Question

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

Answer 1

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`