I've to show for
$$ f\in C^2[a,b], \qquad x_\gamma = a+\gamma(b-a), \qquad \gamma \in[0,1] $$
that
$$ \bigg|\int_a^b f(x)\,dx - f(x_\gamma)(b-a)\bigg| \le \frac{2\gamma^2 - 2\gamma+1}{2} ||f'||_\infty (b-a)^2. $$
So far I have:
\begin{align*} \bigg| \int_a^b f(x)\,dx - f(x_\gamma)(b-a) \bigg| &= | f(\xi) * (b-a) - f(x_\gamma) * (b-a) | && \xi \in (a, b) \\ &= | f(\xi) - f(x_\gamma)| * (b-a) \end{align*}
Therefore, I would have to show that $$ |f(\xi) - f(x_\gamma)| \le \frac{1}{4} ||f'||_\infty(b-a) $$
since $\frac{1}{4} \leqslant (\gamma-\frac{1}{2})^2 + \frac{1}{4} = \frac{2\gamma^2 - 2\gamma+1}{2}$.
Clearly $|f(\xi) - f(x_\gamma)|$ has to be less than $||f||_\infty(b-a)$ because that's the furthest $f(\xi)$ theoretically can "get away" from $f(x_\gamma)$ on that interval. But that does not explain the factor $\frac{1}{4}$.
Could you explain that to me?
Additionally, every advice for a simpler proving method are welcome as well.
Thanks to Kelenner's hint I was able to solve it:
\begin{align*} \bigg|\int_a^b f(x)\,dx - f(x_\gamma)(b-a)\bigg| &= \bigg| \int_a^b f(x) - f(x_\gamma) \,dx \bigg| \\ &\le \int_a^b |f(x) - f(x_\gamma)| \,dx \\ &\le \int_a^b ||f'||_\infty |x - x_\gamma| \,dx \\ &= ||f'||_\infty \int_a^b |x-x_\gamma| \,dx \\ &= ||f'||_\infty \left( \int_a^{x_\gamma} |x-x_\gamma| \,dx + \int_{x_\gamma}^b |x-x_\gamma| \,dx \right) \\ &= ||f'||_\infty \left( - \int_a^{x_\gamma} x-x_\gamma \,dx + \int_{x_\gamma}^b x-x_\gamma \,dx \right) \\ &= ||f'||_\infty \left( \int_{x_\gamma}^b x-x_\gamma \,dx - \int_a^{x_\gamma} x-x_\gamma \,dx \right) \\ &= ||f'||_\infty \left( \int_{x_\gamma}^b x \,dx -\int_{x_\gamma}^bx_\gamma \,dx - \int_a^{x_\gamma} x \,dx + \int_a^{x_\gamma}x_\gamma \,dx \right) \\ &= ||f'||_\infty \left( \frac{1}{2}b^2 - \frac{1}{2}\gamma^2 - (b-x_\gamma)x_\gamma - \frac{1}{2}\gamma^2 + \frac{1}{2}a^2 + (x_\gamma-a)x_\gamma \right) \\ &= ||f'||_\infty \left( \frac{b^2+a^2}{2} - x_\gamma(b+a) + x_\gamma^2 \right) \\ &= ||f'||_\infty \left( \frac{b^2+a^2}{2} - (a+\gamma(b-a))(b+a) + (a+\gamma(b-a))^2 \right) \\ &= ||f'||_\infty \left( \frac{b^2+a^2}{2} - ab - a^2 -\gamma(b^2 - a^2) + a^2 + 2a\gamma(b-a) + \gamma^2(b-a)^2 \right) \\ &= ||f'||_\infty \left( \frac{b^2+a^2-2ab}{2} - \gamma(b^2 - a^2) + 2a\gamma(b-a) + \gamma^2(b-a)^2 \right) \\ &= ||f'||_\infty \left( \frac{(b-a)^2}{2} - \gamma b^2 + \gamma a^2 + 2a\gamma b- 2\gamma a^2 + \gamma^2(b-a)^2 \right) \\ &= ||f'||_\infty \left( \frac{(b-a)^2}{2} - \gamma (b^2 + a^2 - 2ab) + \gamma^2(b-a)^2 \right) \\ &= ||f'||_\infty \left( \frac{(b-a)^2}{2} - \gamma (b-a)^2 + \gamma^2(b-a)^2 \right) \\ &= \frac{2\gamma^2-2\gamma+1}{2}||f'||_\infty(b-a)^2 \end{align*}