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

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Answer 1 · 2015-02-16T20:46:04

The Evaluation Theorem says:
If $f(x)$ is a continuous function and
$(d F(x))/dx = f(x)$
then
$int_a^b f(x) dx$ $= F(b) - F(a)$

For $f(t) = 3t$ ,
$(d F(t))/dt = f(t)$
when $F(t) = (3/2) t^2$
So
$int_o^3 3t dt$

$= F(3) - F(0)$

$= (27/2) - (0) = 13.5$

How do you apply the evaluation theorem to evaluate the integral 3t dt3t dt over the interval [0,3]?