Problem 30 - College Algebra and Calculus: An Applied Approach

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. $$ \int_{0}^{2} \frac{1}{(x-1)^{4 / 3}} d x $$

Step-by-Step Solution

Answer

The improper integral converges and its value is 3.

Step 1: Rewrite the integral as limit

Firstly, rewrite the improper integral as the limit of a proper integral. So the given integral can be rewritten as \[\lim_{{b \to 1^-}} \int_{0}^{b} \frac{1}{(x-1)^{4/3}} dx + \lim_{{a \to 1^+}} \int_{a}^{2} \frac{1}{(x-1)^{4/3}} dx\] Due to symmetry, it is sufficient to investigate one of these limits.

Step 2: Using power rule for integration

Next step, apply the power rule for integral which states that the integral of \(x^n\) with respect to \(x\) is \((1/(n+1))*x^{n+1}\). Here, \(n = -(4/3)\). So, the integral of \((x-1)^{-4/3}\) with respect to \(x\) is \((-3/2)*(x-1)^{-1/3}\). Compute the limit \[\lim_{{b \to 1^-}} -\frac{3}{2}(b-1)^{-1/3}\] at the upper limit \(b\), and the limit \[-\frac{3}{2}(0-1)^{-1/3}\] at the lower limit \(0\). Both of these limits exist and are finite. Therefore, the first integral converges.

Step 3: Repeat step 2 on the other part of integral

Similarly, we compute the limit \[\lim_{{a \to 1^+}} -\frac{3}{2}(2-1)^{-1/3}\] at the upper limit \(2\), and the limit \[\lim_{{a \to 1^+}} -\frac{3}{2}(a-1)^{-1/3}\] at the lower limit \(a\). Both of these limits exist and are finite. Therefore, the second integral also converges.

Step 4: Combining the results

Combine the results of both integrals to get the final result. Both integrals converge, so the improper integral also converges and its value is \[\int_{0}^{2} \frac{1}{(x-1)^{4/3}} dx = (-3/2)(0-1)^{-1/3} - (-3/2)(1-1)^{-1/3}\] at the first part and \[(-3/2)(2-1)^{-1/3} - 0\] at the second part combining these we get \[0 - (-3/2)(1) - (-3/2)(1) + 0 = 3.\]