Calculate $\int_0^2 x^2 e^x dx$ How do I calculate 
$$\int_0^2 x^2 e^x dx$$
Is there a product rule for integration ?
 A: We shall use integration by parts:
$$\int f^{\prime}(x)g(x)dx=f(x)g(x)-\int f(x)g^{\prime}(x)dx$$
We have $$\int x^2 e^x dx=\int x^2 (e^x)^{\prime} dx=x^2e^x-\int (x^2)^{\prime}  e^xdx=x^2e^x-\int2xe^xdx=\\x^2e^x-2\int x (e^x)^{\prime} dx=x^2e^x-2xe^x+2\int e^x=x^2e^x-2xe^x+2e^x+c$$
A: $$f(r) = \int_0^{2} e^{rx} dx = \frac{e^2 - 1}{r}$$
$$f'(r) = \int_0^{2} xe^{rx} dx = -\frac{e^2 - 1}{r^2}$$
$$f''(r) = \int_0^{2} x^2e^{rx} dx = \frac{2e^2 - 2}{r^3}$$
$$f''(1) = \int_0^{2} x^2e^{x} dx = 2e^2 - 2$$
A: The answer of your question is yes, existis. The conection is the fundamental theorem of calculus and produtc ruler diferentiation. We have that
$$
D_x(u(x)\cdot v(x))=v(x)\cdot D_x u(x)+u(x)\cdot D_x v(x)
$$
implies
$$
v(x)\cdot D_x u(x)= D_x(u(x)\cdot v(x)) -u(x)\cdot D_x v(x)
$$
and 
$$
\int^b_a v(x)\cdot D_x u(x)\, dx= \int^b_a D_x(u(x)\cdot v(x))\,dx -\int^b_a u(x)\cdot D_x v(x)\, dx
$$
By Fundamental Theorem of Calculus
$$
\int^b_a v(x)\cdot D_x u(x)\, dx= u(x)\cdot v(x)\bigg|^b_a -\int^b_a u(x)\cdot D_x v(x)\, dx
$$ 
This is the formula of  integration by parts.
\begin{align}
\int^{2}_{0} x^2 e^x dx=
&
\int^{2}_{0} x^2 (e^x)^{\prime} dx
& 
(e^x)^{\prime}=e^x
\\
=
&
x^2e^x\bigg|^{2}_{0}-\int|^{2}_{0} (x^2)^{\prime}  e^xdx
&
\mbox{formula of  integration by parts}
\\
=
&
x^2e^x\bigg|^{2}_{0}-\int^{2}_{0}2xe^xdx
&
(x^2)^{\prime}=2x
\\
=
&
x^2e^x\bigg|^{2}_{0}-2\int^{2}_{0} x (e^x)^{\prime} dx
&
(e^x)^{\prime}=e^x
\\
=
&
x^2e^x\bigg|^{2}_{0}-2xe^x\bigg|^{2}_{0}+2\int^{2}_{0} e^x dx
&
\mbox{formula of  integration by parts}
\\
=
&
x^2e^x-2xe^x+2e^x\bigg|^{2}_{0}
&
\end{align}
A: Hint: Note that $$\int_0^2 x^ne^xdx=\int_0^2 x^nd(e^x)=x^ne^x|_0^2-n\int_0^2 x^{n-1}e^xdx$$
Now continue reducing the power of $x$.
