# The integral of e^x

1. May 18, 2012

### robertjford80

1. The problem statement, all variables and given/known data

I thought sometimes the integral of e^x is xe^x. Under what circumstances is the integral of e^x = xe^x? I think it has something to do with u substitution.

2. May 18, 2012

### Ray Vickson

Under NO circumstances is the integral of e^x equal to x e^x. I cannot imagine why you think that would hold.

RGV

3. May 18, 2012

### robertjford80

4. May 18, 2012

### sharks

The derivative of $e^x$ is: $$e^x .\frac{d(x)}{dx}$$

5. May 18, 2012

### HallsofIvy

Staff Emeritus
The derivative of $xe^x$ is, by the product rule $(x)'e^x+ (x)(e^x)'= 1(e^x)+ x(e^x)= xe^x+ e^x= (x+ 1)e^x$. As Ray Vickson said, the integral of $xe^x$ is NOT equal to itself and neither is the derivative.

The only functions having the property that their derivative is equal to the function itself is a constant times $e^x$.

6. May 18, 2012

### sharks

I don't think anyone could have explained it better. This should resolve your confusion, robertjford80.

7. May 18, 2012

### robertjford80

Here's an example

What's going on here? It clear says that the derivative of

c1e(3/2)x[ = (3/2)c1e(3/2)x

8. May 18, 2012

### SammyS

Staff Emeritus
What's going on with the derivative of c1e(3/2)x[ is mainly the chain rule.

9. May 18, 2012

### robertjford80

so the integral of e^2x is e^2x and the derivative of e^x is e^x but the derivative of e^2x is 2e^2x, is that right?

10. May 18, 2012

### sharks

Correct, except for the integral of $e^{2x}$ which is $\frac{e^{2x}}{2}$

11. May 18, 2012

### Number Nine

<deleted>

12. May 18, 2012

### robertjford80

well, why don't you use the chain rule with e^x which would make it xe^x?

{this referred to number nine's deleted post} i saw it before he deleted it.

13. May 18, 2012

### Villyer

The chain rule is to multiply by the derivative, and the derivative of x is 1.

If it helps, d/dx (ex) = 1 * ex

14. May 18, 2012

### robertjford80

15. May 18, 2012

### Ray Vickson

No, the first statement is not right, and is not what you asked originally. The indefinite integral of exp(a*x) for constant a is (1/a)*exp(a*x) + C; the derivative of exp(a*x) is a*exp(a*x). When a = 1 these both give just exp(x). For a = 2 they give (1/2) exp(2x) and 2 exp(2x), respectively.

RGV

16. May 18, 2012

### sharks

That's exactly what i said before.

17. May 18, 2012

### robertjford80

if you're talking about post 4, then i don't think you provided enough info to convey that

18. May 18, 2012

### sharks

It's obvious that $\frac{dx}{dx}=1$ which gives $e^x .1=e^x$. Unless, you didn't know that, but it's really a basic notion of the principles of differentiation.
You should go over the basic principles, as it should help you to understand $e^x$ and the others more complicated that will follow.

19. May 18, 2012

### robertjford80

if it was obvious i would not have posted the question

20. May 19, 2012

### sankalpmittal

d(ex)/dx = ex

And so

∫ex dx = ___________ ....(i)

Note that integration is just reverse of differentiation.

If you want to verify this , then its simple ! Differentiate the left hand side of equation (i) with respect to x and see if its equal to ex. It will work.

If you want to prove it then analyze it by means of graph of f(x)=ex.

And note if you do this :

d(ex)/d(e) = xex-1

But it can never be xex !

Last edited: May 19, 2012