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How do I calculate $\int_0^2 x^2 e^x dx$

Is there a product rule for integration ?

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    Unfortunately, it seems you're in the position where you have to ask a basic (to people who have studied math for a while) question without providing any work. I imagine for this reason, some people deemed it worthy of a downvote (I didn't downvote). Perhaps next time you're in this situation, explain where the problem came from, your background concerning the relevant subject, and why you find the problem intractable. See [this thread](http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question) for other advice.2012-12-15

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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 $x$.

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    If you call LHS as $I(n)$, then $I(n) = 2^ne^2 - nI(n - 1)$, therefore $I(2) = 4e^2 - 2(2e^2 - I(0)) = 4e^2 - 4e^2 + 2I(0) = 2e^2 - 2$ where $I(0)$ is explicitly evaluated.2012-12-15
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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 $\int x^2 e^x dx=\int x^2 (e^x)^{\prime} dx=x^2e^x-\int (x^2)^{\prime} e^xdx=x^2e^x-\int2xe^xdx=\\x^2e^x-2\int x (e^x)^{\prime} dx=x^2e^x-2xe^x+2\int e^x=x^2e^x-2xe^x+2e^x+c$

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    Why is my question down voted ?2012-12-15
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$f(r) = \int_0^{2} e^{rx} dx = \frac{e^2 - 1}{r}$ $f'(r) = \int_0^{2} xe^{rx} dx = -\frac{e^2 - 1}{r^2}$ $f''(r) = \int_0^{2} x^2e^{rx} dx = \frac{2e^2 - 2}{r^3}$ $f''(1) = \int_0^{2} x^2e^{x} dx = 2e^2 - 2$

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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}

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    @90intuition This is too a standard notation. See any calculus book. Good look.2012-12-15