# Integrating $\int^2_0 xe^{x^2}dx$

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.

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

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.

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