Problem 15 - Essential Calculus

Question

Evaluate the definite integral of the algebraic function. Use a graphing utility to verify your result. $$ \int_{0}^{3}|2 x-3| d x $$

Step-by-Step Solution

Answer

The value of the given integral is 0.

Step 1: Identify the point of discontinuity

The function inside the absolute value is linear, so the point of discontinuity is the zero of \(2x - 3\). Solving for x, we get \(x = 3/2\) or 1.5.

Step 2: Split the integral at the point of discontinuity

The integral should be split into two parts, from 0 to 1.5 and from 1.5 to 3. The absolute value function changes its behavior at those points, becoming \(2x - 3\) when \(x >= 1.5\) and \(-(2x - 3)\) when \(x < 1.5\). Thus, the integral becomes \(\int_{0}^{1.5}-(2x -3) dx + \int_{1.5}^{3}(2x -3) dx\).

Step 3: Evaluate each integral separately

Calculate the integral \(\int_{0}^{1.5}-(2x -3) dx\), which is \(-[x^2 - 3x]_{0}^{1.5} = -(1.5^2 - 3*1.5 - 0) = -0.75\). Similarly, calculate the integral \(\int_{1.5}^{3}(2x -3) dx\), resulting in \([x^2 - 3x]_{1.5}^{3} = (3^2-3*3) - (1.5^2-3*1.5) = 0.75\).

Step 4: Add the results

Adding the results from the two separate integrals gives the final answer \(-0.75 + 0.75 = 0\).