# Integrals of exponentials

1. Oct 23, 2008

### JFonseka

1. The problem statement, all variables and given/known data
Integrate: x.e$$^{2x^{2}}$$

2. Relevant equations

None.

3. The attempt at a solution

I first thought that the coefficient 2 in "2x" would become the denominator:

$$\frac{x.e^{2x^{2}}}{2}$$

and then integrating the x would mean x^2 and division by another two making the answer:

$$\frac{x^{2}.e^{2x^{2}}}{4}$$

But the answer listed in the book doesn't have an x^2, it's only
$$\frac{e^{2x^{2}}}{4}$$

What did I do wrong?

2. Oct 23, 2008

### HallsofIvy

Staff Emeritus
You didn't do anything right! You know, I hope, that the derivative of product is NOT just the product of the derivatives: (fg)'= f'g+ fg', not f'g'. So you can't expect that the integral of a product will just be the product of the integrals: the integral of fg is not just the integral of f times the integral of g.

Here, you need to make a substitution: if u= x2, what is du/dx? What is du in terms of dx?

3. Oct 23, 2008

### JFonseka

It would be 2x.dx of course.

So I guess the substitution you want me to make is u = 2x^{2} ? Therefore du/dx= 4x.dx
And then it would be $$du/4.e^{u}$$

But then I don't see how that becomes the correct answer.

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