How can I use integration by parts to find int_{0}^{5}te^{-t}dt∫50te−tdt?

1 Answer
Gió
Jan 25, 2015

You can do this:
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Using this approach you should get:
int_0^5(te^(-t))dt=-e^-t*t-int(-e^-t)*1dt=∫50(te−t)dt=−e−t⋅t−∫(−e−t)⋅1dt=
=-e^-t*t-e^-t==−e−t⋅t−e−t=
=-e^(-t)(t+1)=−e−t(t+1)
Now you must use your extrema: 0 and 50and5
You get:
-e^-5*(5+1)-[-e^0*(0+1)]=−e−5⋅(5+1)−[−e0⋅(0+1)]=
=1-6e^-5=1−6e−5

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