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echo62's question at Yahoo! Answers (Improper integral)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

How do I evaluate the integral dx/(x^(2/3)) from -3 to 3?

What I did was, so for the integrand f(x)=1/(x^(2/3)) it's only continuous from (0, infinity), it's discontinuous at x=0, so I set up the integral as the limit as b approaches 0 from the right of the same integrand, so my answer was 3*(3)^(1/3) but that's wrong. Please help me?
Here is a link to the question:

Improper integral help? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello echo62,

We have $$\begin{aligned}
\int_0^3\frac{dx}{x^{2/3}}&=\int_0^3x^{-2/3}dx\\&=\lim_{\epsilon \to 0^+}\int_{\epsilon}^3x^{-2/3}dx\\&=\lim_{\epsilon \to 0^+}\left[3x^{1/3}\right]_{\epsilon}^3\\&=3\lim_{\epsilon \to 0^+}(\sqrt [3]{3}-\sqrt [3]{\epsilon})\\&=3\sqrt [3]{3}\end{aligned}$$ On the other hand, using the transformation $x=-t$:
$$\begin{aligned}\int_{-3}^0\frac{dx}{x^{2/3}}&=\int_{3}^0\frac{-dt}{(-t)^{2/3}}\\&=\int_{0}^3\frac{dt}{t^{2/3}}\\&=\int_{0}^3\frac{dx}{x^{2/3}}
\end{aligned}$$ So, $$\int_{-3}^3\frac{dx}{x^{2/3}}=\int_{-3}^0\frac{dx}{x^{2/3}}+\int_0^3\frac{dx}{x^{2/3}}=2\int_0^3\frac{dx}{x^{2/3}}=\boxed{\;6\sqrt [3]{3}\;}$$
 
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