Relative Sizes of Definite Integrals

Theorem
Let $$f$$ and $$g$$ be real functions which are continuous on the closed interval $$\left[{a \,. \, . \, b}\right]$$.

Suppose that $$\forall t \in \left[{a \,. \, . \, b}\right]: f \left({t}\right) \le g \left({t}\right)$$.

Then $$\int_a^b f \left({t}\right) dt \le \int_a^b g \left({t}\right) dt$$.

Proof
Let $$H$$ be a primitive of $$g - f$$ on $$\left[{a \,. \, . \, b}\right]$$.

Then $$\forall t \in \left[{a \,. \, . \, b}\right]: D H \left({t}\right) = g \left({t}\right) - f \left({t}\right) \ge 0$$.

By Derivative of Monotone Function, it follows that $$H$$ is increasing on $$\left[{a \,. \, . \, b}\right]$$.

Thus $$H \left({b}\right) \ge H \left({a}\right)$$.

Hence $$\int_a^b \left({g \left({t}\right) - f \left({t}\right)}\right) dt = H \left({b}\right) - H \left({a}\right) \ge 0$$.

The result follows.