# Find (d/(dx)) ʃ(superscript 2x)(subscript -3x) (e^(t^2)) dt.

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Find `d/(dx)[int_(-3x)^(2x)e^(t^2)dt]`

We have [-3x,2x] a closed interval such that `F(x)=int_(-3x)^(2x)e^(t^2)dt` is continuous. Then `F(x)` is differentiable on (-3x,2x) and `F'(x)=f(x)` for all `x in(-3x,2x)` by the First Fundamental Theorem of Calculus.

Further, by the Second Fundamental Theorem of Calculus we know that `int_a^bf'(x)=F'(b)-F'(a)`

** Note that we must use the chain rule to find `F'(a)` **

So `int_(-3x)^(2x)e^(t^2)dt=e^((2x)^2)(2)-e^((-3x)^2)(-3)`

`=2e^(4x^2)+3e^(4x^2)e^(5x^2)`

`=e^(4x^2)(2+3e^(5x^2))`

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`int_(-3x)^(2x)e^(t^2)dt=e^(4x^2)(2+3e^(5x^2))`

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