# Integration by parts

1. Jan 27, 2007

### aerogurl2

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

Question:
(integration from 1 to 4) e^(x^(1/2))dx

2. Relevant equations

3. The attempt at a solution

So far, i have done the following:

u = e^(x^(1/2))
du = (1/2)(x^(-1/2))e^(x^(1/2))dx

dv = dx
v = x

so after applying the integration by parts formula, I got...

xe^(x^(1/2)) - (integration sign)x(1/2)(x^(-1/2))e^(x^(1/2))
dx

The integration of the second part looks wrong because it seems that I missed a step somewhere and made it more complex. Thank you for helping me.

2. Jan 27, 2007

### Curious3141

Just to clarify, you're supposed to find $$\int_1^4{e^{\sqrt{x}}}dx$$ right?

Start by evaluating the indefinite integral. The first step is to make an obvious substitution like $$x = u^2$$. After you simplify that and put everything in terms of u, you'll find an expression that can more easily be integrated by parts.

3. Jan 27, 2007

### ChaoticLlama

this isn't a by-parts questions.

try letting z = sqrt(x)

4. Jan 27, 2007

### aerogurl2

oh i see now. so after doing the subsitution for sqrt of x = y. I end up with integration of (e^y)(2ydy). which then do integration by parts through u=y and dv = e^y. thus i get x^(1/2)e^(x^(1/2))-e^(x^(1/2)). so after doing the limits it is 2e^2. is that how it is?

Last edited: Jan 27, 2007
5. Jan 27, 2007

### Curious3141

You forgot a factor of 2. EDITED : You changed your post now, the final answer is correct, but the symbolic expression is still missing a factor of two. I'd suggest factoring it and tidying it up to look nicer when you present your answer.

Last edited: Jan 27, 2007
6. Jan 27, 2007

### Gib Z

Hahaha lol, just remember integration by parts questions are for products :)

7. Jan 28, 2007

### aerogurl2

thanks for helping me!!

8. Jan 28, 2007

### Gib Z

Here at physicsforums, We have no lives :P We have nothing else to do :D

9. Jan 28, 2007

### HallsofIvy

Staff Emeritus
Not always. The standard way to integrate log(x) is by parts, letting u= 1and dv= log(x)dx.

10. Jan 28, 2007

### SunGod87

1*log(x) is a product isn't it? :p

11. Jan 28, 2007

### Gib Z

Whats wrong with f(x)=1 :(

lol