Definite Integrations problems

Question

If $f(x)= x^2 e^{x^2}$ then show that $f'(x)= 2xe^{x^2} + 2x^3 e^{x^2}$ and use this result to evaluate $$\int x^3 e^{x^2} \, dx$$

How can I use the result to evaluate the integral?

Answer 1

Hint: $$ \int x^3 e^{x^2} \, dx = \frac{1}{2} \left( \int f'(x)\, dx - \int 2x e^{x^2}\, dx \right). $$ The first integral is trivial. The second one should be solved by a standard trick: notice that $2x = \frac{d}{dx}x^2$, so that if you change variable $u=x^2$, you get...

Answer 2

1) To calculate $f'(x),$ use the product rule, with $u=x^2, \quad v=e^{x^2}$.

2) You know that $$\frac{\mathrm{d}}{\mathrm{d} x} \left[x^2e^{x^2}\right]=2xe^{x^2}+2x^3e^{x^2} \iff x^3e^{x^2} =\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}x}\left[x^2e^{x^2}\right]-xe^{x^2} \quad .$$

Take the integral of both sides with respect to $x$, to give:

$$\int x^3e^{x^2} \mathrm{d}x=\frac{1}{2}\int\frac{\mathrm{d}}{\mathrm{d}x}\left[x^2e^{x^2}\right] \ \mathrm{d}x-\int xe^{x^2} \mathrm{d}x \quad .$$

But the integral and the derivative "cancel out" (give or take an arbitrary constant) by the Fundamental Theorem of Calculus, giving:

$$\color{green}{\int x^3e^{x^2} \mathrm{d}x=\frac{1}{2}\left[x^2e^{x^2}+C_1\right] -\int xe^{x^2} \ \mathrm{d}x} \quad .$$

You should be able to evaluate this now. If you're struggling with $\int xe^{x^2} \mathrm{d}x,$ try letting $t=x^2.$

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Note: when you integrate $xe^{x^2}$, you'll get another arbitrary constant, $C_2$. For simplicity, just create a new arbitrary constant: $$C=\frac{1}{2}C_1+C_2 \quad \text{(that way you've only got one arbitrary constant)}$$

Answer 3

Let $u=x^2$ and $v=e^{x^2}$, then using formula $(uv)'=u'v+v'u$ we get

$$f'(x)=(x^2)'e^{x^2}+(e^{x^2})'x^2=2xe^{x^2} + 2x^3 e^{x^2}$$

where we use chain rules to evaluate $\displaystyle \frac{d(e^{x^2})}{dx}=\frac{d(e^{u})}{du} \cdot \frac{du}{dx}$.

Since $\displaystyle f'(x)=2xe^{x^2} + 2x^3 e^{x^2}$, taking the anti-derivative both of sides we get $$\begin{align} \int f'(x)\,dx&=\int 2xe^{x^2}\,dx+\int2 x^3 e^{x^2}\,dx\\f(x)+C_1&=2\int e^u\,\frac{du}{2} + 2\int x^3 e^{x^2}\,dx\\ x^2e^{x^2} + C_1 &= e^u + C_2 + 2\int x^3 e^{x^2}\,dx\\2\int x^3 e^{x^2}\,dx&=x^2e^{x^2}-e^{x^2} + C_3\\ \int x^3 e^{x^2}\,dx&=\frac{x^2e^{x^2}}{2}-\frac{e^{x^2}}{2} + C\\&= \frac{e^{x^2}}{2} (x^2-1)+C \end{align}$$ where $C_3=C_1-C_2$ and $C=\frac{C_3}{2}$.

Answer 4

$$f'(x)= 2xe^{x^2} + 2x^3 e^{x^2}$$ tells you that $$f(x)= \int(2xe^{x^2} + 2x^3 e^{x^2})\,dx= 2\int xe^{x^2}\,dx + 2\int x^3 e^{x^2}\,dx.$$ You are requested to evaluate$$\int x^3 e^{x^2} \, dx,$$ so you are now missing the expression of $$\int xe^{x^2}\,dx.$$ The latter integral isn't so difficult, as you can observe that the factor $x$ is the derivative of the argument of the exponential, so $$\int xe^{x^2}\,dx=\frac12e^{x^2}.$$ Now, $$2\int x^3 e^{x^2} \, dx=f(x)-2\int xe^{x^2}\,dx.$$