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  • If  <img alt="\mathrm{\int_{0}^{2}\left(\sqrt{2 x}-\sqrt{2 x-x^{2}}\right) d x=\int_{0}^{1}\left(1-\sqrt{1-y^{2}}-\frac{y^{2}}{2}\right) d y+\int_{1}^{2}\left(2-\frac{y^{2}}{2}\right) d y+I}"

If  \mathrm{\int_{0}^{2}\left(\sqrt{2 x}-\sqrt{2 x-x^{2}}\right) d x=\int_{0}^{1}\left(1-\sqrt{1-y^{2}}-\frac{y^{2}}{2}\right) d y+\int_{1}^{2}\left(2-\frac{y^{2}}{2}\right) d y+I}  then \mathrm{I} equals

Option: 1

\mathrm{\int_{0}^{1}\left(1+\sqrt{1-y^{2}}\right) d y}


Option: 2

\mathrm{\int_{0}^{1}\left(\frac{y^{2}}{2}-\sqrt{1-y^{2}}+1\right) d y}


Option: 3

\mathrm{\int_{0}^{1}\left(1-\sqrt{1-y^{2}}\right) d y}


Option: 4

\mathrm{\int_{0}^{1}\left(\frac{y^{2}}{2}+\sqrt{1-y^{2}}+1\right) d y}


Answers (1)

best_answer

\mathrm{\quad \int_{0}^{2} \sqrt{2 x} d x} \\

\mathrm{\text { Let } 2 x=t^{2} \Rightarrow 2 d x=2 t d t} \\

\mathrm{=\int_{0}^{2} t(t d t)} \\

\mathrm{=\int_{0}^2 t^{2} d t}

\mathrm{\text { And } \int_{0}^{2} \sqrt{2 x-x^{2}} d x} \\

\mathrm{= \int_{0}^{2} \sqrt{-\left(x^{2}-2 x\right)} d x} \\

\mathrm{= \int_{0}^{2} \sqrt{1-(x-1)^{2}} d x} \\

\mathrm{\text { Let } x-1=t} \\

\mathrm{= \int_{-1}^{1} \sqrt{1-t^{2}} d t} \\

\mathrm{= 2 \int_{0}^{1} \sqrt{1-t^{2}} d t}  ( even function )

\mathrm{\therefore I= \int_{0}^{2}(\sqrt{2 x}) d x-\int_{0}^{2} \sqrt{2 x-x^{2}} d x-\int_{0}^{1} d y }\\

                  \mathrm{+\int_{0}^{1} \sqrt{1-y^{2}} d y+\int_{0}^{1} \frac{y^{2}}{2} d y-\int_{1}^{2} 2 d y+\int_{1}^{2} \frac{y^{2}}{2} d y}

\mathrm{= \int_{0}^{2} y^{2} d y-2 \int_{0}^{1} \sqrt{1-y^{2}} d y-\int_{0}^{1} d y+\int_{0}^{1} \sqrt{1-y^{2}} d y \\ }

            \mathrm{+\int_{0}^{2} \frac{y^{2}}{2} d y-2 \int_{1}^{2} d y }

\mathrm{=\int_{0}^{2}\left(y^{2}+\frac{y^{2}}{2}\right) d y-\int_{0}^{1} \sqrt{1-y^{2}} d y-\int_{0}^{1} d y\mathrm{-2 \int_{1}^{2} d y} }\\

\mathrm{=\left.\frac{y^{3}}{2}\right|_{0} ^{2}-\int_{0}^{1} \sqrt{1-y^{2}} d y-\left.y\right|_{0} ^{1}-\left.2 y\right|_{1} ^{2}}

\mathrm{=4-\int_{0}^1 \sqrt{{1-y^{2}}} d y-1-2} \\

\mathrm{=1-\int_{0}^{1} \sqrt{1-y^{2}} d y} \\

\mathrm{=\int_{0}^{1}\left(1-\sqrt{1-y^{2}}\right) d y }.

Hence the answer is option 3

Posted by

sudhir.kumar

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