How do I calculate $$\int_0^2 x^2 e^x dx$$
Is there a product rule for integration ?
How do I calculate $$\int_0^2 x^2 e^x dx$$
Is there a product rule for integration ?
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 $x$.
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 $$\int x^2 e^x dx=\int x^2 (e^x)^{\prime} dx=x^2e^x-\int (x^2)^{\prime} e^xdx=x^2e^x-\int2xe^xdx=\\x^2e^x-2\int x (e^x)^{\prime} dx=x^2e^x-2xe^x+2\int e^x=x^2e^x-2xe^x+2e^x+c$$
$$f(r) = \int_0^{2} e^{rx} dx = \frac{e^2 - 1}{r}$$ $$f'(r) = \int_0^{2} xe^{rx} dx = -\frac{e^2 - 1}{r^2}$$ $$f''(r) = \int_0^{2} x^2e^{rx} dx = \frac{2e^2 - 2}{r^3}$$ $$f''(1) = \int_0^{2} x^2e^{x} dx = 2e^2 - 2$$
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}