Question
$$\int_{ - 2}^2 {\left| {x\left( {x - 1} \right)} \right|dx} $$ is :
A.
$$\frac{{11}}{3}$$
B.
$$\frac{{13}}{3}$$
C.
$$\frac{{16}}{3}$$
D.
$$\frac{{17}}{3}$$
Answer :
$$\frac{{17}}{3}$$
Solution :
$$\eqalign{
& I = \int_{ - 2}^0 {\left| {x\left( {x - 1} \right)} \right|dx} + \int_0^1 {\left| {x\left( {x - 1} \right)} \right|dx} + \int_1^2 {\left| {x\left( {x - 1} \right)} \right|dx} \cr
& \,\,\,\,\, = \int_{ - 2}^0 {x\left( {x - 1} \right)dx} + \int_0^1 { - x\left( {x - 1} \right)dx} + \int_1^2 {x\left( {x - 1} \right)dx} \cr
& \,\,\,\,\, = \left[ {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right]_{ - 2}^0 - \left[ {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right]_0^1 + \left[ {\frac{{{x^3}}}{3} - \frac{{{x^2}}}{2}} \right]_1^2 \cr
& \,\,\,\,\, = 0 - \left( { - \frac{8}{3} - 2} \right) - \left( {\frac{1}{3} - \frac{1}{2}} \right) + \left( {\frac{8}{3} - 2} \right) - \left( {\frac{1}{3} - \frac{1}{2}} \right) \cr
& \,\,\,\,\, = \frac{{17}}{3} \cr} $$