# Thread: how do i integrate e^x^1/2

1. ## how do i integrate e^x^1/2

SSIA...
i checked through all theexamples and have no idea how to even begin..i would really appreciate it if someone pointed me in the right direction..thanks

2. Have you considered using U-Substitution? Look it up, If not.

3. Originally Posted by twostep08
SSIA...
i checked through all theexamples and have no idea how to even begin..i would really appreciate it if someone pointed me in the right direction..thanks
How do you integrate $\displaystyle e^{\sqrt{x}}$? Let $\displaystyle z=\sqrt{x}$ and you'll get a computable integral.

4. thats what it looks like, but theres no other x's i could use to get the du.

like u=x^1/2...du=1/2x^(-1/2) dx
but theres no x's besides the exponent to help me---

5. Originally Posted by twostep08
thats what it looks like, but theres no other x's i could use to get the du.

like u=x^1/2...du=1/2x^(-1/2) dx
but theres no x's besides the exponent to help me---
$\displaystyle z^2=x\implies 2z\:dz=dx\implies \int e^{\sqrt{x}}dz\overbrace{\longmapsto}^{z^2=x}2\int z e^z dz$

6. ok...so i kind of understand, but where did that z inside the integral come from---im getting 2 (integral) e^z dz

7. Originally Posted by twostep08
ok...so i kind of understand, but where did that z inside the integral come from---im getting 2 (integral) e^z dz
Because if this was $\displaystyle \int e^{\sqrt{x}}$ I might see where you are coming from but we are integrating $\displaystyle \int e^{\sqrt{x}}\color{red}dx$ so we have to find a replacement for $\displaystyle e^{\sqrt{x}}$ which we do by replacing $\displaystyle x$ with $\displaystyle z^2$ and we have to replace $\displaystyle dx$ by $\displaystyle 2z\:dz$

8. ooh duh. my fault on the bad derivative. im gonna try to work it out from there- well see where i get with integration by parts

9. Originally Posted by twostep08
ooh duh. my fault on the bad derivative. im gonna try to work it out from there- well see where i get with integration by parts
P.S.

Spoiler:

wolframalpha.com

10. Drex why you have to bring it to a single term of x could not just leave it as when doing the sub

$\displaystyle dz = -\frac{1}{2\sqrt{x}}dx$
$\displaystyle 2\sqrt{x} dz = dx$

I never understood when to do things like that

11. sweet Georgia! that seems like the best site in the world. had i known about that sooner, i wouldve been saved many sleepless nights
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