`int_0^2(y - 1)(2y + 1)dy` Evaluate the integral.

`int_0^2 (y-1)(2y+1)dy`

Before evaluating, expand the integrand.

`=int_0^2 (2y^2+y-2y-1)dy`

`=int_0^2(2y^2-y-1)dy`

Then, apply the integral formulas `int x^n dx=x^(n+1)/(n+1)` and `int cdx = cx` .

`=((2y^3)/3-y^2/2-y)|_0^2`

Then, plug-in the limits of  the integral to the resulting function.

`= ((2(2)^3)/3-2^2/2-2)-((2(0)^3)/3-0^2/2-0)`

`=(16/3-4/2-2)-0`

`=16/3-2-2`

`=16/3-4`

`=16/3-12/3`

`=4/3`

Therefore, `int_0^2 (y-1)(2y+1)dy = 4/3` .

`int_0^2 (y-1)(2y+1)dy`

Before evaluating, expand the integrand.

`=int_0^2 (2y^2+y-2y-1)dy`

`=int_0^2(2y^2-y-1)dy`

Then, apply the integral formulas `int x^n dx=x^(n+1)/(n+1)` and `int cdx = cx` .

`=((2y^3)/3-y^2/2-y)|_0^2`

Then, plug-in the limits of  the integral to the resulting function.

`= ((2(2)^3)/3-2^2/2-2)-((2(0)^3)/3-0^2/2-0)`

`=(16/3-4/2-2)-0`

`=16/3-2-2`

`=16/3-4`

`=16/3-12/3`

`=4/3`

Therefore, `int_0^2 (y-1)(2y+1)dy = 4/3` .