# Volumes last week of calc

• physicsed
In summary, the conversation was about solving a problem involving volumes in calculus. The integral \int^{1}_{2} \pi \frac{1}{x} dx was discussed, with a reminder to check the setup in the textbook. The correct antiderivative was determined to be \pi \int^{2}_{1} \frac{1}{x^2} dx = \pi [-\frac{1}{x}] from a power law perspective. The conversation also touched on grammar and ended with a positive note.

#### physicsed

[SOLVED] volumes... last week of calc

i was absent when we went over the volume section.

y= 1/x, x= 1, x= 2, y= 0,; about the x-axis

$$\int^{1}_{2} \pi \frac{1}{x} dx$$

i don't know what's next. can anyone inform me, please

You could actually do the integration, but that would be wrong because you haven't set it up right either. You integrate pi*r^2 where r is the radius of the disk over the volume. I'd suggest checking a few examples in your textbook.

$$\int^{1}_{2} \pi [\frac{1}{x}]^2 dx$$

if you help me with this problem, i am sure i will get the rest of the problems.

It's a power law integral. Like x^n. What's n in this case? What's the antiderivative?

$$\pi \int^{2}_{1} \frac{1}{x^2} dx$$

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Fine start. Now what's the antiderivative of 1/x^2?

$$\pi [ \frac{x^{-3}}{-3}]$$

Beep. Wrong. The antiderivative of x^n is x^(n+1)/(n+1). What's n in this case? Unless that's a careless error because you are paying more attention to texing than thinking, you may have missed more than 'volumes'.

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since it's divided by 1, isn't it negative n
=x^-2

Yes, it is. Can you fix your antiderivative?

(-1/3)x^-3

Beep. Beep. Beep. You goofed it again. What's -2+1? Think this time.

righttt

pi [(1/-1)x^-1]

physicsed said:
righttt

pi [(1/-1)x^-1]

Much better! You've got it now, right? And you promised that meant you could get all the others.

thanks buddy