Integration of log (2 Viewers)

[Coookies]

6. Find the exact area between curve y=1/x, x-axis and lines y=x and x=2 in first quadrant.

9. Find volume of solid formed when curve y=2/sqrt(2x-1) is rotated about x-axis from x=1 to x=5

10. Find area between curve y=lnx, y-axis and lines y=2 and y=4

11. Find exact volume of solid formed when curve y=lnx is rotated about y-axis from y=1 to y=3


I mostly need help with changing it into the right form to be integrated, so if you could help me with that, that would be awesome!

 

[Nooblet94]

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ 6)~A=\int_{1}^{2} (x-\frac{1}{x})dx\\ ~\\ 7)~V=\pi \int_{1}^{5} \frac{4dx}{2x-1}\\ ~\\ 10)~y=\ln x \Rightarrow x=e^y\\ \therefore A=\int_{2}^{4}e^ydy\\ ~\\ 11)~$Similar to above...$\\ V=\pi \int_1^3 e^{2y}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ 6)~A=\int_{1}^{2} (x-\frac{1}{x})dx\\ ~\\ 7)~V=\pi \int_{1}^{5} \frac{4dx}{2x-1}\\ ~\\ 10)~y=\ln x \Rightarrow x=e^y\\ \therefore A=\int_{2}^{4}e^ydy\\ ~\\ 11)~$Similar to above...$\\ V=\pi \int_1^3 e^{2y}" title="\\ 6)~A=\int_{1}^{2} (x-\frac{1}{x})dx\\ ~\\ 7)~V=\pi \int_{1}^{5} \frac{4dx}{2x-1}\\ ~\\ 10)~y=\ln x \Rightarrow x=e^y\\ \therefore A=\int_{2}^{4}e^ydy\\ ~\\ 11)~$Similar to above...$\\ V=\pi \int_1^3 e^{2y}" /></a>

Any questions as to how I got those?

 

[powlmao]

My class hasn't done logs yet, but for the integral of x - 1/x can you just integrate normally and get it?

 

[Aysce]

My class hasn't done logs yet, but for the integral of x - 1/x can you just integrate normally and get it?

Yes, using the fact that when you integrate 1/x you will get ln(x) + c

 

[Timske]

My class hasn't done logs yet, but for the integral of x - 1/x can you just integrate normally and get it?

yes , so it becomes x^2/2 - lnx + c

 

[powlmao]

ZOMG i just re-read it and remember the standard integrals you get.

I am such a derp 1/x = inx + c

 

[Coookies]

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ 6)~A=\int_{1}^{2} (x-\frac{1}{x})dx\\ ~\\ 7)~V=\pi \int_{1}^{5} \frac{4dx}{2x-1}\\ ~\\ 10)~y=\ln x \Rightarrow x=e^y\\ \therefore A=\int_{2}^{4}e^ydy\\ ~\\ 11)~$Similar to above...$\\ V=\pi \int_1^3 e^{2y}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ 6)~A=\int_{1}^{2} (x-\frac{1}{x})dx\\ ~\\ 7)~V=\pi \int_{1}^{5} \frac{4dx}{2x-1}\\ ~\\ 10)~y=\ln x \Rightarrow x=e^y\\ \therefore A=\int_{2}^{4}e^ydy\\ ~\\ 11)~$Similar to above...$\\ V=\pi \int_1^3 e^{2y}" title="\\ 6)~A=\int_{1}^{2} (x-\frac{1}{x})dx\\ ~\\ 7)~V=\pi \int_{1}^{5} \frac{4dx}{2x-1}\\ ~\\ 10)~y=\ln x \Rightarrow x=e^y\\ \therefore A=\int_{2}^{4}e^ydy\\ ~\\ 11)~$Similar to above...$\\ V=\pi \int_1^3 e^{2y}" /></a>

Any questions as to how I got those?

Please explain 6 & 7
& can you do Q11 for me please? I don't know how it gets to be pi/2 e^2(e^4-1)

 

[AAEldar]

ZOMG i just re-read it and remember the standard integrals you get.

I am such a derp 1/x = inx + c

Except its ln not in (lower case L).

 

[powlmao]

Except its ln not in (lower case L).

Using a Ipod to much effort to capitalize

 

[Carrotsticks]

Using a Ipod to much effort to capitalize

But you got the wrong letter anyway =p

 

[AAEldar]

Using a Ipod to much effort to capitalize

That's my point though, there is no capital. It's an L but lower case!

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\int e^{2y} ~ dy = \frac{e^{2y}}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int e^{2y} ~ dy = \frac{e^{2y}}{2}" title="\int e^{2y} ~ dy = \frac{e^{2y}}{2}" /></a>

 

[Coookies]

For example, if its e^6 - e^2, and I take out the e^2 as a common factor, will the e^6 become e^4 or e^3?

 

[Carrotsticks]

e^4 because by log laws:

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=V = \pi\int_{1}^{3}e^{2y}~dy = \pi\left [ \frac{e^{2y}}{2} \right ]_{1}^{3}\\\\ = \pi\left [\left [ \frac{e^{2(3)}}{2} \right ] - \left [\frac{e^{2(1)}}{2} \right ] \right ] = \pi \left [\frac{e^{6}}{2}- \frac{e^{2}}{2} \right ] \\\\ = \frac{\pi(e^6 - e^2)}{2} \\\\ = \frac{\pi*e^2(e^4 - 1)}{2} ~units^3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?V = \pi\int_{1}^{3}e^{2y}~dy = \pi\left [ \frac{e^{2y}}{2} \right ]_{1}^{3}\\\\ = \pi\left [\left [ \frac{e^{2(3)}}{2} \right ] - \left [\frac{e^{2(1)}}{2} \right ] \right ] = \pi \left [\frac{e^{6}}{2}- \frac{e^{2}}{2} \right ] \\\\ = \frac{\pi(e^6 - e^2)}{2} \\\\ = \frac{\pi*e^2(e^4 - 1)}{2} ~units^3" title="V = \pi\int_{1}^{3}e^{2y}~dy = \pi\left [ \frac{e^{2y}}{2} \right ]_{1}^{3}\\\\ = \pi\left [\left [ \frac{e^{2(3)}}{2} \right ] - \left [\frac{e^{2(1)}}{2} \right ] \right ] = \pi \left [\frac{e^{6}}{2}- \frac{e^{2}}{2} \right ] \\\\ = \frac{\pi(e^6 - e^2)}{2} \\\\ = \frac{\pi*e^2(e^4 - 1)}{2} ~units^3" /></a>

 

[Coookies]

Thanks )

Can someone please show me the first few steps for Q 6 & 9?

 

[Timske]

Integrate it and sub the values in .

 

[Aysce]

Integrate it and sub the values in .

No.

Do it graphically and calculate the area. Ie. Sketch the graph, draw out the lines y=x and x=2, find the area bounded by these lines.

Q6 only

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=6. ~ A = \int_{1}^{2} (x - \frac{1}{x}) ~dx = \left [\frac{x^2}{2} - \ln x \right ]_{1}^{2} \\\ = \left [\frac{(2)^2}{2} - \ln (2) \right ] - \left [\frac{(1)^2}{2} - \ln (1) \right ] \\\\ = \frac{4}{2} - \ln (2) - \frac{1}{2} ~, ~ \textup{NOTE}: \ln (1) = 0 \\\\= 2 - \ln (2) - \frac{1}{2} \\\\ = \frac{3}{2} - \ln (2)~ units^2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?6. ~ A = \int_{1}^{2} (x - \frac{1}{x}) ~dx = \left [\frac{x^2}{2} - \ln x \right ]_{1}^{2} \\\ = \left [\frac{(2)^2}{2} - \ln (2) \right ] - \left [\frac{(1)^2}{2} - \ln (1) \right ] \\\\ = \frac{4}{2} - \ln (2) - \frac{1}{2} ~, ~ \textup{NOTE}: \ln (1) = 0 \\\\= 2 - \ln (2) - \frac{1}{2} \\\\ = \frac{3}{2} - \ln (2)~ units^2" title="6. ~ A = \int_{1}^{2} (x - \frac{1}{x}) ~dx = \left [\frac{x^2}{2} - \ln x \right ]_{1}^{2} \\\ = \left [\frac{(2)^2}{2} - \ln (2) \right ] - \left [\frac{(1)^2}{2} - \ln (1) \right ] \\\\ = \frac{4}{2} - \ln (2) - \frac{1}{2} ~, ~ \textup{NOTE}: \ln (1) = 0 \\\\= 2 - \ln (2) - \frac{1}{2} \\\\ = \frac{3}{2} - \ln (2)~ units^2" /></a>

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=9. V = \pi \int_{1}^{5} \frac{4}{2x - 1}~ dx = \pi \left [2\ln (2x -1) \right ]_{1}^{5} \\\\ = \pi\left [\left [ 2\ln(2(5) - 1) \right ] - \left [2\ln(2(1) - 1) \right ] \right ] \\\\ =\pi [2\ln(10 - 1) - 2\ln(2 -1)]\\\\ =\pi2\ln(9)\\\\ = \pi\ln(81) ~units^3" target="_blank"><img src="http://latex.codecogs.com/gif.latex?9. V = \pi \int_{1}^{5} \frac{4}{2x - 1}~ dx = \pi \left [2\ln (2x -1) \right ]_{1}^{5} \\\\ = \pi\left [\left [ 2\ln(2(5) - 1) \right ] - \left [2\ln(2(1) - 1) \right ] \right ] \\\\ =\pi [2\ln(10 - 1) - 2\ln(2 -1)]\\\\ =\pi2\ln(9)\\\\ = \pi\ln(81) ~units^3" title="9. V = \pi \int_{1}^{5} \frac{4}{2x - 1}~ dx = \pi \left [2\ln (2x -1) \right ]_{1}^{5} \\\\ = \pi\left [\left [ 2\ln(2(5) - 1) \right ] - \left [2\ln(2(1) - 1) \right ] \right ] \\\\ =\pi [2\ln(10 - 1) - 2\ln(2 -1)]\\\\ =\pi2\ln(9)\\\\ = \pi\ln(81) ~units^3" /></a>