MHB Solving Integration by Parts: x^3e^x^2

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To solve the integral of x^3e^(x^2) dx, a substitution is made by letting x^2 = t, which simplifies the integral to (t e^t) dt / 2. This allows for the application of integration by parts, where u = x^2 and dv = xe^(x^2) dx. The integration by parts results in the expression (1/2)x^2e^(x^2) - (1/2)e^(x^2) + C. Ultimately, the final solution is (1/2)e^(x^2)(x^2 - 1) + C. This method effectively demonstrates the process of integration by parts for this specific integral.
paulmdrdo1
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any idea how to solve this?

\begin{align*}\displaystyle \int x^3e^{x^{2}}\,dx\end{align*}
 
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Re: integration by parts

paulmdrdo said:
any idea how to solve this?

\begin{align*}\displaystyle \int x^3e^{x^{2}}\,dx\end{align*}

put x^2 = t

so 2 x dx = dt

so x^3 e^(x^2) dx = t e^t dt / 2

now you can differentiate t and integrate e^t thus by parts.
 
Re: integration by parts

Hello, paulmdrdo!

I \;=\; \int x^3e^{x^2}\,dx
We have: .\int x^2\cdot xe^{x^2}dx

By parts: .\begin{Bmatrix}u &=& x^2 && dv &=& xe^{x^2}dx \\ du &=& 2x\,dx && v &=& \tfrac{1}{2}e^{x^2} \end{Bmatrix}

Then: .I \;=\;\tfrac{1}{2}x^2e^{x^2} - \int xe^{x^2}dx

. . . . . I \;=\;\tfrac{1}{2}x^2e^{x^2} - \tfrac{1}{2}e^{x^2} + C

. . . . . I \;=\;\tfrac{1}{2}e^{x^2}(x^2-1) + C
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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