Question

In Exercises \(37-40,\) (a) find the points of discontinuity of the integrand on the interval of integration, and (b) use area to evaluate the integral. $$\int_{-2}^{3} \frac{x}{|x|} d x$$

Short answer

The points of discontinuity are at x=0, and the value of the integral is 5.

Step-by-step solution

Identify the points of discontinuity

The function \(\frac{x}{|x|}\) has a discontinuity at \(x = 0\) because \(|x|\) cannot have values equal to 0. Therefore, the points of discontinuity on the interval of integration [-2, 3] is {0}.

Split the Integral

Since the function exhibits a discontinuity at x=0, split the original integral at this point. Thus, the integral becomes the sum of two integrals: \(\int_{-2}^{0}\frac{x}{|x|}dx+\int_{0}^{3}\frac{x}{|x|}dx\). Note that for \(x < 0\), \(\frac{x}{|x|} = -1\), and for \(x > 0\), \(\frac{x}{|x|} = 1\). Therefore, the integrals can be simplified to \(-\int_{-2}^{0} dx+\int_{0}^{3} dx\).

Compute the integral using area

The integrals can be evaluated using geometric principles because the integrands are constants. The area of the rectangles formed on the intervals are simply the width times the height. For the first integral, the area under the curve from -2 to 0 is \(0--2\), an area of 2. For the second integral, the area under the curve from 0 to 3 is \(3-0\), an area of 3. Summing these areas gives the value of 5.