Evaluate the definite Integral

Question

Evaluate the definite Integral

Answer 1

2x - 1 ≤ 0 when x ≤ 1/2 and 2x - 1 ≥ 0 when x ≥ 1/2.

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

Answer 2

The graph of y = |2x - 1| can be thought of as a piecewise function. When (2x - 1) = 0, there is a dividing line between what happens in the function rule. When (2x - 1) < 0, we drop the negative or, another way to think of it, we make the function -(2x - 1). To the right of the point where (2x - 1) = 0, we just think of the function as y = 2x - 1. The point where (2x - 1) = 0 is x = 1/2

So the absolute value integral ₀∫² | 2x - 1 | dx = ₀∫^1/2 -(2x - 1) dx + _1/2∫² (2x - 1) dx

₀∫^1/2 -(2x - 1) dx = - ₀∫^1/2 (2x - 1) dx = _1/2∫⁰ (2x - 1) dx = x² - x evaluated between 1/2 and zero = (0² - 0) - ((1/2)² - 1/2) = 0 - (1/4 - 1/2) = -(-1/4) = 1/4

_1/2∫² (2x - 1) dx = x² - x evaluated between 2 and 1/2 = (2² - 2) - ((1/2)² - 1/2) = (2) - (-1/4) = 2 + 1/4 = 9/4

So the entire integral ₀∫² | 2x - 1 | dx = 1/4 + 9/4 = 10/4 = 5/2