Problem 38 - Essential Calculus

Question

Evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{0}^{1} \frac{x-1}{x+1} d x $$

Step-by-Step Solution

Answer

The definite integral of the function (x-1)/(x+1) from 0 to 1 is equal to 1 - ln 2.

Step 1: Finding the antiderivative

First, find the antiderivative (indefinite integral) of the function \(\frac{x-1}{x+1}\). Because the function is a rational function, we can separate it into two parts and perform the integration separately. Hence, \(\int \frac{x-1}{x+1} dx = \int \left(\frac{x}{x+1} - \frac{1}{x+1} \right) dx = \int dx - \int \frac{1}{x+1} dx\). The first term is a straightforward integral, while for the second term, we can use elementary integrals to find its integral which is \( \ln |x+1| \). Therefore, the antiderivative is \(x - \ln |x + 1| + C\), with C representing the constant of integration.

Step 2: Applying the Fundamental Theorem of Calculus

Once we have found the antiderivative, we can evaluate the definite integral from 0 to 1. This involves substituting x = 1 and x = 0 into the antiderivative and calculating the difference. \( \int_{0}^{1} \frac{x-1}{x+1} dx = (1 - \ln |1 + 1|) - (0 - \ln |0 + 1|) = 1 - \ln 2 - ln 1 = 1 - \ln 2\).

Step 3: Verifying the result with a graphing utility

To confirm the validity of the solution, enter the function (x-1)/(x+1) into a graphical calculator and find the area under the curve from 0 to 1. The result should be approximately equal to 1 - \ln 2.