What Is the Correct Anti-Derivative of e^(x/2)?

[in the rye]

Homework Statement

I am working on an integration by parts problem, and in order to work it I need to figure out the anti-derivative of e^x/2. We've covered basic integration concepts, the definite/indefinite integral, u-sub, and integration by parts. Now, examining the derivative, I expect the anti-derivative to be 2e^x/2, but I can't show my work on how I get here. I tried u-sub and I just get the same function.

 

[BvU]

No. This one requires a trick.

And the integral with finite bounds is the error function

 

[blue_leaf77]

No. This one requires a trick.

And the integral with finite bounds is the error function

I think the OP is not dealing with a Gaussian function.

 

[BvU]

Oh boy, misread. And answered too quickly - again. Sorry.

Dear Rye,

if you differentiate $e^{x\over 2}$ you get ${1\over 2}e^{x\over 2}$ so the $2e^{x\over 2}$ you found is indeed a primitive of $e^{x\over 2}$.

You state you get the same function. Should n't be the case. If $e^y$ is a primitive of $e^y$ then the substitution $y = {x\over 2}$ and the chain rule give $$ {d \; e^y \over dx} = {d\; e^y \over dy} {dy\over dx} = {d \; e^y \over dy} \; {1\over 2} = {1\over 2} \; e^y \;$$

(but maybe I have no idea what u-subbing is ...)

 

[Ray Vickson]

Homework Statement

I am working on an integration by parts problem, and in order to work it I need to figure out the anti-derivative of e^x/2. We've covered basic integration concepts, the definite/indefinite integral, u-sub, and integration by parts. Now, examining the derivative, I expect the anti-derivative to be 2e^x/2, but I can't show my work on how I get here. I tried u-sub and I just get the same function.

What is your problem? You say you used u-substitution and got the same function, which is fine, since your answer is correct!