How do you use part I of the Fundamental Theorem of Calculus to find the derivative of: #f(x)= int(((1/3)t^2)-1)^7dt# from x to 3?

1 Answer
Apr 6, 2018

Answer:

# f'(x) = - ( 1/3x^2-1)^7 #

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:

# f'(x) # where #f(x) = int_x^3 ( 1/3t^2-1)^7 \ dt #

ie, we want:

# f'(x) = d/dx int_x^3 ( 1/3t^2-1)^7 \ dt #

We must change the limits of integration to get the function into the correct form:

# f'(x) = d/dx (-int_3^x ( 1/3t^2-1)^7 \ dt )#
# \ \ \ \ \ \ \ \ \ = -d/dx int_3^x ( 1/3t^2-1)^7 \ dt #

And so we can directly apply the FTOC, giving:

# f'(x) = - ( 1/3x^2-1)^7 #