Question

How do you compute the value of `int t^2+3tdt` of [1,4]?

Answer 1

I found: `87/2`

Explanation:

We first inegrate and then apply the limits:
`int(t^2+3t)dt=` separate:
`=intt^2dt+int3tdt=` use standard integration rules:
`=t^3/3+3t^2/2`
let us now use the estremes evaluating our function at `t=4` and `t=1`:

For `t=4`
`4^3/3+3*4^2/2=64/3+24=136/3`
For `t=1`
`1^3/3+3*1^2/2=(2+9)/6=11/6`

Now we need to subtract the values obtained:

`136/3-11/6=(272-11)/6=261/6=87/2`

So:`int_1^4(t^2+3t)dt=87/2`