Problem 38 - 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. $$ \int_{1}^{3} \frac{2}{(x-2)^{8 / 3}} d x $$

Step-by-Step Solution

Answer

The integral converges and is equal to 0.

Step 1: Rewriting the integral

Rewrite the fraction inside the integral as \(2 (x - 2)^{-8 / 3}\) to emphasize the power law property that will be used in the integration step.

Step 2: Identify Singular Point

The singular point for this integral is \(x = 2\), where the denominator becomes zero. This point divides the interval of integration in two, [1, 2) and (2, 3]. The integral should be therefore separated into two integrals each over one of these intervals and each one should be evaluated separately.

Step 3: Integration

Now, perform the integration using the power rule for integrals, which states that the integral of \(x^n\) equals \(\frac{{x^{n+1}}}{{n+1}}\) for \(n \neq -1\). Applying the power rule, the integral of \((x-2)^{-8/3}\) is \(-\frac{3}{5}(x-2)^{-5/3}\). Apply this rule on each of the two integrals.

Step 4: Evaluate The Integral

Substitute the limits of integration into the result to calculate the integral over each interval. For the first interval [1, 2), substitute \(x=2\) and then \(x=1\), you will obtain \(-\frac{3}{5}\left[(2-2)^{-5/3} - (2-1)^{-5/3}\right]\) which equals \(\frac{3}{5}\). Repeat the substitution for the second interval (2, 3], you will find \(-\frac{3}{5}\left[(3-2)^{-5/3} - (2-2)^{-5/3}\right]\) which equals \(-\frac{3}{5}\). Adding these results gives the total integral.