# Derivative of integral

1. Jan 27, 2007

### asdifnlg

S will be my integral sign. Find the derivative of this:

S(from -x to x) e^(-(t)^2) dt

I found it to be 2e^(-x^2)
My teacher says it is 4xe^(-x^2), or maybe a negative in front of the 4 (I forgot), which is also what the math book she got it out of says.

I don't really agree with this solution.. Which is right? (I graphed it and it looked to be what I got for an answer, but I may have done something wrong) I got it counted wrong for a test and I tried to explain my way, but my teacher said it wasn't the right way :[ Another thing I thought was wrong with the books answer is that the integral is always increasing.. so wouldn't the derivative always be positive? (which would count the book's answer wrong because for a negative number for x, it says the function would be decreasing at that point)

2. Jan 27, 2007

### d_leet

If the problem is exactly as you have stated it then you are correct, however if it were x2 in place of both "x"s then your teacher would be correct. So are you sure you copied the problem correctly?

3. Jan 27, 2007

### Curious3141

I think your solution is correct, based on the fundamental theorem of calculus. Using the symmetry of the function, note that :

$$\int_0^x{e^{-t^2}}dt = \int_{-x}^0{e^{-t^2}}dt$$

So essentially you're evaluating,

$$2\frac{d}{dx}\int_0^x{e^{-t^2}}dt$$

which comes to $$2e^{-x^2}$$ by the F.T.C.

4. Jan 27, 2007

### Curious3141

No. If the bounds were positive and neg. x^2, the answer would be $$4xe^{-x^4}$$.

5. Jan 27, 2007

### d_leet

Yes you're right, I'm sorry about that, I forgot about the x2 in the integral for some reason in an attempt to slightly justify the teacher's answer.

6. Jan 28, 2007

### asdifnlg

The way I wrote it is the exact way the problem was worded on the test. I thought my way was right, and the teacher said that it would equal, like curious said, 2 S(0 to x) e^(-t^2), but then he said to do the derivative of that, you would need to plug in the x for t, then take the derivative of the e^-x^2, or something like that, which I didn't really agree with.

7. Jan 28, 2007