How do you find the derivative of #G(x)=int (e^(t^2))dt# from #[1,x]#?

2 Answers
Jun 7, 2017

You should get #e^(x^2)#.


You can use Leibniz's integral rule.

#(dG(x))/(dx) -= d/(dx) [int_(g(x))^(h(x)) f(x,t)dt]#

#= int_(g(x))^(h(x)) (delf)/(delx)dt + f(x,h(x))(dh)/(dx) - f(x,g(x))(dg)/(dx)#

In this case, #g(x) = 1# and #h(x) = x#, so:

  • #(dg)/(dx) = 0#
  • #(dh)/(dx) = 1#
  • #f(x,h(x)) = f(x,t = x) = e^(x^2)#
  • #f(x,g(x)) = f(x,t = 1) = e^(1^2) = e#
  • #(delf)/(delx) = (del)/(delx)[e^(t^2)]_x = 0#

Therefore:

#color(blue)((dG)/(dx)) = cancel(int_(1)^(x) 0dt)^(0) + e^(x^2)cdot1 - cancel(ecdot0)#

#= color(blue)(e^(x^2))#

Jun 7, 2017

Answer:

# d/dx \ int_1^x e^(t^2) \ dt = e^(x^2) #

Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.

The FTOC tells us that:

# d/dx \ int_a^x \ f(t) \ dt = f(x) # for any constant #a#

(ie the derivative of an integral gives us the original function back).

We are asked to find:

# d/dx \ int_1^x e^(t^2) \ dt #

And this integral is in the correct form for the FTOC to be applied, giving:

# d/dx \ int_1^x e^(t^2) \ dt = e^(x^2) #