Solving Integration by Parts Problem from 1 to 4

aerogurl2
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Homework Statement


Hi, I've been having trouble solving the following problem, please help me.

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


Homework Equations





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.
 
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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. :smile:
 
this isn't a by-parts questions.

try letting z = sqrt(x)
 
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?
 
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aerogurl2 said:
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 e^2. is that how it is?

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.
 
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Hahaha lol, just remember integration by parts questions are for products :)
 
thanks for helping me!
 
Here at physicsforums, We have no lives :P We have nothing else to do :D
 
Gib Z said:
Hahaha lol, just remember integration by parts questions are for products :)
Not always. The standard way to integrate log(x) is by parts, letting u= 1and dv= log(x)dx.
 
  • #10
1*log(x) is a product isn't it? :p
 
  • #11
Whats wrong with f(x)=1 :(

lol
 
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