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### 2 Answers

- AlanLv 71 month agoFavorite Answer
Integration by parts says

∫,udv = uv - ∫vdu

so pick u and v

dv = (5)e^(5x) dx

v = e^(5x)

u = (1/5) x

du = 1/5 dx

Into the formula

= e^5x( x/5) - ∫ (1/5)e^(5x) dx

= (1/5) xe^(5x) - (1/25)e^(5x)

then add bounds 0 to 1

= (1/5)e^(5) - (1/25)e^(5) - ( 0 - (1/25) )

= (5/25 - 1/25) e^(5) + 1/25

Exact answer

= (4/25)e^5 + 1/25

Approximate answer

= 23.78610546

- husoskiLv 71 month ago
After you've practiced these a while, substitution is u=x, dv=e^(ax)dx will practically write itself. That leads to an easy du = 1, v = (1/a) e^(ax).

∫ x e^(ax) dx = ∫ u dv = uv - ∫ v du

= x (1/a) e^(ax) - ∫ (1/a) e^(ax) dx

= (x/a) e^(ax) - (1/a²) e^(ax) + C

= (ax - 1)/a² * e^(ax) + C

The rest should be easy, right? Plug a=5, evaluate at x=0 and x=1, then subtract.

For "by-parts" to work, look for something as u that will get simpler when you take the derivative. Small positive powers of x are easy things to try for that. You also want something left over in dv that you can easily integrate, and something like e^(ax) dx is a common choice.