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

<p>Given,</p> <p>$$\int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $$</p> <p>Now,</p> <p>$L.H.S. = \int_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx}$</p> <p>$= \sqrt 2 \int_0^2 {{x^{1/2}}dx - \int_0^2 {\sqrt { - ({x^2} - 2x)} dx} }$</p> <p>$$ = \sqrt 2 \left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2 - \int_0^2 {\sqrt { - \left[ {{{(x - 1)}^2} - 1} \right]} dx} $$</p> <p>$$ = \sqrt 2 \times {2 \over 3}\left[ {{2^{{3 \over 2}}}} \right] - \int_0^2 {\sqrt {1 - {{(x - 1)}^2}} dx} $$</p> <p>[We know,</p> <p>$$\sqrt {{a^2} - {x^2}} = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{\sin ^{ - 1}}\left( {{x \over a}} \right) + C$$]</p> <p>$$ = {{2\sqrt 2 \times 2\sqrt 2 } \over 3} - \left[ {{{x - 1} \over 2}\sqrt {1 - {{(x - 1)}^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{{x - 1} \over 1}} \right)} \right]_0^2$$</p> <p>$$ = {8 \over 3} - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {\left( {{1 \over 2}} \right)\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}( - 1)} \right)} \right]$$</p> <p>$$ = {8 \over 3} - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) + {1 \over 2}{\sin ^{ - 1}}\left( {\sin {{3\pi } \over 2}} \right)$$</p> <p>$$ = {8 \over 3} - {1 \over 2}\,.\,{\pi \over 2} + {1 \over 2}\,.\,{{3\pi } \over 2}$$</p> <p>$= {8 \over 3} - {\pi \over 4} + {{3\pi } \over 4}$</p> <p>$= {8 \over 3} - {{2\pi } \over 4}$</p> <p>$= {8 \over 3} - {\pi \over 2}$</p> <p>Now in R.H.S.,</p> <p>$\int_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy}$</p> <p>$$ = \int_0^1 {dy - \int_0^1 {\sqrt {1 - {y^2}} - \int_0^1 {{{{y^2}} \over 2}dy} } } $$</p> <p>$$ = \left[ y \right]_0^1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1 - \left[ {{{{y^3}} \over 6}} \right]_0^1$$</p> <p>$$ = 1 - \left[ {\left( {{1 \over 2}\sqrt {1 - 0} + {1 \over 2}{{\sin }^{ - 1}}(1)} \right) - \left( {0 + {{\sin }^{ - 1}}(0)} \right)} \right] - {1 \over 6}$$</p> <p>$$ = 1 - {1 \over 2}{\sin ^{ - 1}}\left( {\sin {\pi \over 2}} \right) - {1 \over 6}$$</p> <p>$= 1 - {1 \over 2} \times {\pi \over 2} - {1 \over 6}$</p> <p>$= {5 \over 6} - {\pi \over 4}$</p> <p>Now,</p> <p>$\int_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy}$</p> <p>$= \left[ {2y - {{{y^3}} \over 6}} \right]_1^2$</p> <p>$$ = \left[ {\left( {4 - {8 \over 6}} \right) - \left( {2 - {1 \over 6}} \right)} \right]$$</p> <p>$= 4 - {8 \over 6} - 2 + {1 \over 6}$</p> <p>$= 2 - {7 \over 6}$</p> <p>$= {5 \over 6}$</p> <p>$\therefore$ $R.H.S. = {5 \over 6} - {\pi \over 4} + {5 \over 6} + I$</p> <p>As $L.H.S. = R.H.S.$</p> <p>$\Rightarrow {8 \over 3} - {\pi \over 2} = {{10} \over 6} - {\pi \over 4} + I$</p> <p>$$ \Rightarrow {{10} \over 6} - {8 \over 3} + I = - {\pi \over 2} + {\pi \over 4}$$</p> <p>$\Rightarrow {{ - 6} \over 6} + I = - {\pi \over 4}$</p> <p>$\Rightarrow I = 1 - {\pi \over 4}$</p> <p>From option (c)</p> <p>$\int_0^1 {(1 - \sqrt {1 - {y^2}} )dy}$</p> <p>$= \left[ y \right]_0^1 - \int_0^1 {(\sqrt {1 - {y^2}} )dy}$</p> <p>$$ = 1 - \left[ {{y \over 2}\sqrt {1 - {y^2}} + {1 \over 2}{{\sin }^{ - 1}}\left( {{y \over 1}} \right)} \right]_0^1$$</p> <p>$$ = 1 - \left[ {{1 \over 2}\sqrt {1 - 1} + {1 \over 2}{{\sin }^{ - 1}}(1) - \left( {0 + {1 \over 2}{{\sin }^{ - 1}}(0)} \right)} \right]$$</p> <p>$= 1 - \left[ {{1 \over 2} \times {\pi \over 2} - 0} \right]$</p> <p>$= 1 - {\pi \over 4} = I$</p> <p>$\therefore$ Option (c) is the right answer.</p> <p>Note :</p> <p>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.</p>