# Difference of definite integrals inequality

Could you help me how prove that for any $\mathcal{C}^1$ function we have:

$$\left|\int_{a} ^{\frac{a+b}{2}}f(x) d x - \int_{\frac{a+b}{2}} ^bf(x)dx\right| \le \frac{(b-a)^2}{4} \cdot \max _{x \in [a,b]} |f'(x)|.$$

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You were missing an integrand, so I edited. Is this what you meant? –  1015 Jun 6 '13 at 18:54
Yes, this is exactly what I meant. Could you help me with that? –  Hagrid Jun 6 '13 at 18:57
You already got an excellent hint. Note that the inequality is optimal, as $f(x)=x$ achieves the equality. –  1015 Jun 6 '13 at 19:02
A small comment: if you replace $\max$ with $\sup$ in the inequality, then both the statment and the answer below are still valid when $f$ is only differentiable, i.e. the continuity of $f'$ is not necessary. –  23rd Jun 6 '13 at 19:11
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## 1 Answer

Hint: Using mean value theorem for $$\left|\int_{a} ^{\frac{a+b}{2}}f(x) d x - \int_{\frac{a+b}{2}} ^bf(x)dx\right| =\left|\int_{a} ^{\frac{a+b}{2}}\big[f(x)-f(x+\frac{b-a}{2})\big] d x \right|.$$

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Could you explain to me why this equality holds? –  Hagrid Jun 6 '13 at 19:10
@Hagrid $\int_{\frac{a+b}{2}}^b f(x)dx =\int_{a} ^{\frac{a+b}{2}}f(x+\frac{b-a}{2})d x$ –  Ma Ming Jun 6 '13 at 19:13
I see that. Why do we need floor function? –  Hagrid Jun 6 '13 at 19:20
@Hagrid I do not know why it becomes floor, I imputed simply square brackets. –  Ma Ming Jun 6 '13 at 19:24
Aha. Thank you. Is it equal to $|F(\frac{a+b}{2})-F(a) - F(\frac{a+b}{2} - \frac{b-a}{2})+F(a + \frac{b-a}{2})|$?, where $F'(x) = f(x)$ –  Hagrid Jun 6 '13 at 19:36
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