Use reduction formulas in Exercise 64 to evaluate the integral:
[Hint: First make a substitution]
I think this is the reduction formula they are talking about:
I think the substitution that I have to make is making the e^(3x) to e^u, but then I get confused where to put du.
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Yes, you are on the right track. You can use the substitution u = 3x, which then means that
du = 3dx (so, dx = du/3) and
`x^2 = (u/3)^2 = u^2/9`
When you plug all this into original integral, it will become
`int u^2/9 e^u (du)/3` = `1/27 int u^2 e^u du`
Now you can apply the reduction formula with n = 2, which comes from the integration by part:
`int u^2 e^u du = u^2e^u - 2int ue^u du` + C ( C here and everywhere below is an arbitrary constant)
Now, apply the reduction formula again to the second integral, with n = 1:
`int ue^u du = ue^u - int e^udu = ue^u - e^u` +C
Putting this back into the first reduction formula results in
`int u^2 e^udu = u^2e^u - 2(ue^u - e^u) = u^2e^u - 2ue^u + 2e^u` +C
Now, recall that the original integral (before the substitution) was 1/27 of the integral in terms of u. Substituting u = 3x back, we get
`1/27 ((3x)^2 e^(3x) - 2(3x)e^(3x) + 2e^(3x))` +C
Multiplying this out results in
`x^2/3 e^(3x) - 2/9e^(3x) + 2/27e^(3x)` + C.
This is the final result for the given integral.
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