Integration of log (1 Viewer)

[Nooblet94]

Have a look at this

The area you're wanting is the one enclosed by the 2 lines and the line x=2. Now, to find that area we first find the area underneath the line that's on top and then subtract the line that's on the bottom from it.

If you still don't get it I'd recommend actually drawing a diagram to convince yourself why that's the case. If you don't get it after that you should ask your teacher because it's far easier to explain this kind of thing in person

Off-topic, but Timske, what's with the FLA in your sig?

 

[Coookies]

I need help with how you actually got to the first bit (the A= bit)

 

[Nooblet94]

I need help with how you actually got to the first bit (the A= bit)

Alright, let's say you wanted to find the area underneath y=x. You'd integrate y=x, correct?

Similarly, if you want to find the area underneath y=1/x, you'd integrate y=1/x.

Now, we're wanting to find the area between the two curves, but if you visualise it, you'll see that the area between the two curves is the area underneath y=1/x subtracted from the area underneath y=x, i.e. you would take the integral of x-1/x.

 

[Carrotsticks]

<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>

It's actually preferable to simplify the ln as far as possible:

 

[Timske]

It's actually preferable to simplify the ln as far as possible:

Oo Cheers

 

[Coookies]

The answer is actually 2piln9 lol

 

[Coookies]

I have one more question

Find the indefinite integral
3x/x^2+7

I know its easy but I can't do it lol

 

[qwerty44]

I have one more question

Find the indefinite integral
3x/x^2+7

I know its easy but I can't do it lol

Because the differential of x^2+7 is 2x, you have to make the numerator 2x. To do that we take 3 out of the integral, then take out a half which leaves the numerator as 2x.

 

[RealiseNothing]

The answer is actually 2piln9 lol

 

[Coookies]

Yeah but the book doesn't simplify it that much

 

[Coookies]

Because the differential of x^2+7 is 2x, you have to make the numerator 2x. To do that we take 3 out of the integral, then take out a half which leaves the numerator as 2x.

lol thanks!
I was a bit stumped about the 3
& you're a prelim student! I feel so dumb hahaa

 

[Coookies]

This is a different topic but this is another really easy question.

Find the arc length if radius = 1.5m and the angle subtended is 0.43

 

[Carrotsticks]

Yeah but the book doesn't simplify it that much

Just because the book doesn't, does not mean your answer is incorrect.

A general rule of thumb is to leave your answer in the most factorised/simplest way possible and believe me 4 pi ln(3) is MUCH simpler than 2 pi ln(9).

 

[qwerty44]

This is a different topic but this is another really easy question.

Find the arc length if radius = 1.5m and the angle subtended is 0.43

dont feel dumb, im doing accelerated maths, so we are in the same boat


Arc length=r * (theta) (where theta is in radians)

=1.5 * 0.43

=0.645m

 

[RealiseNothing]

This is a different topic but this is another really easy question.

Find the arc length if radius = 1.5m and the angle subtended is 0.43

 

[qwerty44]

Just because the book doesn't, does not mean your answer is incorrect.

A general rule of thumb is to leave your answer in the most factorised/simplest way possible and believe me 4 pi ln(3) is MUCH simpler than 2 pi ln(9).

It also makes it a lot easier to work with later on (if needed).

 

[RealiseNothing]

It also makes it a lot easier to work with later on (if needed).

Also something that is completely simplified is a lot more aesthetic in my opinion. When i see something not full simplified, it doesn't look right.

 

[Coookies]

Just because the book doesn't, does not mean your answer is incorrect.

A general rule of thumb is to leave your answer in the most factorised/simplest way possible and believe me 4 pi ln(3) is MUCH simpler than 2 pi ln(9).

I know, I know, I was just saying what the book had
I didn't say anything was wrong

 

[Coookies]

dont feel dumb, im doing accelerated maths, so we are in the same boat


Arc length=r * (theta) (where theta is in radians)

=1.5 * 0.43

=0.645m

Oh good, I actually felt really dumb for a moment there lol!
I guess I'm average and you're just smart!

How come 0.43 = theta?
Do you just assume that because theres no symbol or anything

 

[qwerty44]

Yes because it is radians.

Radians are usually presented with a ^c (0.43^c), but the ^c is usually left out. So if there is no ^o for degrees, then it is radians.