What is the correct derivative of the integral S(from -x to x) e^(-(t)^2) dt?

[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)

 

[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?

 

[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.

 

[Curious3141]

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?

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

 

[d_leet]

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

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.

 

[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.

 

[d_leet]

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.

And you rightfully shouldn't. Your method and answer were correct.