Need help w/ Riemann Sums+Antiderivatives please

May 2010
6
0
usa
Can someone please explain to me how to find the volume of y=(log(x))+2 on the interval [.2 ; 7] ?

And then also help with finding the approximate volume using Riemann sums?

Any help would greatly be appreciated. My calculus book doesn't explain how to find volume with log functions
 
Oct 2009
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Can someone please explain to me how to find the volume of y=(log(x))+2 on the interval [.2 ; 7] ?

And then also help with finding the approximate volume using Riemann sums?

Any help would greatly be appreciated. My calculus book doesn't explain how to find volume with log functions

Your question makes no sense: you give a one-variable function and you ask for its volume...what volume?!? You could ask about the area enclosed between the function's graph and the x-axis, or even about the volume of the 3-dimensional body created by revolving that graph around the x-axis or any other line, but just as it is the question makes no sense.

Tonio
 
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chiph588@

MHF Hall of Honor
Sep 2008
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Champaign, Illinois
Can someone please explain to me how to find the volume of y=(log(x))+2 on the interval [.2 ; 7] ?

And then also help with finding the approximate volume using Riemann sums?

Any help would greatly be appreciated. My calculus book doesn't explain how to find volume with log functions
I'm going to assume by volume you mean \(\displaystyle 2 \) dimensional volume, aka area.


Are you familiar with integration by parts?

If so take \(\displaystyle u=\log(x),\;\; dv=dx \implies du=\frac{dx}{x},\;\;v=x \)

Thus \(\displaystyle \int \log(x)dx = x\log(x)-\int x\frac{dx}{x} = x\log(x)-x+C \)
 
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May 2010
6
0
usa
My apologies. I meant for that function to be rotated around the x-axis. Hope that clarifies.

No, I can't say I am familiar with integration by parts. My teacher said that I may need to find the opposite by switching x and y and finding a new equation involving ln but she didn't fully explain.
 
Jul 2007
894
298
New Orleans
Can someone please explain to me how to find the volume of y=(log(x))+2 on the interval [.2 ; 7] ?

And then also help with finding the approximate volume using Riemann sums?

Any help would greatly be appreciated. My calculus book doesn't explain how to find volume with log functions
Are you talking about the area? Use the change of base formula and you get

\(\displaystyle \int (\ln{x} - \ln{10} + 2)dx\)

Use integration by parts for the first term
 
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chiph588@

MHF Hall of Honor
Sep 2008
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429
Champaign, Illinois
My apologies. I meant for that function to be rotated around the x-axis. Hope that clarifies.

No, I can't say I am familiar with integration by parts. My teacher said that I may need to find the opposite by switching x and y and finding a new equation involving ln but she didn't fully explain.
Oh, in that case we get \(\displaystyle V = \pi\int_{.2}^7 (\log(x)-2)^2dx = \pi\int_{.2}^7 (\log^2(x)-4\log(x)+4)dx \).

Would you be able to evaluate this integral?
 
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May 2010
6
0
usa
With my calculator yes lol. But I need to show work. I haven't been able to find any help with integrating log
 

chiph588@

MHF Hall of Honor
Sep 2008
1,163
429
Champaign, Illinois
My apologies. I meant for that function to be rotated around the x-axis. Hope that clarifies.

No, I can't say I am familiar with integration by parts. My teacher said that I may need to find the opposite by switching x and y and finding a new equation involving ln but she didn't fully explain.
Let's try it this way then:

\(\displaystyle y=\log(x)+2\implies x=e^{y-2} \) and our new interval is \(\displaystyle [\log(.2)+2,\;\log(7)+2] \)

Now we want to rotate our new equation around the y-axis.

Can you manage to do this?
 
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May 2010
6
0
usa
So I start with \(\displaystyle V = \pi\int_{log(.2)+2}^{log(7)+2} (e^{y-2}
)dy\) right?

Then that becomes \(\displaystyle V = \pi\ ((y-2)(e^{y-2})) \) from log(.2)+2 to log(7)+2?

Then just plug in the top and bottom and subtract?
 

chiph588@

MHF Hall of Honor
Sep 2008
1,163
429
Champaign, Illinois
So I start with \(\displaystyle V = \pi\int_{log(.2)+2}^{log(7)+2} (e^{y-2}
)dy\) right?

Then that becomes \(\displaystyle V = \pi\ ((y-2)(e^{y-2})) \) from log(.2)+2 to log(7)+2?

Then just plug in the top and bottom and subtract?
No, there is a different formula for rotating about the y-axis.
 
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