Problem 57 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. \(\int_{1}^{\infty} \frac{d x}{x(x+1)}<\infty\).

Step-by-Step Solution

Answer

False. The improper integral \(\int_{1}^{\infty} \frac{d x}{x(x+1)}\) is divergent, meaning that it does not have a finite value.

Step 1: Perform partial fraction decomposition

The integral looks like: \[ \int_{1}^{\infty} \frac{1}{x(x+1)} \, dx \:. \] Let's perform partial fraction decomposition to simplify the integrand: \[ \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \:. \] We can rewrite the equation as: \[ 1 = A(x+1) + Bx \:. \] By setting x = 0, we get: \[ 1 = A \implies A =1 \:. \] By setting x = -1, we get: \[ 1 = -B \implies B = -1 \:. \] Thus, \[ \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1} \:. \]

Step 2: Evaluate the integral

Now, we can rewrite the integral as: \[ \int_{1}^{\infty} \frac{1}{x(x+1)} \, dx = \int_{1}^{\infty} \left(\frac{1}{x} - \frac{1}{x+1}\right) \, dx \:. \] Evaluate the difference of two integrals: \[ \int_{1}^{\infty} \left(\frac{1}{x} - \frac{1}{x+1}\right) \, dx = \int_{1}^{\infty} \frac{1}{x} \, dx - \int_{1}^{\infty} \frac{1}{x+1} \, dx \:. \] Now, let's compute each integral separately. The first one: \[ \int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{b\to\infty} \left[ \ln |x| \right]_1^b = \lim_{b\to\infty} (\ln b - \ln 1) = \infty. \] The second one: \[ \int_{1}^{\infty} \frac{1}{x+1} \, dx = \lim_{b\to\infty} \left[ \ln |x+1| \right]_1^b = \lim_{b\to\infty} (\ln (b+1) - \ln 2) = \infty. \] Since both integrals go to infinity, their difference also diverges: \[ \int_{1}^{\infty} \frac{1}{x(x+1)} \, dx = \infty - \infty = \text{diverges} \:. \] Therefore, the given statement is:

Step 3: Answer

False. The improper integral \(\int_{1}^{\infty} \frac{d x}{x(x+1)}\) is divergent, meaning that it does not have a finite value.