Question

Using the definition of improper integrals,evaluate the integral ,or show that it diverges
http://i.stack.imgur.com/a7ud0.png

Answer 1

Answer on Question #50338 – Math – Integral Calculus

1. Using the definition of improper integrals, evaluate the integral, or show that it diverges.

$$\int_{c}^{1} \frac{1}{\sqrt[3]{x}} dx$$

Solution.

1) If $c > 0$, then the integral is proper.

$$\int_{c}^{1} \frac{dx}{3/x} = \left. \frac{3}{2} x^{2/3} \right|_{c}^{1} = \frac{3}{2} \left(1 - c^{2/3}\right).$$

2) If $c = 0$, then $\int_{0}^{1} \frac{dx}{3/x} = \lim_{\varepsilon \to +0} \int_{\varepsilon}^{1} \frac{dx}{3/x} = \lim_{\varepsilon \to +0} \left( \frac{3}{2} x^{2/3} \right|_{e}^{1} = \lim_{\varepsilon \to +0} \left( \frac{3}{2} \left(1 - \varepsilon^{2/3}\right) \right) = \frac{3}{2} \left(1 - 0^{2/3}\right) = \frac{3}{2}$.

3) If $c < 0$, then $\int_{c}^{1} \frac{dx}{3/x} = \lim_{\varepsilon \to +0} \left( \int_{c}^{-\varepsilon} \frac{dx}{3/x} + \int_{c}^{1} \frac{dx}{3/x} \right) = |t = -x| = \lim_{\varepsilon \to +0} \left( \int_{c}^{c} \frac{dt}{3/t} + \int_{c}^{1} \frac{dx}{3/x} \right) =$

$$= \lim_{\varepsilon \to +0} \left[ \left. \frac{3}{2} t^{2/3} \right|_{-c}^{-\varepsilon} + \left( \frac{3}{2} x^{2/3} \right)_{e}^{1} \right] = \frac{3}{2} \lim_{\varepsilon \to +0} \left[ \varepsilon^{2/3} - (-c)^{2/3} + 1 - \varepsilon^{2/3} \right] = \frac{3}{2} \lim_{\varepsilon \to +0} \left[ 1 - (-c)^{2/3} \right] = \frac{3}{2} \left[ 1 - (-c)^{2/3} \right].$$

Answer: $\frac{3}{2} \left(1 - c^{2/3}\right)$, if $c > 0$; $\frac{3}{2} \left[1 - (-c)^{2/3}\right]$, if $c \leq 0$.

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