Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 91217 by mhmd last updated on 28/Apr/20

Commented by Prithwish Sen 1 last updated on 28/Apr/20

$${\int }_0^{\mathrm{n}\pi }\mid \mathrm{cosx}\mid \mathrm{dx}=\mathbf{n}{\int }_0^{\mathit{\pi }}\mid \mathrm{cosx}\mid \mathrm{dx}=\mathrm{n}[{\int }_0^{\frac{\mathit{\pi }}{2}}\mathbf{cosxdx}+{\int }_{\frac{\mathit{\pi }}{2}}^{\mathit{\pi }}(-\mathrm{cosx})\mathrm{dx}]$$$$=2\mathbf{n}\mathbf{n}=\mathbf{posituve}\mathbf{integer}.\mathbf{please}\mathbf{check}$$

Commented by MJS last updated on 28/Apr/20

$$2n$$$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\mid \cos x\mid dx=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\cos xdx=1$$$$\Rightarrow \mathrm{the}\mathrm{area}\mathrm{beneath}\mid \cos x\mid \mathrm{equals}1\mathrm{for}\mathrm{each}$$$$\mathrm{interval}[\frac{k\pi }{2};\frac{(k+1)\pi }{2}];k\in \mathbb{Z}\Rightarrow \mathrm{the}\mathrm{area}\mathrm{in}\mathrm{each}$$$$\mathrm{interval}[k\pi ;(k+1)\pi ];k\in \mathbb{Z}\mathrm{equals}2.\mathrm{we}\mathrm{have}$$$$n\mathrm{intervals}\Rightarrow \mathrm{answer}\mathrm{is}2n$$

Commented by abdomathmax last updated on 28/Apr/20

$${\int }_0^{n\pi }\mid codx\mid dx=\sum _{k=0}^{n-1}{\int }_{k\pi }^{(k+1)\pi }\mid cosx\mid dx(chx=k\pi +t$$$$=\sum _{k=0}^{n-1}{\int }_0^{\pi }\mid {(-1)}^{k}cost\mid dt=\sum _{k=0}^{n-1}{\int }_0^{\pi }\mid cost\mid dt$$$$but{\int }_0^{\pi }\mid cost\mid dt={\int }_0^{\frac{\pi }{2}}costdt-{\int }_{\frac{\pi }{2}}^{\pi }costdt$$$$=1-(-1)=2\Rightarrow {\int }_0^{n\pi }\mid cosx\mid dx=2\sum _{k=0}^{n-1}\left(1\right)$$$$=2n$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com