Question

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} d x $$

Short answer

The improper integral \(\int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} dx\) diverges.

Step-by-step solution

Define the Improper Integral as Limit

Set the improper integral as the limit of a proper integral. Here, instead of the upper limit of integration being infinite, replace it with \(t\), where \(t\) will later approach infinity. Hence, \(\int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{\sqrt[3]{x}} dx\).

Evaluate the Integral

Next, compute the integral without considering the limit: \(\int_{1}^{t} \frac{1}{\sqrt[3]{x}} dx\). This is a basic power rule problem in integral calculus, where the integral of \(x^n\) is \(\frac{x^{n+1}}{n+1}\). The integral becomes \(3 (t^{\frac{2}{3}} - 1)\).

Applying the Limit

Now, apply the limit: \(\lim_{t \to \infty} 3 (t^{\frac{2}{3}} - 1)\). As \(t\) approaches infinity, \(t^{\frac{2}{3}}\) also approaches infinity. Thus, the limit is infinite.

Determining Convergence or Divergence

When the value of the limit is infinite, the integral diverges. Therefore, the improper integral \(\int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} dx\) diverges, and there exists no finite area under the curve from 1 to infinity for the function \(f(x) = \frac{1}{\sqrt[3]{x}}\).