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Chapter 4: Problem 15

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

Short Answer

Expert verified
The value of the given integral is 0.

Step by step solution

01

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.
02

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 \(\int_{0}^{1.5}-(2x -3) dx + \int_{1.5}^{3}(2x -3) dx\).
03

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 \([x^2 - 3x]_{1.5}^{3} = (3^2-3*3) - (1.5^2-3*1.5) = 0.75\).
04

Add the results

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

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