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. $$\int_{-5}^{6} \frac{9-x^{2}}{x-3} d x$$

Short answer

The points of discontinuity of the integrand on the interval of integration is x = 3. The integral \(\int_{-5}^{6} \frac{9-x^{2}}{x-3} dx\) equals -50.778.

Step-by-step solution

Finding points of discontinuity

To find the points of discontinuity, you have to set the denominator of the integrand equal to zero and solve for 'x', because a function will be discontinuous where the denominator is zero. Here, setting \(x - 3 = 0\) yields \(x = 3.\) Now, check if this value lies in the interval of integration \(-5 \leq x \leq 6\). Yes, it does. So, the point of discontinuity within the interval of integration is \(x = 3.\)

Evaluating the integral using area

The integral can be split into two intervals due to the discontinuity at x = 3: \(\int_{-5}^{3} \frac{9-x^{2}}{x-3} dx + \int_{3}^{6} \frac{9-x^{2}}{x-3} dx.\) The given integrand's anti-derivative with respect to 'x' is \(-x - \frac{9}{x - 3} + c\). Therefore, by applying the Fundamental Theorem of Calculus, we calculate the area under the curve by evaluating the anti-derivative at the endpoints of the intervals and subtracting the two results. After calculations, it is found that the integral equals -50.778.