Question

How do you find `\int _ { 0} ^ { 3} ( x ^ { 2} + x + 1) d x`?

Answer 1

`16.5`

Explanation:

Let's take apart the integral.

`int (x^2+x+1) dx = intx^2 dx + int x dx + int 1 dx`
`=x^3/3+x^2/2+x+C`

Then, let's substitute the upper and lower bounds back in.

`3^3/3+3^2/2+3-0`
`=9+9/2+3=16.5`

Answer 2

`int_0^3 \ x^2+x+1 \ dx = 33/2`

Explanation:

Integrating each term, we have:

`int_0^3 \ x^2+x+1 \ dx = [x^3/3+x^2/2+x]_0^3`
`" " = (9+9/2+3) - (0)`2
`" " = 33/2`