Problem 37 - Essential Calculus

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

Step-by-Step Solution

Answer

The integral converges, and its value is 0.

Step 1: 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.

Step 2: 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 \(-3\).

Step 3: 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.

Step 4: Add the results

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