Question

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

Answer 1

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

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

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

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

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

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