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2022-09-30

Need help with the following question about finding volume of improper integral

Find the exact volume of the solid created by rotating the region bounded by $f(x)=\frac{1}{\sqrt{x}\mathrm{ln}x}$ and the x-axis on the interval $[2,\infty )$. State the method of integral used.

My issue specifically is that I have no clue how to integrate the integral because I am using the disk method and the integral ends up being

${\int}_{2}^{\mathrm{\infty}}\pi \cdot \frac{1}{x(\mathrm{ln}(x){)}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx$

How would I go about solving this because I am unsure how to integrate that integral.

Find the exact volume of the solid created by rotating the region bounded by $f(x)=\frac{1}{\sqrt{x}\mathrm{ln}x}$ and the x-axis on the interval $[2,\infty )$. State the method of integral used.

My issue specifically is that I have no clue how to integrate the integral because I am using the disk method and the integral ends up being

${\int}_{2}^{\mathrm{\infty}}\pi \cdot \frac{1}{x(\mathrm{ln}(x){)}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx$

How would I go about solving this because I am unsure how to integrate that integral.

bequejatz8d

Beginner2022-10-01Added 6 answers

Step 1

I see $x(\mathrm{ln}x{)}^{2}$ in the denominator, which suggests that the antiderivative involves $\frac{1}{\mathrm{ln}x}$. And indeed:

${\int}_{2}^{\mathrm{\infty}}\pi \cdot \frac{1}{x(\mathrm{ln}(x){)}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx=\pi {[-\frac{1}{\mathrm{ln}x}]}_{2}^{\mathrm{\infty}}$

Step 2

Since $\underset{x\to \mathrm{\infty}}{lim}\frac{1}{\mathrm{ln}x}=0$, we get

$\pi {[-\frac{1}{\mathrm{ln}x}]}_{2}^{\mathrm{\infty}}=\pi (0-(-\frac{1}{\mathrm{ln}2}))=\frac{\pi}{\mathrm{ln}2}$

I see $x(\mathrm{ln}x{)}^{2}$ in the denominator, which suggests that the antiderivative involves $\frac{1}{\mathrm{ln}x}$. And indeed:

${\int}_{2}^{\mathrm{\infty}}\pi \cdot \frac{1}{x(\mathrm{ln}(x){)}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx=\pi {[-\frac{1}{\mathrm{ln}x}]}_{2}^{\mathrm{\infty}}$

Step 2

Since $\underset{x\to \mathrm{\infty}}{lim}\frac{1}{\mathrm{ln}x}=0$, we get

$\pi {[-\frac{1}{\mathrm{ln}x}]}_{2}^{\mathrm{\infty}}=\pi (0-(-\frac{1}{\mathrm{ln}2}))=\frac{\pi}{\mathrm{ln}2}$

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