# How do you apply the evaluation theorem to evaluate the integral 3t dt over the interval [0,3]?

Feb 16, 2015

The Evaluation Theorem says:
If $f \left(x\right)$ is a continuous function and
$\frac{d F \left(x\right)}{\mathrm{dx}} = f \left(x\right)$
then
${\int}_{a}^{b} f \left(x\right) \mathrm{dx}$ $= F \left(b\right) - F \left(a\right)$

For $f \left(t\right) = 3 t$,
$\frac{d F \left(t\right)}{\mathrm{dt}} = f \left(t\right)$
when $F \left(t\right) = \left(\frac{3}{2}\right) {t}^{2}$
So
${\int}_{o}^{3} 3 t \mathrm{dt}$

$= F \left(3\right) - F \left(0\right)$

$= \left(\frac{27}{2}\right) - \left(0\right) = 13.5$