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- ?Lv 73 months agoFavourite 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
= (4/25)e^5 + 1/25
- husoskiLv 73 months 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.