# Definite Integral: Exponential

1. Sep 18, 2010

### michonamona

1. The problem statement, all variables and given/known data
$$\int^{\infty}_{0} e^{-y} dy = 1$$

2. Relevant equations

3. The attempt at a solution

Why is this equality true? I understand that the integral and derivative of e^y is always e^y, but I can't make out why this definite integral is equal to 1. I graphed it using a graphing application and can see why it converges to a finite number, but I can't work it out step by step.

Thanks, I appreciate your help.

M

2. Sep 18, 2010

### Staff: Mentor

How about this integral? Can you evaluate it?

$$\int_0^b e^{-y} dy$$
If you can get that, then just take the limit as b approaches infinity.

BTW, this should have been posted in the Calculus & Beyond section.

3. Sep 18, 2010

### HallsofIvy

Staff Emeritus
Start by letting u= -x.

4. Sep 18, 2010

### michonamona

ok, suppose u=-y, then

$$\int^{\infty}_{0} e^{u} dy = e^{u} ] ^{\infty}_{0} = e^{-y} ] ^{\infty}_{0}$$

By the fundemental theorem of calculus

$$e^{-y} ]^{\infty}_{0} = 0 - e^0 = -1$$

Ok, so where in my solution did I go wrong?

Last edited: Sep 18, 2010
5. Sep 18, 2010

### snipez90

Um, I wouldn't make that substitution. Mark44's suggestion is the easiest way.

6. Sep 18, 2010

### michonamona

I got it. Thank you!

7. Sep 19, 2010

### Staff: Mentor

Here's where you went wrong: If u = - y, then du = -dy. You just replaced dy with du.

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