Evaluate the definite integral from -2 to 3 of the absolute value of x^2-x

Question

Definite Integral

Answer 1

x² - x = x(x-1) = 0 when x = 0 or x = 1.

The graph of y = x² - x is a parabola opening downward with x-intercepts at x = 0 and x = 1.

Between x = 0 and x = 1, the graph is above the x-axis. So, x²-x > 0. l x²-x l = x² - x

When x < 0 or x > 1, the graph is below the x-axis. So, x² - x < 0. l x²-x l = -(x² - x) = x - x²

∫_{(-2 to 3)} l x² - x l dx = ∫_{(-2 to 0)} (x - x²)dx + ∫_{(0 to 1)} (x² - x)dx + ∫_{(1 to 3)} (x - x²)dx

Evaluate the integrals.