Question-11019

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} {x e}^{- 2 x + 4} d x = - \frac{1}{2} ({[{x e}^{- 2 x + 4}]}_{2}^{n} - \int_{2}^{n} e^{- 2 x + 4} d x)

Answered by geovane10math last updated on 07/Mar/17

\int_{2}^{n} x \cdot e^{- 2 x + A} dx = F (n) - F (2)

\int x \cdot e^{- 2 x + A} dx = \int x \cdot e^{- 2 x} \cdot e^{A} dx =

= e^{A} \int x \cdot e^{- 2 x} dx

- 2 x = u \Rightarrow \frac{d u}{d x} = - 2 \Rightarrow d x = - \frac{d u}{2}

e^{A} \int - \frac{u}{2} \cdot e^{u} du = e^{A} \cdot \frac{1}{2} \cdot \int u \cdot e^{u} du =

= \frac{e^{A}}{2} \int u \cdot e^{u} du

\int u \cdot e^{u} du

u = s, e^{u} du = d t

\int s dt = s t - \int t ds

\int u \cdot e^{u} du = u (e^{u} + c_{1}) - \int [e^{u} + c_{1}] du

\int u \cdot e^{u} du = {u e}^{u} + c_{1} u - (e^{u} + c_{1} u + C)

\int {u e}^{u} du = {u e}^{u} + c_{1} u - e^{u} - c_{1} u - C

\int {u e}^{u} du = e^{u} (u - 1) - C

\frac{e^{A}}{2} [e^{u} (u - 1) - C] = \frac{e^{A}}{2} [e^{- 2 x} (- 2 x - 1 - C)]

\int x \cdot e^{- 2 x + A} dx = - \frac{e^{A}}{2} [e^{- 2 x} (2 x + 1 + C)]

F (n) - F (2) =

= - \frac{e^{A}}{2} [e^{- 2 n} (2 n + 1 + C)] - (- \frac{e^{A}}{2} [(5 + C)])

= - \frac{e^{A}}{2} [e^{- 2 n} (2 n + 1 + C)] + \frac{e^{A}}{2} (5 + C)