I need to prove that $$\int^1_0 f(x)dx=f(0)+\frac{1}{2}f'(a)$$ where $f$ is continuous and differentiable in $[0,1]$ and $a\in(0,1)$.

I know that as $f$ is continuous in $[0,1]$ there exists $c\in (0,1)$ such that $\int^1_0 f(x)dx=f(c)$. Now maybe I should show that $f(c)=f(0)+\frac{1}{2}f'(a)$ ? If I should I don't know how to do it.
The other approach that I had, was going by the definition of $\int^1_0 f(x)dx$ to see if I could somehow get a similar result to the right side of the equation above, it didn't seem to lead anywhere.
Do you have any idea how to approach it?


First define $F = \int_0^x f(t) dt$, so now we can rewrite the question as

$F(1) - F(0) = F'(0) + \frac{1}{2}F''(a)$, or if we move over the $F(0)$ term,

$F(1) = F(0) + F'(0) + \frac{1}{2}F''(a)$.

which is a special case of the second degree MacLaurin polynomial with Lagrange error term

$F(x) = F(0) + xF'(0) + x^2\frac{1}{2}F''(a)$

Which is equivalent to the above expression when $x=1$. This is a case of the Lagrange remainder theorem.

So by citing the theorem, we are done.

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    $\begingroup$ The hyperlink is broken. $\endgroup$ – user499203 Nov 20 '17 at 2:42

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