Use the fundamental theroem of calculus to find
d/dx integral from (-3x)^(2x) e^t^2 dt
(-3x) is suppose to be on the bottom and (2x) is suppose to be on the top for the integral part.
Also, please split the integral to two integrals of which will have only one limit that varies. And then use the chain rule and apply the Theorem of Calculus.
We haven't learned much about the fundamental therem of calculus yet, so don't expand on it too much.
You should use the following substitution such that:
`t^2 = u => 2t dt = du => dt = (du)/2sqrtu`
You need to change the variable such that:
`int e^(t^2) dt = int e^u*(1/(2sqrtu)) du`
You should use integration by parts using the following formula, such that:
`intf'*g =fg - int fg'`
`f = e^u =>f' = e^u du`
`g' = (1/(2sqrtu)) du => g = sqrt u`
`int e^u*(1/(2sqrtu)) du = e^u sqrt u - int e^u*sqrt u du`
Substituting back `t^2` for u yields:
`int_(-3x)^(2x) e^(t^2) dt = e^(t^2)*t|_(-3x)^(2x) - int_(-3x)^(2x) e^(t^2)*t dt`
If you continue to solve `int_(-3x)^(2x) e^(t^2)*t dt ` using parts, since there are no other methods indicated in this case, you will enter in an infinite loop, hence, you will need to use the imaginary error function.
Hence, evaluating the given integral yields `int_(-3x)^(2x) e^(t^2) dt = e^(t^2)*t|_(-3x)^(2x) - int_(-3x)^(2x) e^(t^2)*t dt.`