4
$\begingroup$

Well what I was thinking was to integrate the indefinite integral first.

$u=x^2$, $x=\sqrt u$

$du=2xdx = 2\sqrt {u} dx$

$dx= \frac{1}{2\sqrt{u}}du$

$\int xe^{x^2} dx = \int \sqrt{u}\frac{1}{2\sqrt{u}} du =\frac{1}{2}\int e^u du = \frac{1}{2}e^u =\frac{1}{2}e^{x^2} +C$

Now I can evaluate $\frac{1}{2}e^{x^2}\Big|_0^2= \frac{1}{2} e^{4} -\frac{1}{2} e^0 =\frac{1}{2}e^4-1$

so my answer should be $$\frac{1}{2}e^4-1$$

Is this correct? It's been a while since I've done stuff like this.

$\endgroup$
3
  • $\begingroup$ Yes, it is correct. $\endgroup$ Jun 15 '19 at 19:56
  • 1
    $\begingroup$ If you're not sure whether your antiderivative is correct, differentiate it. If you get $x e^{x^2}$, it's correct. $\endgroup$ Jun 15 '19 at 19:58
  • $\begingroup$ You're missing a pair of parentheses in the evaluation. $\endgroup$
    – Bernard
    Jun 15 '19 at 20:06
2
$\begingroup$

Also, one might set

$g(x) = e^{x^2}; \tag 1$

then

$g'(x) = 2xe^{x^2}; \tag 2$

then

$\displaystyle \int_0^2 xe^{x^2} \; dx = \dfrac{1}{2} \int_0^2 g'(x) \; dx = \dfrac{1}{2}(g(2) - g(0))$ $= \dfrac{1}{2}(e^{2^2} - e^0) = \dfrac{1}{2} (e^4 - 1) = \dfrac{1}{2}(e^4 - 1) = \dfrac{1}{2}e^4 - \dfrac{1}{2}. \tag{3}$

If one wants to use indefinite integrals, we write

$\displaystyle \int xe^{x^2} \; dx = \dfrac{1}{2} \int g'(x) \; dx = \dfrac{1}{2}g(x) + C = \dfrac{1}{2}e^{x^2} + C, \tag 4$

and then proceed to take

$g(2) - g(0) = \dfrac{1}{2}e^4 - \dfrac{1}{2}; \tag 5$

the constant of integration $C$ of course has been cancelled out of this expression.

$\endgroup$
3
$\begingroup$

You mean $\frac12 e^4-\frac12$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.