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[Ngoc Vu]

Tính tích phân suy rộng $\int_{1}^{2}\frac{1}{\left (x-2 \right )^{\frac{4}{3}}}$

Edited by perfectstrong, 10-02-2023 - 15:33.
Tiêu đề & LaTeX

[Le Tuan Canhh]

$\int_{1}^{2}\frac{1}{(x-2)\frac{4}{3}} dx=\lim_{t\rightarrow 2^{-}}\int_{1}^{t}\frac{1}{(x-2)^{\frac{4}{3}}}dx=\lim_{t\rightarrow 2^{-}}[-3[(x-2)^{\frac{-1}{3}}]|^{t} _{1}]= \lim_{t\rightarrow 2^{-}}[\frac{-3}{\sqrt[3]{t-2}}+\frac{3}{\sqrt[3]{1-2}}]=+\infty$

CTTQ: $\int x^{a}dx=\frac{1}{1+a}.x^{a+1}+C$

Edited by Le Tuan Canhh, 10-02-2023 - 19:31.