Question

Evaluate the integral using integration by parts where possible.

Integral symbol 1(at top) -1 (at bottom) (3x^2 +x)e^-x dx

So I know this is the tabular method, and so far I have :

u=(3x^2+x)

derivative of U: Anti-deriv. of Dv:

3x^2+x e^-x

=6x -e^-x


=6 e^-x

=0

I dont know how to do the rest!

Answer 1

\[\int\limits_{-1}^{1}\frac{3x^2+x}{e^x} dx\]
Select 3x^2+x as your u and e^-x as your dv
u = 3x^2+x
du = 6x + 1 dx (take the derivative to find du)
v = -e^-x (take the antiderivative to find v)
dv = e^-x

IBP states that...
\[\int\limits u dv = uv - \int\limits v du\]
Using our values for u, du, v and dv, we find that...
\[\frac{-(3x^2+x)}{e^x} -\int\limits \frac{-(6x+1)}{e^x} dx\]
You will see we will have to use IBP once more, but we are getting closer to the solution.
u = 6x+1
du = 6 dx
v = e^-x
dv = -e^-x dx
du = 6x + 1 dx (take the derivative to find du)
v = -e^-x (take the antiderivative to find v)
dv = e^-x
Now...
\[\frac{-(3x^2+x)}{e^x}-(\frac{6x+1}{e^x}- 6\int\ \frac{1}{e^x}dx)\]
You will find through the integration of the final e^-x and a bit of algebra that...
\[\int\limits\frac{3x^2+x}{e^x} dx=\frac{-(3x^2+7x+7)}{e^x} + C \]
Now using this, we can find the definite integral
Evalutate -(3x^2+7x+7)e^-x at 1, then take that value and subtract away -(3x^2+7x+7)e^-x evaluated at -1

That should give you -17/e + 3e which is about 1.900894985...

Answer 2

Thank you so much!!! Just one question, how do you know how many times you have to use IBP?

Answer 3

Until the integrand has been simplified to a point where you can integrate without IBP.

I stopped using IBP when I got to the integration of 1/e^x or e^-x because its a fairly simple integration problem, being -e^-x