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Let $f(x)$ be defined and continuous on an interval $[\frac{\pi}{2}, \pi]$. Using the mean value theorem of integrals prove or disprove the following.

$$\int_{\frac{\pi}{2}}^{\pi} f(x)\cos x dx= - \int_{\frac{\pi}{2}}^{\pi} f(x)dx$$

The mean value theorem states that if $f(x)$ and $g(x)$ are integrable on an interval $[a,b]$, $g(x) \leq 0 \lor g(x) \geq 0$ $(\forall x \in [a,b])$ and $\exists m \in [a, b], m=\inf_{x \in [a, b]}(f(x))$ and $\exists M \in [a, b], M=\sup_{x \in [a, b]}(f(x))$ then $\exists f(c) \in [m, M]$ such that $\int_{a}^{b} f(x)g(x) dx = f(c)\int_{a}^{b} g(x) dx$.

For the given functions i don't have the information for the sign of $f(x)$ on a given interval so that must be the function which is originally $f(x)$ in the theorem and $\cos x$ holds as it is $\cos x \leq 0, \forall x \in [\frac{\pi}{2}, \pi]$. $f(x)$ is continuous which means it's integrable which means it's bounded and has it's min and max values on the interval but the expression doesn't hold as the function that remains in the integral must be the one defined as $g(x)$ in the theorem.

Is this enough to disprove it?

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  • $\begingroup$ No, all you've shown is that a particular attempt at proving it with MVT doesn't work. $\endgroup$ – saulspatz Jun 26 '18 at 15:44
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You cannot neither prove nor disprove it. The equality holds for some functions, like $f(x)=0$, and does not hold for others, like $f(x)=1$. More generally, the equality is equivalent to $$ \int_{\pi/2}^\pi f(x)(1+\cos x)\,dx=0. $$ Since $1+\cos x\ge0$, a necessary condition is that $f(x)=0$ for some $x\in(\pi/2,\pi)$.

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  • $\begingroup$ +1 but I think the question intends to ask, "Is this statement true for all such $f$," so you have disproved it. $\endgroup$ – saulspatz Jun 26 '18 at 16:06

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