Why do Wolframalpha and Symbolab say $$\int_0^8\frac1{\sqrt[3]x}dx=6$$ when the function is not continuous in the given interval?
2026-04-26 11:50:33.1777204233
Glitch in computing $\int_0^8\frac{dx}{\sqrt[3]x}$?
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When an integrand is continuous except at the left endpoint of integration, it is an improper integral, and by definition it means the quantity $$ \int_0^8 \frac1{\sqrt[3]x}\,dx = \lim_{a\to0^+} \int_a^8 \frac1{\sqrt[3]x}\,dx. $$ (The limit might or might not exist.)