Problem 24 - College Algebra and Calculus: An Applied Approach

Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{0}^{27} \frac{5}{\sqrt[3]{x}} d x $$

Step-by-Step Solution

Answer

The improper integral converges and its value is \(13.5\).

Step 1: Rewrite the function

First, rewrite the function in a form that allows easier integration. The integrand, \( \frac{5}{\sqrt[3]{x}} \) can be written as \(5x^{-1/3}\).

Step 2: Application of the Power Rule for Integration

Apply the power rule, which states: \[ \int x^{n} dx = \frac{x^{n+1}}{n+1}+C\], where \(n ≠ -1\). For our function, \[ \int_{0}^{27} x^{-1/3}\ dx = \left[ \frac{3}{2}x^{2/3} \right]_0^{27}\]. We get the expression \( \frac{3}{2} \times (27)^{2/3} - \frac{3}{2} \times (0)^{2/3}\).

Step 3: Evaluation of the Definite Integral

Compute the values to get the definite integral. We find that \((27)^{2/3} = 9\) and \((0)^{2/3} = 0\). So, the integral becomes \( \frac{3}{2} \times 9 - \frac{3}{2} \times 0 = 13.5\). Since the integral is a real number, it converges and its value is 13.5.