determine whether the improper integral converges or diverges. if it converges, determine the value of the…

Explanation:

Step1: Identify the singularity

The integrand $\frac{4}{\sqrt[3]{(x - 4)^2}}$ has a singularity at $x = 4$ in the interval $[1,6]$. We split the integral into two - integrals: $\int_{1}^{6}\frac{4}{\sqrt[3]{(x - 4)^2}}dx=\int_{1}^{4}\frac{4}{\sqrt[3]{(x - 4)^2}}dx+\int_{4}^{6}\frac{4}{\sqrt[3]{(x - 4)^2}}dx$.

Step2: Evaluate $\int\frac{4}{\sqrt[3]{(x - 4)^2}}dx$

Let $u=x - 4$, then $du=dx$. So $\int\frac{4}{\sqrt[3]{(x - 4)^2}}dx = 4\int u^{-\frac{2}{3}}du$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $4\int u^{-\frac{2}{3}}du=4\times\frac{u^{-\frac{2}{3}+1}}{-\frac{2}{3}+1}+C=12u^{\frac{1}{3}}+C = 12(x - 4)^{\frac{1}{3}}+C$.

Step3: Evaluate $\int_{1}^{4}\frac{4}{\sqrt[3]{(x - 4)^2}}dx$

$\lim_{a\rightarrow4^{-}}\int_{1}^{a}\frac{4}{\sqrt[3]{(x - 4)^2}}dx=\lim_{a\rightarrow4^{-}}\left[12(x - 4)^{\frac{1}{3}}\right]{1}^{a}=\lim{a\rightarrow4^{-}}\left(12(a - 4)^{\frac{1}{3}}-12(1 - 4)^{\frac{1}{3}}\right)=0+12\sqrt[3]{3}$.

Step4: Evaluate $\int_{4}^{6}\frac{4}{\sqrt[3]{(x - 4)^2}}dx$

$\lim_{b\rightarrow4^{+}}\int_{b}^{6}\frac{4}{\sqrt[3]{(x - 4)^2}}dx=\lim_{b\rightarrow4^{+}}\left[12(x - 4)^{\frac{1}{3}}\right]{b}^{6}=\lim{b\rightarrow4^{+}}\left(12(6 - 4)^{\frac{1}{3}}-12(b - 4)^{\frac{1}{3}}\right)=12\sqrt[3]{2}-0$.

Step5: Find the value of the original integral

$\int_{1}^{6}\frac{4}{\sqrt[3]{(x - 4)^2}}dx=\int_{1}^{4}\frac{4}{\sqrt[3]{(x - 4)^2}}dx+\int_{4}^{6}\frac{4}{\sqrt[3]{(x - 4)^2}}dx=12\sqrt[3]{3}+12\sqrt[3]{2}\approx12\times1.4422+12\times1.2599=17.3064 + 15.1188=32.4252$.

Answer:

$32.4252$