Find F'(x)?

Question

If F(x)=∫ sqrt(1+t^3)dt from 1 to x^2, find F'(x).

 

Answer: 2x*sqrt(1+x^6)

 

Please show all your work.

Answer 1

The Second Fundamental Theorem of Calculus states:

 

d/dx (∫{a,x} f(t) dt) = f(x)

 

Using the chain rule:

 

d/dx (∫{a,u(x)} f(t) dt) = (f(x))(du/dx)

 

So if F(x) = ∫{1,x^2} √(1+t^3) dt, then

 

F'(x) = (√(1+t^3))(2x) = 2x√(1+t^3)

Answer 2

Use chain rule

 

                   F'(x) = dF(x)/dx = (dF(x)/dx²)(dx²/dx) = 2x (dF(x)/dx²)  (1)

 

When you differentiate the integral with variable upper limit it gives you the function being integrated.

 

Thus (because the lower limit doesn't play any role being a constant number)

 

              dF(x)/dx² = √(1 + t³)  where t = x^{2                         (2)}

 

If you combine (1) and (2) you will obtain your answer:

 

                       F'(x) = 2x√(1+x⁶)  

 

Comment. There is no need to integrate and then differentiate by x.

 

If you want to type x⁶ instead of x^6 simply type x6, highlight only 6 and click on the icon (x²).

 

Good luck!

 

 

 

Answer 3

                 X^2                                         X^2

   F( X) =  ∫     √( 1 + t^3) dt = √ (1 + t^3 ) du/dx       = 2X √(1+ X^6) - ( 1- 1) = 2X √( 1 + X^6)

               1                                              1

 

    Note: Integration and differentiation are inverse of each other.