evaluate the improper integral or show that it diverges.\n int_{0}^{4} \frac{d x}{(21-7 x)^{\frac{1}{3}}}…

Explanation:

Step1: Substitute ( u = 21 - 7x )

Differentiate ( u ) with respect to ( x ): ( du=-7dx), so ( dx =-\frac{1}{7}du ). When ( x = 0), ( u = 21); when ( x = 4), ( u = 21-7\times4=21 - 28=-7). The integral (\int_{0}^{4}\frac{dx}{(21 - 7x)^{\frac{1}{3}}}) becomes (-\frac{1}{7}\int_{21}^{-7}u^{-\frac{1}{3}}du=\frac{1}{7}\int_{-7}^{21}u^{-\frac{1}{3}}du).

Step2: Integrate ( u^{-\frac{1}{3}} )

Using the power - rule for integration (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)), for (n=-\frac{1}{3}), (\int u^{-\frac{1}{3}}du=\frac{u^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+C=\frac{3}{2}u^{\frac{2}{3}}+C).

Step3: Evaluate the definite integral

(\frac{1}{7}\left[\frac{3}{2}u^{\frac{2}{3}}\right]{-7}^{21}=\frac{3}{14}\left(u^{\frac{2}{3}}\right|{-7}^{21}). (u^{\frac{2}{3}}=\sqrt[3]{u^{2}}), so (\frac{3}{14}\left(\sqrt[3]{21^{2}}-\sqrt[3]{(-7)^{2}}\right)=\frac{3}{14}\left(\sqrt[3]{441}-\sqrt[3]{49}\right)=\frac{3}{14}\times2\sqrt[3]{49}=\frac{3}{7}\sqrt[3]{49}).

Answer:

A. The improper integral converges and (\int_{0}^{4}\frac{dx}{(21 - 7x)^{\frac{1}{3}}}=\frac{3}{7}\sqrt[3]{49})