left(\sqrt{2 x}-\sqrt{2 x-x^{2}}\right) d x=$ $\int\limits_{0}^{1}\left(1-\sqrt{1-y^{2}}-\frac{y^{2}}{2}\right) d y+\int\limits_{1}^{2}\left(2-\frac{y^{2}}{2}\right) d y+I$  then $I=\dots\dots\dots$
  • A
    $\int\limits_{0}^{1}\left(1+\sqrt{1-y^{2}}\right) d y$
  • B
    $\int\limits_{0}^{1}\left(\frac{y^{2}}{2}-\sqrt{1-y^{2}}+1\right) d y$
  • $\int\limits_{0}^{1}\left(1-\sqrt{1-y^{2}}\right) d y$
  • D
    $\int\limits_{0}^{1}\left(\frac{ y ^{2}}{2}+\sqrt{1- y ^{2}}+1\right) d y$

Answer

Correct option: C.
$\int\limits_{0}^{1}\left(1-\sqrt{1-y^{2}}\right) d y$
c
$LHS =\int\limits_{0}^{2}\left(\sqrt{2 x }-\sqrt{2 x - x ^{2}}\right) dx =\frac{8}{3}-\frac{\pi}{2}$

$RHS =\int\limits_{0}^{1}\left(1-\sqrt{1- y ^{2}}-\frac{ y ^{2}}{2}\right) dy +\int\limits_{1}^{2}\left(2-\frac{ y ^{2}}{2}\right) dy + I$

$I +\frac{5}{3}-\frac{\pi}{4}$

So, $I=1-\frac{\pi}{4}=\int\limits_{0}^{1}\left(1-\sqrt{1-y^{2}}\right) d y$

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