Question

True or False? Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \int_{-1}^{2} \frac{1}{x} d x=[\ln |x|]_{-1}^{2}=\ln 2-\ln 1=\ln 2 $$

Short answer

The statement is false because the integral from -1 to 2 of the function 1/x is undefined due to the discontinuity of the function at x=0 in the interval of integration.

Step-by-step solution

Inspect the Expression

First, note that the integral is a definite integral from -1 to 2 of the function \(1/x\). The antiderivative of \(1/x\) is \(\ln|x|\), which is valid only when \(x > 0\). Check that the interval of integration does not contain the discontinuity, \(x=0\).

Compute the Integral Correctly

The interval of integration from -1 to 2 does contain the discontinuity \(x=0\) of the function \(1/x\), so the integral of \(1/x\) from -1 to 2 is undefined. Therefore, the given statement is false. We should split the integral at the discontinuity to compute it correctly. Thus, the correct computation of the integral would be \(\int_{-1}^{0} \frac{1}{x} d x + \int_{0}^{2} \frac{1}{x} d x\). Notice that \(\int_{-1}^{0} \frac{1}{x} d x\) would evaluate to \( \ln|-1| - \ln|0|\), which is not defined correctly.

Find Example to Show Falsehood

Since -1 to 0 contains 0 in its domain, the logarithm function is discontinuous here. Therefore, attempting to evaluate the integral of 1/x from -1 to 2 is not valid and the expression for the integral given in the exercise is false.