sue asked in Science & MathematicsMathematics · 3 months ago

# integration by part?

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Lv 7
3 months ago

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

= 23.78610546

• 3 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.

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