Author Topic: Nagisa Maths Thread (Read 26733 times) Tweet Share

Nagisa · « **Reply #15 on:** December 24, 2012, 01:32:25 am »

Quote from: FlorianK on December 24, 2012, 01:26:19 am

TT, why do you destroy my working out with a line ?

he did have working, just didnt wanna show it. that's why he said "why?". to see if anyone knew why

Nagisa · « **Reply #16 on:** December 24, 2012, 02:45:17 am »

find the volume bounded by the curve $x^2+(y-1)^2 = 1$ about the y-axis.

TrueTears · « **Reply #17 on:** December 24, 2012, 02:56:56 am »

$\int_0^2 \pi\left[1-(y-1)^2] dy$

I'll let you think about why.

Nagisa · « **Reply #18 on:** December 24, 2012, 03:02:43 am »

find the volume bounded by the curves $y=x^3$ , $y = 0$ , $x=1$ about $y=1$

TrueTears · « **Reply #19 on:** December 24, 2012, 03:19:41 am »

Spoiler

$\int_0^1 2\pi (1-y)(1-y^{\frac{1}{3}})dy$

Planck's constant · « **Reply #20 on:** December 25, 2012, 09:56:38 am »

Quote from: TrueTears on December 24, 2012, 12:42:40 am

This volume can be found very easily by the following computation:

$\int_1^3 2\pi(x-1)(4x-x^2-3) dx = \frac{8\pi}{3}$

Why?

I had not seen this approach before, but I get it now

We are taught to visualise solids of revolution as disks of thickness dy stacked one on top of each other. But this approach suggests that we can visualise the same solid of revolution as cylinders of thickness dx one inside of each other.
I can see that this approach simplifies problems such as this one.
I like.

Nagisa · « **Reply #21 on:** December 25, 2012, 11:45:59 pm »

Evaluate the integral of $\int (\ln x)^2 \ dx$

Hancock · « **Reply #22 on:** December 25, 2012, 11:58:30 pm »

Using integration by parts.

$\int(log_ex)^2 \ dx$

Let

$u = (log_ex)^2$ and $\frac{dv}{dx} = 1$
$\frac{du}{dx} = \frac{2log_e(x)}{x}$ and $v = x$

$\int {u\frac{dv}{dx}} \ dx$ = $uv - \int {v\frac{du}{dx}}dx$

$\int(log_ex)^2 \ dx = x(log_ex)^2 - \int {x\frac{2log_e(x)}{x}}dx$

$\int(log_ex)^2 \ dx = x(log_ex)^2 - 2\int {log_e(x)}dx$

Using a similar idea, with $u = log_e(x)$ and $\frac{dv}{dx} = 1$

$\int {log_e(x)}dx = xlog_e(x) - x + c$

$\int(log_ex)^2 \ dx = x(log_e(x))^2 - 2(xlog_e(x) - x) + C = x(log_e(x))^2 - 2xlog_e(x) + 2x + C$

TrueTears · « **Reply #23 on:** December 25, 2012, 11:59:10 pm »

Spoiler

$\int (\ln x)^2 \ dx$

$\int u dv = uv - \int v du$

Let $u = ln(x)$ and $dv = ln(x) dx$

So $du = \frac{1}{x} dx$ and $v = x ln(x) - x$

Note to work out $v$ we have $\frac{dv}{dx} = ln(x)$ and we need to integrate $ln(x)$ , but to integrate ln(x) we need to use integration by parts again, so $\int ln(x) dx$ .

In this case let $u = ln(x)$ and $dv = dx$ and you should get $\int ln(x) dx = xln(x) - x$

Now back to the original question: $\int (\ln x)^2 \ dx = ln(x)\left[xln(x) - x\right] - \int ln(x) - 1 dx = ln(x)\left[xln(x) - x\right] -\left[xln(x) - x - x\right] = x\left[ \left(ln(x)\right)^2 -2ln(x) +2\right]$

Phy124 · « **Reply #24 on:** December 26, 2012, 12:05:54 am »

Spoiler

Using integration by parts:

$\int u \frac{dv}{dx}dx = uv - \int v \frac{du}{dx} dx$

(You can "cancel out" dx's and write dv and du, but for some reason I prefer not to)

$\frac{dv}{dx}=1$

$v = x$

$u = \log_e(x)^2$

$\frac{du}{dx}=\frac{2 \log_e(x)}{x}$

$\int \log_e(x)^2dx = x\log_e(x)^2 - \int x \frac{2 \log_e(x)}{x} dx= x \log_e(x)^2 - 2 \int \log_e(x) dx + C$

Need to work out $\int \log_e(x) dx$

$\frac{dv}{dx}=1$

$v = x$

$u = \log_e(x)$

$\frac{du}{dx}=\frac{1}{x}$

$\int \log_e(x) dx = x \log_e(x) - \int x\frac{1}{x} dx = x \log_e(x) - x + C$

$\int \log_e(x)^2dx = xlog_e(x)^2 - \int x \frac{2 \log_e(x)}{x} dx= x\log_e(x)^2 - 2(x \log_e(x) - x) = x(\log_e(x)^2 - 2\log_e(x) +2) + C$

aw beaten again

wow double beaten too haha fuark, gotta lift my game

Quote from: Hancock on December 26, 2012, 12:39:56 am

Couldn't remember how to print fractions and integrals. FUUUUUUUUU

I shouldn't have walked off to get food mid-way through

Planck's constant · « **Reply #25 on:** December 26, 2012, 12:37:04 am »

I am prepared to call this a three-way dead heat .

I am thinking of buying a 'Voice to Latex' app to stay competitive in 2013. It's a tough gig

Hancock · « **Reply #26 on:** December 26, 2012, 12:39:56 am »

Couldn't remember how to print fractions and integrals. FUUUUUUUUU

b^3 · « **Reply #27 on:** December 26, 2012, 12:46:52 am »

Quote from: argonaut on December 26, 2012, 12:37:04 am

I am prepared to call this a three-way dead heat .

I am thinking of buying a 'Voice to Latex' app to stay competitive in 2013. It's a tough gig

Worst thing is when you see the post, latex it up and it turns out that someone posted 30 seconds or so after you initially saw the post.... so all the latexing for nothing...

I'm not refreshing AN as much as I used to..... so hard to actually get in on questions nowdays when I've got other projects to be working on too

Nagisa · « **Reply #28 on:** December 26, 2012, 01:13:19 am »

Evaluate $\int e^{2x}\sin{3x} \ dx$ another one im havin trouble with lolz

Hancock · « **Reply #29 on:** December 26, 2012, 01:26:50 am »

$\int e^{2x}\sin{3x} \ dx$

$Im(\int e^{2x}e^{3ix} \ dx)$

$Im(\int e^{(2+3i)x}\ dx)$

$Im(\frac{1}{2+3i}e^{(2+3i)x})$

$Im(\frac{2-3i}{13}e^{2x}e^{3ix})$

$Im(\frac{2-3i}{13}e^{2x}(cos(3x) + isin(3x))$

$\frac{e^{2x}}{13}Im((2-3i)(cos(3x) + isin(3x))$

$\frac{e^{2x}}{13}(2sin(3x)-3cos(3x)) + C$

Lolz, forgot the +C.

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Author Topic: Nagisa Maths Thread (Read 26733 times) Tweet Share

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