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Question Number 17219 by Arnab Maiti last updated on 02/Jul/17

$${\int }_0^{2\mathrm{a}}\mathrm{xy}\mathrm{dx}=?\mathrm{where}{\mathrm{x}}^2-{\mathrm{y}}^2={\mathrm{a}}^2\mathrm{and}\mathrm{y}⩾0$$

Commented by Arnab Maiti last updated on 02/Jul/17

$$\mathrm{I}\mathrm{had}\mathrm{an}\mathrm{overlook}.\mathrm{It}\mathrm{was}{\int }_{\mathrm{a}}^{2\mathrm{a}}\mathrm{xy}\mathrm{dx}=?$$$$\mathrm{The}\mathrm{given}\mathrm{answer}\mathrm{is}\sqrt{3}{\mathrm{a}}^3$$

Answered by ajfour last updated on 02/Jul/17

$$\mathrm{I}={\int }_{\mathrm{a}}^{2\mathrm{a}}\mathrm{xydx}=\frac{1}{2}{\int }_{\mathrm{a}}^{2\mathrm{a}}\sqrt{{\mathrm{x}}^2-{\mathrm{a}}^2}\left(2\mathrm{xdx}\right)$$$$\mathrm{let}{\mathrm{x}}^2-{\mathrm{a}}^2=\mathrm{t}\mathrm{x}=\mathrm{a}\Rightarrow \mathrm{t}=0$$$$\mathrm{and}\mathrm{x}=2\mathrm{a}\Rightarrow \mathrm{t}={3\mathrm{a}}^2$$$$\mathrm{then}\mathrm{dt}=2\mathrm{xdx}$$$$\mathrm{I}=\frac{1}{2}{\int }_0^{{3\mathrm{a}}^2}\sqrt{\mathrm{t}}\mathrm{dt}=\frac{1}{2}\times \frac{{2\mathrm{t}}^{3/2}}{3}{\mid }_0^{{3\mathrm{a}}^2}$$$$=\sqrt{3}{\mathrm{a}}^3.$$

Commented by Arnab Maiti last updated on 02/Jul/17

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.$$

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