/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Determine whether the improper i... [FREE SOLUTION] | 91Ó°ÊÓ

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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

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

Step by step solution

01

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, \(\int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{\sqrt[3]{x}} dx\).
02

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 \(3 (t^{\frac{2}{3}} - 1)\).
03

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.
04

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 \(f(x) = \frac{1}{\sqrt[3]{x}}\).

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