# 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^2-3x)] 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^2-3x)^2dv/dx-e^(-2^2)du/dx
=(2x-3)e^(x^2-3x)^2-0*e^(-2^2)
(2x-3)e^(x^2-3x)^2=0 to find the critical point which would be at 3/2.

Thank you!!

1. Steve

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