B bobers New member Joined Nov 29, 2008 Messages 10 Mar 23, 2009 #1 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)
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)
F fasteddie65 Full Member Joined Nov 1, 2008 Messages 360 Mar 23, 2009 #2 Re: Second fundamental Theorem of Integral bobers said: 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) Click to expand... 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.
Re: Second fundamental Theorem of Integral bobers said: 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) Click to expand... 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.
