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Question Number 113110 by gopikrishnan last updated on 11/Sep/20

$${\stackrel{1}{\int }}_0\sqrt{x(x-1)dx}$$

Commented by 1549442205PVT last updated on 11/Sep/20

$$\mathrm{The}\mathrm{function}\sqrt{\mathrm{x}(\mathrm{x}-1)}\mathrm{is}\mathrm{defined}\mathrm{on}$$$$\mathrm{set}\mathrm{X}=(-\mathrm{\infty },0]\cup [1,+\mathrm{\infty })\mathrm{and}{\mathrm{isn}}^{\prime }\mathrm{t}$$$$\mathrm{defined}\mathrm{on}(0,1),\mathrm{so}{\stackrel{1}{\int }}_0\sqrt{x(x-1)dx}$$$${\mathrm{don}}^{\prime }\mathrm{t}\mathrm{exist}$$$$\mathbf{You}\mathbf{should}\mathbf{see}\mathbf{your}\mathbf{question}\mathbf{again}$$

Commented by gopikrishnan last updated on 11/Sep/20

$$oksir$$

Answered by MJS_new last updated on 11/Sep/20

$$\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{mean}\int \sqrt{x(x-1)}dx$$$$\int \sqrt{x(x-1)}dx=$$$$[t=\sqrt{\frac{x}{x-1}}\to dx=-2\sqrt{x{(x-1)}^3}dt]$$$$=-2\int \frac{t^2}{{(t^2-1)}^3}dx=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{method}\right]$$$$=\frac{t(t^2+1)}{4{(t^2-1)}^2}+\frac{1}{4}\int \frac{dt}{t^2-1}=$$$$=\frac{t(t^2+1)}{4{(t^2-1)}^2}+\frac{1}{8}\ln \frac{t-1}{t+1}=$$$$=\frac{1}{4}(2x-1)\sqrt{x(x-1)}+\frac{1}{4}\ln (\sqrt{x}-\sqrt{x-1})+C$$$$\Rightarrow \underset{0}{\stackrel{1}{\int }}\sqrt{x(x-1)}dx=\frac{\pi }{8}\mathrm{i}$$

Commented by gopikrishnan last updated on 11/Sep/20

$$Thankusir$$

Answered by abdomsup last updated on 11/Sep/20

$$ithinkis{\int }_0^1\sqrt{x(1-x)}dx$$$$wedothechangement\sqrt{x}=t\Rightarrow$$$$I={\int }_0^{1}t\sqrt{1-t^2}\left(2t\right)dt$$$$=2{\int }_0^1{t}^2\sqrt{1-t^2}dt$$$${=}_{t=sin\theta }2{\int }_0^{\frac{\pi }{2}}{sin}^2\theta cos\theta cos\theta d\theta$$$$=2{\int }_0^{\frac{\pi }{2}}{\left(\frac{1}{2}sin\left(2\theta \right)\right)}^{2}d\theta$$$$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{1-cos\left(4\theta \right)}{2}d\theta$$$$=\frac{\pi }{8}-\frac{1}{4}{\left[\frac{1}{4}sin\left(4\theta \right)\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{8}-0$$$$\Rightarrow I=\frac{\pi }{8}$$

Commented by gopikrishnan last updated on 11/Sep/20

$$Thankusir$$

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