# use the fundamental Theorem of Calculus to find `d/(dx) int_(-3x)^(2x) e^(t^2) dt`

### 1 Answer | Add Yours

We want `d/(dx) int_(-3x)^(2x) e^(t^2) dt`

Now `(int_a^b e^(t^2) dt)^2 = (int_a^b e^(x^2) dx)(int_a^b e^(y^2) dy) = int_0^((b-a)^2) e^(x^2+y^2) dxdy `

Converting to polar coordinates

`(int_a^b e^(t^2) dt)^2 = int_0^(2pi) theta d theta int_0^((b-a)^2) re^(r^2) dr = int_0^(2pi)(d theta)/2e^(r^2)|_0^((b-a)^2) = (theta/2e^(r^2)|_0^((b-a)^2))|_0^(2pi)`

`= pi e^(r^2)|_0^((b-a)^2) = pie^((b-a)^4)`

Therefore `d/(dx) int_(-3x)^(2x) e^(t^2) dt = d/(dx) sqrt(pi)e^(1/2(5x)^4) = (2)(5^4)x^3sqrt(pi)e^(1/2(5x)^4)`

**answer**

**Sources:**