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Question Number 11019 by ridwan balatif last updated on 07/Mar/17

Answered by bahmanfeshki last updated on 07/Mar/17

$$I={\int }_2^n{xe}^{-2x+4}dx=-\frac{1}{2}({\left[{xe}^{-2x+4}\right]}_2^n-{\int }_2^n{e}^{-2x+4}dx)$$

Answered by geovane10math last updated on 07/Mar/17

$${\int }_2^{n}x\cdot e^{-2x+A}\mathrm{dx}=F\left(n\right)-F\left(2\right)$$$$\int x\cdot e^{-2x+A}\mathrm{dx}=\int x\cdot e^{-2x}\cdot e^A\mathrm{dx}=$$$$=e^A\int x\cdot e^{-2x}\mathrm{dx}$$$$-2x=u\Rightarrow \frac{du}{dx}=-2\Rightarrow dx=-\frac{du}{2}$$$$e^A\int -\frac{u}{2}\cdot e^u\mathrm{du}=e^A\cdot \frac{1}{2}\cdot \int u\cdot e^u\mathrm{du}=$$$$=\frac{e^A}{2}\int u\cdot e^u\mathrm{du}$$$$\int u\cdot e^u\mathrm{du}$$$$u=s,e^u\mathrm{du}=dt$$$$\int s\mathrm{dt}=st-\int t\mathrm{ds}$$$$\int u\cdot e^u\mathrm{du}=u(e^u+c_1)-\int [e^u+c_1]\mathrm{du}$$$$\int u\cdot e^u\mathrm{du}={ue}^u+c_{1}u-(e^u+c_{1}u+C)$$$$\int {ue}^u\mathrm{du}={ue}^u+c_{1}u-e^u-c_{1}u-C$$$$\int {ue}^u\mathrm{du}=e^u(u-1)-C$$$$\frac{e^A}{2}[e^u(u-1)-C]=\frac{e^A}{2}\left[e^{-2x}(-2x-1-C)\right]$$$$\int x\cdot e^{-2x+A}\mathrm{dx}=-\frac{e^A}{2}\left[e^{-2x}(2x+1+C)\right]$$$$F\left(n\right)-F\left(2\right)=$$$$=-\frac{e^A}{2}\left[e^{-2n}(2n+1+C)\right]-(-\frac{e^A}{2}\left[(5+C)\right])$$$$=-\frac{e^A}{2}\left[e^{-2n}(2n+1+C)\right]+\frac{e^A}{2}(5+C)$$$$$$

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