Does the converse of the following theorem hold?

Question

If $f (x) \leq g(x)$ for all $x \in [a, b]$, then $$ \int_a^b f (x) \,dx \leq \int_a^b g(x)\, dx.$$

Any suggestions and solution will be appreciated.

Answer 1

The converse statement is false in general. For example, let $f(x) = x$, $g(x) = \dfrac{3}{4}$. Then

$$\int\limits_0^1 f(x) = \dfrac{1}{2} < g(x)$$ but $f(1) > g(1)$.

Geometrically this means that the area under $f(x)$ is less than the area under $g(x)$ in $[0, 1]$, but there obviously exists a point, where $f(x)$ is greater than $g(x)$ in this interval.

Answer 2

If you mean the converse, then it is false:

Take $f(x)=\sin x$ and $g(x)=\cos x$ and the interval $[0,\frac{\pi}{2}]$. Then $$\int_0^{\frac{\pi}{2}} \sin x \, dx=1=\int_0^{\frac{\pi}{2}}\cos x \, dx$$

But $\sin \frac \pi 2 =1 \gt \cos \frac \pi 2 =0$