Use part I of the Fundamental Theorem of Calculus to find the derivative of F ( x ) = ∫ 5 x sin ( t 5 ) d t

Question

Use part I of the Fundamental Theorem of Calculus to find the derivative of 

The lower limit is x and the upper is 5.

I am not sure how to do this because I though the answer was cosx^5 which was wrong so I changed my answer to -cosx^5 which was also wrong...

Answer 1

The FTC says that if F(x) = a∫x f(t) dt then F'(x) = f(t) a is a constant

So in your case F(x) = x ∫ 5 sin (t^5) dt. The x is the lower integral limit and to apply the FTC we need to switch it to the upper one.

a ∫ x ..... = - x∫ a so F(x) = x ∫ 5 sin (t^5) dt = - 5 ∫ x sin (t^5) dt

Now we can apply the FTC, giving F'(x) = - sin (x^5)

Mike