Question

How do you find the definite integral for: `(3x^2+2x+3) dx` for the intervals `[0, 6]`?

Answer 1

`int_0^6(3x^2+2x+3)d x=?`

`int_0^6(3x^2+2x+3)d x=|3*x^3/3+2*x^2/2+3x|_0^6`

`int_0^6(3x^2+2x+3)d x=|x^3+x^2+3x|_0^6`

`int_0^6(3x^2+2x+3)d x=[(6^3+6^2+3*6)-(0^3+0^2+3*0)]`

`int_0^6(3x^2+2x+3)d x=[(216+36+18)-(0)]`

`int_0^6(3x^2+2x+3)d x=270`