Definite Integral: Exponential

[michonamona]

Homework Statement

\int^{\infty}_{0} e^{-y} dy = 1

Homework Equations

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

 

[Mark44]

Homework Statement

\int^{\infty}_{0} e^{-y} dy = 1

Homework Equations

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

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.

 

[HallsofIvy]

Start by letting u= -x.

 

[michonamona]

ok, suppose u=-y, then\int^{\infty}_{0} e^{u} dy = e^{u} ] ^{\infty}_{0} = e^{-y} ] ^{\infty}_{0}

By the fundamental theorem of calculus

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

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

 

[snipez90]

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

 

[michonamona]

I got it. Thank you!

 

[Mark44]

ok, suppose u=-y, then
\int^{\infty}_{0} e^{u} dy = e^{u} ] ^{\infty}_{0} = e^{-y} ] ^{\infty}_{0}

By the fundamental theorem of calculus

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

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

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