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. $$ \int_{0}^{27} \frac{5}{\sqrt[3]{x}} d x $$

Short answer

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

Step-by-step solution

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 \(5x^{-1/3}\).

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 \( \frac{3}{2} \times (27)^{2/3} - \frac{3}{2} \times (0)^{2/3}\).

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.