Text Compute the following improper integrals: (a) ∫1+∞1x3 dx\displaystyle\int_1^{+\infty}\frac{1}{x^{3}}\,dx∫1+∞x31dx (b) ∫011x dx\displaystyle\int_0^{1}\frac{1}{\sqrt{x}}\,dx∫01x1dx Solution (a) ∫1tx−3dx=12−12t2→12\displaystyle\int_1^{t}x^{-3}dx=\tfrac12-\tfrac{1}{2t^{2}}\to\tfrac12∫1tx−3dx=21−2t21→21 as t→+∞t\to+\inftyt→+∞. (b) ∫ε1x−1/2dx=2−2ε→2\displaystyle\int_{\varepsilon}^{1}x^{-1/2}dx=2-2\sqrt{\varepsilon}\to2∫ε1x−1/2dx=2−2ε→2 as ε→0+\varepsilon\to0^{+}ε→0+. a=12;b=2\boxed{a=\tfrac12;\quad b=2}a=21;b=2