Author Topic: Integration (Read 10621 times) Tweet Share

d0minicz · « **Reply #30 on:** April 07, 2009, 03:39:40 pm »

Ye NE2000 is right

Mao · « **Reply #31 on:** April 07, 2009, 10:33:27 pm »

Quote from: TrueTears on April 07, 2009, 12:43:26 pm

$\int{sin^{-1}(x)}dx$ $= \frac{1}{\sqrt{1^2-x^2}} = \frac{1}{\sqrt{1-x^2}}$

$\int{cos^{-1}(x)}dx$ $= \frac{-1}{\sqrt{1^2-x^2}} = \frac{-1}{\sqrt{1-x^2}}$

That is incorrect, integrating inverse circular functions require the use of 'integration by parts'. This is not in the syllabus of SM.

However, as 'fun' activity because you love math, you might want to try differentiating $x\sin^{-1}(x)$ , and hence find an antiderivative of $\sin^{-1}(x)$

d0minicz · « **Reply #32 on:** April 08, 2009, 06:05:56 pm »

Find the volume of the region bounded by $y= 4x^2$ , $x=0$ , $y=16$ , about $y=16$ .
thx.

kamil9876 · « **Reply #33 on:** April 08, 2009, 06:15:37 pm »

Revoling around y=16 eh. We're not used to that, however we can turn this into a situation that we are used to, namely y=0. The way we do that is by translating the graph 16 units down so that the axis of rotation is y=0.

So the graph u want to deal with now is:
$<br />y=4x^2 - 16$

Now you know one of the terminals is x=0. The other is found by subbing y=16 into the OLD equation or subbing in y=0 into the NEW equation(the latter is a bit more complicated to justify, it's basically due to translation of 16 units down)

NE2000 · « **Reply #34 on:** April 09, 2009, 03:39:01 pm »

Quote from: Mao on April 07, 2009, 10:33:27 pm

Quote from: TrueTears on April 07, 2009, 12:43:26 pm
$\int{sin^{-1}(x)}dx$ $= \frac{1}{\sqrt{1^2-x^2}} = \frac{1}{\sqrt{1-x^2}}$

$\int{cos^{-1}(x)}dx$ $= \frac{-1}{\sqrt{1^2-x^2}} = \frac{-1}{\sqrt{1-x^2}}$

That is incorrect, integrating inverse circular functions require the use of 'integration by parts'. This is not in the syllabus of SM.

However, as 'fun' activity because you love math, you might want to try differentiating $x\sin^{-1}(x)$ , and hence find an antiderivative of $\sin^{-1}(x)$

Is it $x\sin^{-1}(x) + \sqrt{1-x^2}} + c$ (although the question wasn't directed at me, couldn't help but try)

And what's the basic idea behind integration by parts?

shinny · « **Reply #35 on:** April 09, 2009, 04:16:20 pm »

Quote from: NE2000 on April 09, 2009, 03:39:01 pm

Quote from: Mao on April 07, 2009, 10:33:27 pm
Quote from: TrueTears on April 07, 2009, 12:43:26 pm
$\int{sin^{-1}(x)}dx$ $= \frac{1}{\sqrt{1^2-x^2}} = \frac{1}{\sqrt{1-x^2}}$

$\int{cos^{-1}(x)}dx$ $= \frac{-1}{\sqrt{1^2-x^2}} = \frac{-1}{\sqrt{1-x^2}}$

That is incorrect, integrating inverse circular functions require the use of 'integration by parts'. This is not in the syllabus of SM.

However, as 'fun' activity because you love math, you might want to try differentiating $x\sin^{-1}(x)$ , and hence find an antiderivative of $\sin^{-1}(x)$

Is it $x\sin^{-1}(x) + \sqrt{1-x^2}} + c$ (although the question wasn't directed at me, couldn't help but try)

And what's the basic idea behind integration by parts?

It's a manipulation of the product rule basically (reverse product rule perhaps), whilst the substitution technique is basically reverse chain rule. You might even be able to just figure it out yourself on how to do it actually.

Mao · « **Reply #36 on:** April 09, 2009, 08:32:22 pm »

$\frac{d}{dx} (uv) = u'v + v'u$

$\implies uv + C = \int u'v + v'u\; dx$

$\implies \int u'v\; dx = uv - \int v' u\; dx$

With this, you can integrate the product of two functions, such as $\int xe^x\; dx$
The art of mastering integration by parts is choosing what is u and what is v, which you learn in first year. In this case, let $u' = e^x$ and $v = x$

NE2000 · « **Reply #37 on:** April 10, 2009, 06:27:50 pm »

Thanks Mao

d0minicz · « **Reply #38 on:** April 11, 2009, 02:41:54 pm »

Find the volume of the region bounded by $y= x^3$ , $y=0$ , $x=2$ about $x=2$ .
Also, waht is the Disk Formula?
thansk

dcc · « **Reply #39 on:** April 11, 2009, 03:22:15 pm »

~~incorrect mathematics~~

disregard that, misread question

d0minicz · « **Reply #40 on:** April 11, 2009, 03:27:26 pm »

Hey the answer says $\frac {16\pi}{5}$
are they wrong?
also i need help on questions where they ask ; "revolving around x=3" , instead of "revolving around the x-axis"
how would i approach these? thanks.

TrueTears · « **Reply #41 on:** April 11, 2009, 03:31:24 pm »

Quote from: d0minicz on April 11, 2009, 03:27:26 pm

Hey the answer says $\frac {16\pi}{5}$
are they wrong?
also i need help on questions where they ask ; "revolving around x=3" , instead of "revolving around the x-axis"
how would i approach these? thanks.

check here: http://vcenotes.com/forum/index.php/topic,9192.msg120718.html#msg120718

ed_saifa · « **Reply #42 on:** April 11, 2009, 03:49:52 pm »

[IMG]http://img11.imageshack.us/img11/3424/ccf1104200900000.th.jpg[/img]

d0minicz · « **Reply #43 on:** April 11, 2009, 04:14:23 pm »

Find the volume of region within the parabola $x= 9-y^2$ and between $y=x-7$ and the y-axis, about the y-axis.
thank you

ed_saifa · « **Reply #44 on:** April 11, 2009, 05:09:02 pm »

[IMG]http://img4.imageshack.us/img4/3424/ccf1104200900000.th.jpg[/img]

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