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. $$\int_{-3}^{4} \frac{x^{2}-1}{x+1} d x$$

Short answer

The point of discontinuity of the integrand on the interval of integration is -1. The integral of the function over the interval [-3, 4] is -2.

Step-by-step solution

Identify Points of Discontinuity

Points of discontinuity exist where the denominator of the fraction is equal to 0. We solve \(x+1 = 0\) for x, which results in x = -1. This is the only point of discontinuity in the function.

Simplify the Integrand

The integrand \(\frac{x^{2}-1}{x+1}\) can be simplified by separating it into two terms, \(x-1\), when dividing \(x^{2}-1\) by \(x+1\). Thus, the integral can be rewritten as \(\int_{-3}^{4} (x - 1) dx\) (note that we should exclude x=-1 since it's a point of discontinuity of the original function).

Integrate

Now, integrate the simplified function. The integral \(\int_{-3}^{4} (x - 1) dx\) splits into two terms, \(\int_{-3}^{4} x dx - \int_{-3}^{4} dx\), and becomes \(\frac{x^2}{2} - x\) evaluated from -3 to 4.

Evaluate the Definite Integral

Substitute the limit values into the antiderivative: \( [\frac{(4)^{2}}{2} - 4] - [\frac{(-3)^{2}}{2} + 3] = 4 - 6 = -2\)