evaluate the following integral or state that it diverges.\n int_{3}^{4}\frac{1}{(x - 3)^{\frac{3}{2}}}dx…

Explanation:

Step1: Identify the improper - integral type

The integral $\int_{3}^{4}\frac{1}{(x - 3)^{\frac{3}{2}}}dx$ is an improper integral of Type 2 (infinite discontinuity) since the integrand $f(x)=\frac{1}{(x - 3)^{\frac{3}{2}}}$ has a vertical asymptote at $x = 3$. We rewrite it as a limit: $\lim_{a\rightarrow3^{+}}\int_{a}^{4}\frac{1}{(x - 3)^{\frac{3}{2}}}dx$.

Step2: Integrate the function

Use the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\int\frac{1}{(x - 3)^{\frac{3}{2}}}dx=\int(x - 3)^{-\frac{3}{2}}dx$. Let $u=x - 3$, then $du=dx$. So $\int(x - 3)^{-\frac{3}{2}}dx=\frac{(x - 3)^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}+C=\frac{(x - 3)^{-\frac{1}{2}}}{-\frac{1}{2}}+C=-2(x - 3)^{-\frac{1}{2}}+C$.

Step3: Evaluate the definite integral with the limit

$\lim_{a\rightarrow3^{+}}\int_{a}^{4}\frac{1}{(x - 3)^{\frac{3}{2}}}dx=\lim_{a\rightarrow3^{+}}\left[-2(x - 3)^{-\frac{1}{2}}\right]{a}^{4}=\lim{a\rightarrow3^{+}}\left(-2(4 - 3)^{-\frac{1}{2}}+2(a - 3)^{-\frac{1}{2}}\right)=\lim_{a\rightarrow3^{+}}\left(-2 + \frac{2}{\sqrt{a - 3}}\right)$. As $a\rightarrow3^{+}$, $\frac{2}{\sqrt{a - 3}}\rightarrow+\infty$.

Answer:

B. The improper integral diverges.