Question 194 of 246MediumMCQ
If \[\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } \], then I equals
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Official solution
Given,
\[\int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } \]
Now,
\[L.H.S. = \int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx} \]
\[ = \sqrt 2 \int_0^2 {{x^{1/2}}dx - \int_0^2 {\sqrt { - ({x^2} - 2x)} dx} } \]
\[ = \sqrt 2 \left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2 - \int_0^2 {\sqrt { - \left[ {{{(x - 1)}^2} - 1} \right]} dx} \]
\[ = \sqrt 2 \times {2 \over 3}\left[ {{2^{{3 \over 2}}}} \right] - \int_0^2 {\sqrt {1 - {{(x - 1)}^2}} dx} \]
[We know,
\[\sqrt {{a^2} - {x^2}} = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}\left( {{x \over a}} \right) + C\]]
\[ = {{2\sqrt 2 \times 2\sqrt 2 } \over 3} - \left[ {{{x - 1} \over 2}\sqrt {1 - {{(x - 1)}^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{{x - 1} \over 1}} \right)} \right]_0^2\]
\[ = {8 \over 3} - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {\left( {{1 \over 2}} \right)\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}( - 1)} \right)} \right]\]
\[ = {8 \over 3} - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) + {1 \over 2}{\sin ^{ - 1}}\left( {\sin {{3\pi } \over 2}} \right)\]
\[ = {8 \over 3} - {1 \over 2}\,.\,{\pi \over 2} + {1 \over 2}\,.\,{{3\pi } \over 2}\]
\[ = {8 \over 3} - {\pi \over 4} + {{3\pi } \over 4}\]
\[ = {8 \over 3} - {{2\pi } \over 4}\]
\[ = {8 \over 3} - {\pi \over 2}\]
Now in R.H.S.,
\[\int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy} \]
\[ = \int_0^1 {dy - \int_0^1 {\sqrt {1 - {y^2}} - \int_0^1 {{{{y^2}} \over 2}dy} } } \]
\[ = \left[ y \right]_0^1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1 - \left[ {{{{y^3}} \over 6}} \right]_0^1\]
\[ = 1 - \left[ {\left( {{1 \over 2}\sqrt {1 - 0} + {1 \over 2}{{\sin }^{ - 1}}(1)} \right) - \left( {0 + {{\sin }^{ - 1}}(0)} \right)} \right] - {1 \over 6}\]
\[ = 1 - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) - {1 \over 6}\]
\[ = 1 - {1 \over 2} \times {\pi \over 2} - {1 \over 6}\]
\[ = {5 \over 6} - {\pi \over 4}\]
Now,
\[\int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy} \]
\[ = \left[ {2y - {{{y^3}} \over 6}} \right]_1^2\]
\[ = \left[ {\left( {4 - {8 \over 6}} \right) - \left( {2 - {1 \over 6}} \right)} \right]\]
\[ = 4 - {8 \over 6} - 2 + {1 \over 6}\]
\[ = 2 - {7 \over 6}\]
\[ = {5 \over 6}\]
\[\therefore\] \[R.H.S. = {5 \over 6} - {\pi \over 4} + {5 \over 6} + I\]
As L.H.S.=R.H.S.
\[ \Rightarrow {8 \over 3} - {\pi \over 2} = {{10} \over 6} - {\pi \over 4} + I\]
\[ \Rightarrow {{10} \over 6} - {8 \over 3} + I = - {\pi \over 2} + {\pi \over 4}\]
\[ \Rightarrow {{ - 6} \over 6} + I = - {\pi \over 4}\]
\[ \Rightarrow I = 1 - {\pi \over 4}\]
From option (c)
\[\int_0^1 {(1 - \sqrt {1 - {y^2}} )dy} \]
\[ = \left[ y \right]_0^1 - \int_0^1 {(\sqrt {1 - {y^2}} )dy} \]
\[ = 1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1\]
\[ = 1 - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {0 + {1 \over 2}{{\sin }^{ - 1}}(0)} \right)} \right]\]
\[ = 1 - \left[ {{1 \over 2} \times {\pi \over 2} - 0} \right]\]
\[ = 1 - {\pi \over 4} = I\]
\[\therefore\] Option (c) is the right answer.
Note :
There is no way to guess which option is correct. You have to check all the options to see which give value equal to I.
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