Question

Can anyone help me with this question?
it's additional mathematics (Integration)​

Answer 1

`$A=-\int\limits_0^1 {(x-1)(x+1)(3-x)} \, dx +\int\limits_1^3 {(x-1)(x+1)(3-x)} \, dx=`

`$=-\int\limits_0^1 {(x^2-1)(3-x)} \, dx +\int\limits_1^3 {(x^2-1)(3-x)} \, dx=`

`$=-\int\limits_0^1 {(3x^2-x^3-3+x)} \, dx +\int\limits_1^3 {(3x^2-x^3-3+x)} \, dx=`

`$=\int\limits_0^1 {(x^3-3x^2-x+3)} \, dx -\int\limits_1^3 {(x^3-3x^2-x+3)} \, dx=`

`=\Bigg[(x^4)/(4)-3(x^3)/(3)-(x^2)/(2)+3x\Bigg]_0^1-\Bigg[(x^4)/(4)-3(x^3)/(3)-(x^2)/(2)+3x\Bigg]_1^3=\\\\\\=\left((1)/(4)-3(1)/(3)-(1)/(2)+3\right)-\left[\left((3^4)/(4)-3(3^3)/(3)-(3^2)/(2)+3\cdot3\right)-\left((1)/(4)-3(1)/(3)-(1)/(2)+3\right)\right]=\\\\\\=\left((1)/(4)-1-(2)/(4)+3\right)-\left[\left((81)/(4)-27-(18)/(4)+9\right)-\left((1)/(4)-1-(2)/(4)+3\right)\right]=`

`=1(3)/(4)-\left(-2(1)/(4)-1(3)/(4)\right)=1(3)/(4)-\left(-4\right)=1(3)/(4)+4=5(3)/(4)=\boxed{5.75}`