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#### pacovf

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« Reply #225 on: January 15, 2015, 03:46:40 pm »
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You could also prove that all the solutions to y' = y are proportional to each other, and then prove that exp is a solution. Then you find that 0 is also a solution. You don't even need to define ln to prove this.

PPE ~317: geez people you're like a pack of hungry wolves.
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#### sudgy

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« Reply #226 on: January 15, 2015, 03:56:19 pm »
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You could also prove that all the solutions to y' = y are proportional to each other, and then prove that exp is a solution. Then you find that 0 is also a solution. You don't even need to define ln to prove this.

PPE ~317: geez people you're like a pack of hungry wolves.

Well, there are like 317 different ways to solve differential equations...
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### sudgy

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« Reply #227 on: February 11, 2015, 03:36:21 pm »
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Is there any website like wolframalpha that will give me a step-by-step solution to whatever I put in without making me pay money?
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### scott_pilgrim

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« Reply #228 on: February 11, 2015, 03:40:37 pm »
+3

But wolfram has the best step-by-step solutions!

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#### Ozle

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« Reply #229 on: February 11, 2015, 03:48:33 pm »
+3

Post it here, i'll do it for you!
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#### sudgy

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« Reply #230 on: February 11, 2015, 03:53:26 pm »
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But wolfram has the best step-by-step solutions!

But you have to pay money...
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### pacovf

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« Reply #231 on: February 11, 2015, 03:59:03 pm »
+1

Pretty sure scotty was being sarcastic there, or at the very least facetious.
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#### Witherweaver

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« Reply #232 on: February 11, 2015, 04:00:16 pm »
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Pretty sure scotty was being sarcastic there, or at the very least facetious.

Maybe even sarcetious.  Or facastic.
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#### sudgy

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« Reply #233 on: February 11, 2015, 04:13:20 pm »
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Also, that example is really easy...
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### Witherweaver

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« Reply #234 on: February 11, 2015, 04:25:35 pm »
+1

Is there any website like wolframalpha that will give me a step-by-step solution to whatever I put in without making me pay money?

Step 1: Integrate by parts
Step 2: Profit
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#### pacovf

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« Reply #235 on: February 11, 2015, 04:29:36 pm »
+2

Personally, I prefer the Feynman method:

1. You write down the problem.
2. You think very hard.
3. You write down the answer.
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#### sudgy

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« Reply #236 on: February 11, 2015, 04:33:38 pm »
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Is there any website like wolframalpha that will give me a step-by-step solution to whatever I put in without making me pay money?

Step 1: Integrate by parts
Step 2: Profit

Personally, I prefer the Feynman method:

1. You write down the problem.
2. You think very hard.
3. You write down the answer.

Already tried both of these for the past hour or two.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### pacovf

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« Reply #237 on: February 11, 2015, 04:40:11 pm »
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I don't know of any free automatic solver, sorry...

In the meantime, you could try posting the problem that is bothering you. Ozle has already offered to write the proof down!
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#### sudgy

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« Reply #238 on: February 11, 2015, 06:38:06 pm »
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∫sqrt(1-x2) dx

I've tried a bajillion methods.  I know the answer from wolframalpha, but still can't figure it out.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### Witherweaver

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« Reply #239 on: February 11, 2015, 06:45:01 pm »
+2

∫sqrt(1-x2) dx

I've tried a bajillion methods.  I know the answer from wolframalpha, but still can't figure it out.

Well, sqrt{1-x^2} is naturally the length of a leg of a right triangle if the other leg is x and the hypotenuse is 1.  So draw that triangle.  Call one of the angles theta.  Make a natural substitution (i.e., x = trigfunction(theta)).  Take differentials and plug it into the integral.  Then you have an integral of trig functions, and you can do that.

Edit: The integrand itself, \sqrt{1-x^2}, is also a convenient trig function of theta.  (Also, fixed a typo above.)
« Last Edit: February 11, 2015, 06:46:13 pm by Witherweaver »
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#### liopoil

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« Reply #240 on: February 11, 2015, 06:50:14 pm »
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Disclaimer: I am currently taking Calc. A, so fair chance I have no clue what I'm talking about

-1/3x(1 - x2)3/2 + c

Thinking about what gives sqrt(1 - x2) when you take it's derivative... this is what I get. First I increased the exponent on the outside by one, then multiplied by the reciprocal of the inside function (1 - x2). Then I multiplied by the reciprocal of the new outer exponent, and added the constant of integration.

PPE: It seems witherweaver knows what he's talking about and I don't. That is an integral sign, right? Mind explaining where I err? Something tells me it shouldn't be this simple.
« Last Edit: February 11, 2015, 06:51:51 pm by liopoil »
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#### sudgy

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« Reply #241 on: February 11, 2015, 06:56:45 pm »
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-1/3x(1 - x2)3/2 + c

Differentiating this involves a product rule, making its derivative 1/3 sqrt(1-x2) (4 x2-1), not sqrt(1-x2)
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### liopoil

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« Reply #242 on: February 11, 2015, 06:58:39 pm »
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-1/3x(1 - x2)3/2 + c

Differentiating this involves a product rule, making its derivative 1/3 sqrt(1-x2) (4 x2-1), not sqrt(1-x2)
ah yes, thank you... not so simple when the power on the inside is greater than 1. And wolfram disagrees with me too, so looks like I have no clue.
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#### heron

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« Reply #243 on: February 11, 2015, 07:01:52 pm »
+1

Okay hint: sqrt(1 - x^2) is hard to integrate, darn. Good thing we have techniques for getting around that, namely substitution. What sort of substitution could you make so that sqrt(1 - x^2) turns into something nicer?

Alternative strategy: Apply geometry.
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#### sudgy

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« Reply #244 on: February 11, 2015, 07:10:48 pm »
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Okay hint: sqrt(1 - x^2) is hard to integrate, darn. Good thing we have techniques for getting around that, namely substitution. What sort of substitution could you make so that sqrt(1 - x^2) turns into something nicer?

I tried all sorts of things like this, to no avail.  I tried u = 1 - x2, but that makes du = -2x dx, making it even more confusing.  I tried crazy forms of integration by parts but that didn't work either.

Quote
Alternative strategy: Apply geometry.

I just did what Witherweaver said and it worked.  Yay!
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### Witherweaver

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« Reply #245 on: February 11, 2015, 07:48:40 pm »
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"Geometry" mean "trigonometry" here
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#### silverspawn

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« Reply #246 on: February 11, 2015, 08:20:54 pm »
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I actually have a math exam coming up in a few days. So, I'll also try this (without using geometry)

So, there is this formula which I think works

which you have to read from the right side. So you have
$f(x) = \sqrt{1-x^2}$

you define $g(x) = sin(x)$ and

$f(g(x)) * g'(x) = \sqrt{1-sin^2(x)} * cos(x) = \sqrt{cos^2(x))} * cos(x) = cos^2(x)$

which according to my list of common integrals I made to use in the exam is the derivation of

$\frac{x+\frac{sin(2x)}{2}}{2}$

and so

$\int_{b}^{a} \sqrt{1-x^2} = \int_{\arcsin b}^{\arcsin a} \frac{x + \frac{sin(2x)}{2}}{2}$

edit: ehhm that's of course not an integral anymore on the right side

so... is that correct?

« Last Edit: February 11, 2015, 08:28:37 pm by silverspawn »
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#### heron

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« Reply #247 on: February 11, 2015, 09:11:39 pm »
+1

Okay hint: sqrt(1 - x^2) is hard to integrate, darn. Good thing we have techniques for getting around that, namely substitution. What sort of substitution could you make so that sqrt(1 - x^2) turns into something nicer?

I tried all sorts of things like this, to no avail.  I tried u = 1 - x2, but that makes du = -2x dx, making it even more confusing.  I tried crazy forms of integration by parts but that didn't work either.

Quote
Alternative strategy: Apply geometry.

I just did what Witherweaver said and it worked.  Yay!

You never tried sinu = x though did you
In calc AB class, the teacher never told us that you could do u-substitutions like that, but when you think about it, it is a natural extension of the technique.

As for geometry, I just meant use the area under a curve definition of the integral. It's not too difficult to find the area of those bits of circle.
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#### heron

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« Reply #248 on: February 11, 2015, 09:18:06 pm »
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For additional integration practice, here is a problem:

Given two randomly selected points in a unit square, what is the probability that the distance between them is greater than 1/2?
(2015 AMC A, modified)

The original problem specified that the points lie on the perimeter of the square, which makes the problem significantly easier, and allows you to use geometry rather than calculus.

Edit: Also, sqrt(1 - x^2) would be one of the integrals you would need to evaluate in the original problem.
« Last Edit: February 11, 2015, 09:50:38 pm by heron »
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#### sudgy

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« Reply #249 on: February 11, 2015, 10:19:54 pm »
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As for geometry, I just meant use the area under a curve definition of the integral. It's not too difficult to find the area of those bits of circle.

Well, I knew that, I was trying to find a better reason though.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm
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