# Homework Help: Definite integral

1. Dec 1, 2008

### zeno23

1. The problem statement, all variables and given/known data
Evaluate the definite integral.

2. Relevant equations
$$\int_{1}^{2} ( 2e^{-4x} -\frac{1}{x^2} ) dx$$

Answer given by the book: $$\frac{1}{2}(e^{-4}-e^{-8}-1)$$

3. The attempt at a solution

u = 4x; x = u/4; du = 4 dx; dx = du/4;

$$\frac{1}{2}\int e^{-u} du - 4\int u^{-2} du$$

$$\frac{1}{2}\int_{1}^{2} e^{-4x} dx - 4\int_{1}^{2} \frac{1}{16x^2} dx$$

... and I continue down this path to wind up with:

$$\frac{1}{2e^8}-\frac{1}{2e^4}-\frac{21}{32}$$

In other words, I've apparently no idea what I'm doing, and would appreciate any help.

2. Dec 1, 2008

### Staff: Mentor

Split the integral into two integrals:
$$\int_{1}^{2} 2e^{-4x} dx - \int_{1}^2 x^{-2}} dx$$

It looks like you are letting your substitution slop over into the 2nd part of your integral.

For your substitution use u = -4x, so du = -4 dx.

After you have found an antiderivative for the 1st integral, undo your substitution and evaluate the antiderivative at x = 2 and x = 1.

Alternatively, you can write new limits of integration for the 1st integral by converting the two values of x into values of u.

3. Dec 1, 2008

### zeno23

Wait, I got it. It should just be e^u, so e^(-4*2). /facepalm

Thanks again.

----------------

Thanks for the response. I've managed to arrive at the proper answer, but I don't grasp the logic completely. I start with:

$$\int_{1}^{2} 2e^{-4x}$$

I set u=-4x and dx=(-1/4) du, so then:
$$- \frac{1}{2} \int e^u du = -\frac{ e^{ \frac{u^2}{2} } }{2}$$.

Putting this back into terms of X, I get: $$-\frac{e^{\frac{(-4x)^2}{2}}}{2}$$

But I it looks like it should be $$-\frac{e^{-4\frac{x^2}{2}}}{2}$$. When I use this, I get the proper answer. Since u=(-4x), why do we multiply x^2 by -4 in the final step rather than (-4x)^2 ?

Last edited: Dec 1, 2008
4. Dec 1, 2008

### Staff: Mentor

$$\int e^u du = e^u + C$$