Let f(x) = integral from (-2) to (x^2-3x) of e^(t^2) dt. At what value of x is f(x) a minimum?
Notice, because of our excellent continuity, we have:
dt/dx = 2x-3
and
d/dx = d/dt * dt/dx
and we have:
df/dt = e^(t^2)
therefore:
df/dx = d(∫_{(-2)}^{(x}^{^2-3x)} df/dt dt ) / dx
= {d[∫(df/ dt ) dt] / dt |_{(-2)}^{(x^2-3x)} }* dt/dx
But the derivative of the integral by the same variable is the original thing, as long as we are careful accounting for taking the derivative of the constant term t = -2. This is just:
= {(df/ dt )|^{(x^2-3x)} }* dt/dx
substituting and evaluating we get:
= {(e^(t^2) )|t= (x^2-3x) } * (2x-3)
= e^((x^2-3x)^2) * (2x-3)
this has a critical point precisely when x = 3/2. If x < 3/2 this expression is negative. If x > 3/2 this expression is positive, so we have a minimum.
Sometimes doing the computations just gets in the way.