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Question Number 56107 by Joel578 last updated on 10/Mar/19

$${\int }_{-1}^0\mid x\sin \left(\pi x\right)\mid dx$$

Commented by maxmathsup by imad last updated on 10/Mar/19

$$letI={\int }_{-1}^0\mid xsin\left(\pi x\right)\mid dxchangement\pi x=-tgive$$$$I={\int }_{\pi }^0\mid \frac{-t}{\pi }sin(-t)\mid \frac{-dt}{\pi }=\frac{1}{{\pi }^2}{\int }_0^{\pi }tsintdtbyparts$$$${\pi }^{2}I={[-tcost]}_0^{\pi }+{\int }_0^{\pi }costdt=\pi +{\left[sint\right]}_0^{\pi }=\pi \Rightarrow I=\frac{1}{\pi }$$

Answered by mr W last updated on 10/Mar/19

$$I=\int x\sin \left(x\pi \right)dx$$$$=-\frac{1}{\pi }\int xd\left(\cos \pi x\right)$$$$=-\frac{1}{\pi }[x\cos \left(\pi x\right)-\int \cos \left(\pi x\right)dx]$$$$=-\frac{1}{\pi }[x\cos \left(\pi x\right)-\frac{1}{\pi }\int \cos \left(\pi x\right)d\left(\pi x\right)]$$$$=\frac{\sin \left(\pi x\right)}{{\pi }^2}-\frac{x\cos \left(\pi x\right)}{\pi }+C$$$${\int }_{-1}^0\mid x\sin \left(\pi x\right)\mid dx$$$$={\int }_0^{1}x\sin \left(\pi x\right)dx$$$$={[\frac{\sin \left(\pi x\right)}{{\pi }^2}-\frac{x\cos \left(\pi x\right)}{\pi }]}_0^1$$$$=\frac{1}{\pi }$$

Answered by MJS last updated on 10/Mar/19

$$\int \mid x\sin \pi x\mid dx=\mathrm{sign}\left(x\sin \pi x\right)\int x\sin \pi xdx=$$$$=\mathrm{sign}\left(x\sin \pi x\right)(\frac{1}{{\pi }^2}\sin \pi x-\frac{1}{\pi }x\cos \pi x)$$$$\underset{-1}{\stackrel{0}{\int }}\mid x\sin \pi x\mid dx=\frac{1}{\pi }$$

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