Calculus
posted by ashley .
Can someone check my work on this problem? I'm studying for the AP test and I would really appreciate any help.
Let f(x) = the definite integral from [2,(x^23x)] of e^(t^2) dt
At what value of x is f(x) a minimum?
a. For no value of x
b. 1/2
c. 3/2
d. 2
e. 3
The answer I got was 3/2. Here's my process:
h'(x)=e^(x^23x)^2dv/dxe^(2^2)du/dx
=(2x3)e^(x^23x)^20*e^(2^2)
(2x3)e^(x^23x)^2=0 to find the critical point which would be at 3/2.
Thank you!!

your logic and your answer are correct.
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