Question
Consider the improper integral $$\int_{1}^{\infty} x^{-2 / 3} d x$$a. Evaluate $I(b)=\int_{1}^{b} x^{-2 / 3} d x$b. Show that $$\lim _{b \rightarrow \infty} I(b)=\infty$$ thus proving that the given improper integral is divergent.
Step 1
This is a simple power rule integration problem. The power rule states that $\int x^n dx = \frac{1}{n+1}x^{n+1} + C$ for $n \neq -1$. Here, $n = -2/3$. Show more…
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