Second fundamental Thm: F(x) = int f(t)dt on interval [1, x]

[bobers]

F(x)=int f(t)dt on interval [1,x]
f(t)=int (((1+u^4)^1/2)/u)du on interval [1,t^2]

i need to find the expression for F'(x) and F"(x)

 

[fasteddie65]

Re: Second fundamental Theorem of Integral

F(x)=int f(t)dt on interval [1,x]
f(t)=int (((1+u^4)^1/2)/u)du on interval [1,t^2]

i need to find the expression for F'(x) and F"(x)

If F(x) = int f(t) dt on interval [1,x], then F'(x) = f(x), and F"(x) = f'(x).

If f(t) = int (((1 + u^4)^1/2)/u) du on interval [1,t^2], then f'(x) = (1 + t^8)^(1/2)/t^2 • 2t = 2(1 + t^8)^(1/2)/t

Does that help?
You have to use the Chain Rule, because the upper limit of integration is not just t, it is t^2.

I hope I understood your question.