How to Evaluate the Integral of a Derivative at a Given Point

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Samuelb88
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Evaluate the integral... ?:(

Homework Statement


Suppose that the function f and g and their derivatives have the following values at x=0, x=1.

[tex]f(0)=1, f(1)=3[/tex]
[tex]f'(0)=5, f'(1)=\frac{1}{3}\right)[/tex]
[tex]g(0)=1, g(1)=-4[/tex]
[tex]g'(0)=\frac{1}{3}\right),g'(1)=-\frac{8}{3}\right)[/tex]

Evaluate the integral:

[tex]\frac{d}{dx}\right)(f^(^3^)(x^1^/^2))|[/tex] x=1

Homework Equations





The Attempt at a Solution


f(1)=g(1)

I know how to evaluate definite integrals and indefinate too, but i don't understand what it means by "evaluate the integral" in the question? I only see a derivative.

To my understanding...

[tex]\frac{d}{dx}\right)(f^(^3^)(x^1^/^2))|[/tex] x=1 ...
[tex]= f^(^4^)(x^1^/^2)[/tex]

So...
[tex]\frac{d}{dx}\right)(f^(^3^)(x^1^/^2))=\frac{d^4y}{dx^4}\right)[/tex]

And multiplying the differential dx and integrating the integrand [tex]f^(^4^)[/tex] will give you [tex]f^(^3^)(x^1^/^2)[/tex] so somehow I am suppose to integrate [tex]f^(^3^)(x^1^/^2)[/tex] until I get to f`(x) which should equal 5 or -1/3?

I honestly don't understand the how to even start what the question is asking me. I just transferred to a different school and the professor already taught basic definite and indefinite integration in calculus I which I never learned in my calculus I class.

?:|
 
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What does g have to do with it? Are you sure you have stated the problem correctly?
 


Yes, I just read the question again and it's word for word. It's has a part A and B, but part b has no function g either. *shrug*