Improper Integrals — Question 2

Question 2

Determine whether the integral converges. If it does, evaluate it. $$\int_0^1x^{-2/3}\,dx$$

Question 2 – Solution

Step 1: Identify the improper endpoint. The integrand is undefined at $x=0$, so replace the lower bound with $\varepsilon>0$: $$I=\lim_{\varepsilon\to0^+}\int_\varepsilon^1x^{-2/3}\,dx.$$ Step 2: Integrate. $$\int x^{-2/3}\,dx=\frac{x^{1/3}}{1/3}=3x^{1/3}.$$ Step 3: Evaluate the limit. $$\begin{align*} I&=\lim_{\varepsilon\to0^+}[3x^{1/3}]_\varepsilon^1\\ &=\lim_{\varepsilon\to0^+}(3-3\varepsilon^{1/3})=3. \end{align*}$$ $$\boxed{3\text{ (convergent)}}$$