Question

Find the value of \(\displaystyle\int_{-1}^{2}|x^{3}-x|\,dx\).

Hint:

Find the sign of x^3-x on each subinterval and integrate piecewise.

Answer 1

Step 1: Finding where x^3 - x changes sign:
\(x^{3}-x=x(x-1)(x+1)\), which is zero at \(x=-1,0,1\). On \([-1,0]\): positive test at \(x=-0.5\) gives \((-0.5)^3-(-0.5)=0.375>0\). On \([0,1]\): test \(x=0.5\) gives \(-0.375<0\). On \([1,2]\): test \(x=1.5\) gives \(1.875>0\).

Step 2: Splitting the integral accordingly:
\(\displaystyle\int_{-1}^{2}|x^3-x|dx=\int_{-1}^{0}(x^3-x)dx-\int_{0}^{1}(x^3-x)dx+\int_{1}^{2}(x^3-x)dx\).

Step 3: Using the antiderivative F(x) = x^4/4 - x^2/2:
\(F(0)=0,\ F(-1)=\tfrac14-\tfrac12=-\tfrac14,\ F(1)=\tfrac14-\tfrac12=-\tfrac14,\ F(2)=4-2=2\).

Step 4: Evaluating each piece:
Piece 1: \(F(0)-F(-1)=0-(-\tfrac14)=\tfrac14\). Piece 2: \(-[F(1)-F(0)]=-[-\tfrac14-0]=\tfrac14\). Piece 3: \(F(2)-F(1)=2-(-\tfrac14)=\tfrac94\).

Final Answer:
Total \(=\tfrac14+\tfrac14+\tfrac94=\tfrac{11}{4}\).\[ \boxed{\dfrac{11}{4}} \]