Hello,
I think you want integrate x^2 e^{-5x}x2e−5x, not x^2 e^{-5}xx2e−5x...
Use two integrations by parts :
\int_0^x u(t) v'(t) dt = [u(t)v(t)]_0^x - \int_0^x u'(t) v(t) dt.
You can prove that if you write (uv)' = u'v+uv' and if you integrate between 0 and x.
Let F be the function define by F(x) = \int_0^x t^2 e^{-5t} dt.
Take u(t) = t^2 and v(t) =-\frac{1}{5}e^{-5t}. So you have
u'(t) = 2t and v'(t) = e^{-5t}
Apply the integration by parts formula :
F(x) = [-\frac{1}{5} t^2 e^{-5t} ]_0^x +\frac{2}{5} \int_0^x te^{-5t}dt
Use integration by parts again to calculate \int_0^x te^{-5t}dt. Take
u(t) =t and v(t) = -\frac{1}{5}e^{-5t}. So you have
u'(t) = 1 and v'(t) = e^{-5t}, and
\int_0^x te^{-5t}dt = [-\frac{1}{5}t e^{-5t}]_0^x + \frac{1}{5}\int_0^x e^{-5t}dt = [-\frac{1}{5}t]_0^x + \frac{1}{5}[-\frac{1}{5}e^{-5t}]_0^x
You can simplify :
\int_0^x te^{-5t}dt =-\frac{1}{5}xe^{-5x} - \frac{1}{25}(e^{-5x}-1)
Finally, plug that in the above expression :
F(x) = -\frac{1}{5}x^2e^{-5x} - \frac{2}{5}(\frac{1}{5}xe^{-5x} + \frac{1}{25}(e^{-5x}-1))
Primitives (or antiderivatives) of x^2e^{-5x} are
-(\frac{1}{5}x^2 + \frac{2}{25}x + 2/125)e^{-5x} + c
where c \in RR.
