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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. $$ \int_{0}^{2} \frac{1}{\sqrt[3]{x-1}} d x $$

Short Answer

Expert verified
The integral converges, and its value is 0.

Step by step solution

01

Establish the two bounds

Since there's a discontinuity at \(x=1\), the problem breaks down into two integrals: \( \int_{0}^{1} \frac{1}{\sqrt[3]{1-x}} d x \) and \( \int_{1}^{2} \frac{1}{\sqrt[3]{x-1}} d x \) We should evaluate both of these.
02

Evaluate the first integral

Let's evaluate the first integral \( \int_{0}^{1} \frac{1}{\sqrt[3]{1-x}} d x \). This is an indefinite integral, but as x approaches 1 from the left, the denominator approaches 0, making the integral improper. A possible solution for this is to use the limit definition of an improper integral: \( \lim_{a \to 1^-} \int_{0}^{a} \frac{1}{\sqrt[3]{1-x}} d x \). The antiderivative is \(-3(1-x)^{2/3}\) by using the power rule, substituting and simplifying, this limit comes out to be \(-3\).
03

Evaluate the second integral

Now, let's evaluate the second integral \( \int_{1}^{2} \frac{1}{\sqrt[3]{x-1}} d x \). Similar to the steps we took to solve the first integral, we'll work with the limit definition of this improper integral: \( \lim_{b \to 1^+} \int_{b}^{2} \frac{1}{\sqrt[3]{x-1}} d x \). Also, the antiderive of \( \frac{1}{\sqrt[3]{x-1}} \) is \(3(x-1)^{2/3}\), substitituting and simplifying, the limit comes out to be 3.
04

Add the results

The final step is to add both results together: \(-3+3=0\)

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