SOLUTION: The answer that I have come up with is: (e -1).2: The problem is what are the exact evaluation of, (integral sign), 2xe^(x^2) dx over the intervals (0, 1). I feel this is

Algebra ->  Rational-functions -> SOLUTION: The answer that I have come up with is: (e -1).2: The problem is what are the exact evaluation of, (integral sign), 2xe^(x^2) dx over the intervals (0, 1). I feel this is      Log On


   

Question 127249: The answer that I have come up with is: (e -1).2:
The problem is what are the exact evaluation of, (integral sign), 2xe^(x^2) dx over the intervals (0, 1). I feel this is not correct, can you verify that for me please?
Also please, I have determined that In x is not the correct completion of the formula, (integral sign), (1/x). Is that correct, or should I do more research on this subject?
Thank you all, for your God given talent that helps us less fortunate.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
int%28+2xe%5E%28x%5E2%29%2C+dx%2C+0%2C+1+%29 Start with the given integral

Let u=x%5E2. So this means that du=2xdx. So the x%5E2 gets replaced with u and the 2xdx (you have to group these together) gets replaced with du

Also, since the first endpoint is 0, this means that x=0. But we want this in terms of u. So just plug in x=0 into u=x%5E2 to get u=0. Do the same thing with the other endpoint to get u=1

So we now have:
int%28+e%5Eu%2C+du%2C+0%2C+1+%29


So we now have:
e%5Eu%2BC Take the integral of e%5Eu to get e%5Eu. Remember int%28+e%5Eu%2C+du%29=e%5Eu%2BC


Now let's evaluate the endpoints


%28e%5E1%2BC%29-%28e%5E0%2BC%29 Plug in the endpoints and subtract


e-1 Combine like terms and simplify



So int%28+2xe%5E%28x%5E2%29%2C+dx%2C+0%2C+1+%29=e-1



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Are you trying to evaluate int%28+1%2Fx%2C+dx%29 ?


The integral of 1%2Fx is ln%28x%29%2BC. In other words, int%28+1%2Fx%2C+dx%29=ln%28x%29%2BC