Functions Defined by Integrals

Question

Let F(x) = Int[0,2x] tan(t^2)dt. Use your calculator to find F"(1).

Answer 1

F(x) = ∫_{(from 0 to 2x)} tan(t²) dt

 

By the Fundamental Theorem of Calculus, F'(x) = tan((2x)²)(2x)'

 

                                                                     = 2tan(4x²)

 

So, F"(x) = 2sec²(4x²)(8x) = 16xsec²(4x²)

 

Therefore, F"(1) = 16sec²(4) ≈ 37.449