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Question Number 170290 by muneer0o0 last updated on 19/May/22

Commented by Engr_Jidda last updated on 19/May/22

$$thequestionisnotclear$$

Answered by Mathspace last updated on 21/May/22

$$iftheQis{\int }_0^2{\int }_x^{\sqrt{4-x^2}}{\left(\frac{x}{y}\right)}^{2}dydx$$$$\Rightarrow I={\int }_0^2\left({\int }_x^{\sqrt{4-x^2}}\frac{1}{y^2}dy\right)x^{2}dx$$$$={\int }_0^2\left({[-\frac{1}{y}]}_x^{\sqrt{4-x^2}}\right)x^{2}dx$$$$={\int }_0^2{x}^2(\frac{1}{x}-\frac{1}{\sqrt{4-x^2}})dx$$$$={\int }_0^{2}xdx-{\int }_0^2\frac{x^2}{\sqrt{4-x^2}}dx$$$$={\left[\frac{x^2}{2}\right]}_0^2-{\int }_0^2\frac{x^2}{\sqrt{4-x^2}}dx$$$$=2+{\int }_0^2\frac{4-x^2-4}{\sqrt{4-x^2}}dx$$$$=2+{\int }_0^2\sqrt{4-x^2}dx+4{\int }_0^2\frac{dx}{\sqrt{4-x^2}}$$$$wehave$$$${\int }_0^2\sqrt{4-x^2}dx{=}_{x=2sin\theta }{\int }_0^{\frac{\pi }{2}}2cos\theta \left(2cos\theta \right)d\theta$$$$=4{\int }_0^{\frac{\pi }{2}}{cos}^2\left(\theta \right)d\theta =2{\int }_0^{\frac{\pi }{2}}(1+cos\left(2\theta \right))d\theta$$$$=\pi +{\left[sin\left(2\theta \right)\right]}_0^{\frac{\pi }{2}}=\pi$$$${\int }_0^2\frac{dx}{\sqrt{4-x^2}}{=}_{x=2t}{\int }_0^1\frac{2dt}{2\sqrt{1-t^2}}$$$$={\left[arcsint\right]}_0^1=\frac{\pi }{2}\Rightarrow$$$$I=2+\pi +2\pi =2+3\pi$$

Commented by Tawa11 last updated on 08/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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