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Say we have a function $f:[a,b] \to \mathbb{R}$ and we assume that this function is Riemann integrable. This implies that $f$ is bounded, say by $M$. Then $|f| \le M$.

Now consider $F:[a,b] \to \mathbb{R}$ where $F(x) = \int_{a}^{x}f(t)dt$. Because $f$ is bounded by $M$ do we have that $F(x)$ is bounded by $M(x-a)$? I feel this result might not be true, but there is a similar result for bounding an integral, does anyone have any ideas?

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Let $|f| \le M$ then

$$ \int_a^x |f(y)|dy \le \int_a^x Mdy = M\int_a^x dy = M(x-a)$$

and $$\left \lvert \int_a^x f(y)dy \right \rvert \le \int_a^x |f(y)|dy.$$

So what can we conclude?

Remark: You have written $\int_a^xf(x)dx$ but I am taking the assumption you didn't mean to have the same variable as your upper limit of integration as well as the variable you are integrating with respect to.

oliverjones
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  • I've edited the variable to be $t$ rather than $x$, thank you for pointing this out. We can conclude that $|\int_{a}^{x}f(t)dt| \le M(x-a)$ so the integral is bounded by $M(x-a)$. Is this correct? –  Mar 27 '22 at 22:27
  • Yes that is correct – oliverjones Mar 27 '22 at 22:28
  • Thank you for the help. Is this true in general? I.e. we require no other conditions other than Riemann integrability? –  Mar 27 '22 at 22:49
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    Yes that is correct for Riemann integrable functions. Cf. https://math.stackexchange.com/a/610088/191011 for more information – oliverjones Mar 27 '22 at 22:56
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    @HungaryGoose The theorem that you really need to learn and study and apply is this one which is more general and which is for sure in your text: If $f,g:[a,b]\to\mathbb R$ are both Riemann integrable and if $f(x)\leq g(x)$ at every point, then $\int_a^bf(t),dt\leq \int_a^b g(t),dt$. Learn how to prove it too. – B. S. Thomson Mar 27 '22 at 23:06
  • @B.S.Thomson I wasn't aware of this, thank you this is very useful. –  Mar 27 '22 at 23:45
  • Since f is Riemann integrable, F is continuous on the compact interval and hence it’s bounded. – Lawrence Mano Mar 27 '22 at 23:51
  • @LawrenceMano True that $F$ is bounded, but the issue here was finding an estimate for that bound, not whether a bound exists. It is also true that every Riemann integable function is bounded, but for this problem we needed to know more, that $|f(x)|\leq M$. – B. S. Thomson Mar 28 '22 at 03:05