Prove that for every $b>a$ there exist $c\in[a,b]$ such as: $\int_a^b f(t)\,dt=\frac{1}{2}f'(c)(b-a)^2$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a derivative function such as $f(a)=0$ for some $a\in\mathbb{R}$.

Prove that for every $b>a$ there exist $c\in[a,b]$ such as:

$$\int_a^b f(t) \, dt=\frac{1}{2}f'(c)(b-a)^2$$

My attempt:

I know there exist $d\in[a,b]$ such as: $\int_a^b f(t)\,dt=f(d)(b-a)$, and using MVT there exist $e\in[a,b]$ such as $\frac{f(b)-f(a)}{b-a}=\frac{f(b)}{b-a} = f'(e)$. But I am stuck here.

Answer 1

Note that $$F(x)=F(a)+F'(a)(x-a)+ \frac {1}{2}F''(c)(x-a)^2$$

Let $$F(x)= \int_{a}^{x}f(t)dt$$

$$ \int_{a}^{b}f(t)dt = f(a)(b-a) + \frac {1}{2}f'(c)(b-a)^2 $$

since $f(a)=0,$ we get $$ \int_{a}^{b}f(t)dt = \frac {1}{2}f'(c)(b-a)^2 $$

Answer 2

Hint. Apply Taylor's theorem to $F(x) = \int_{a}^{x}f(t)dt$.