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November 15th, 2010, 01:45 PM   #1
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2 questions.

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November 15th, 2010, 02:30 PM   #2
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Re: 2 questions.

I take it that you now know integration. If you integrate a function from a to b, then you find the area under the function f between the points a and b. What do you think the limits of integration are here?
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November 15th, 2010, 02:35 PM   #3
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Re: 2 questions.

2.) We have

Using Newton's method, we have the recursive formula:

, where



We can easily see the root lies between 1 and 2, so choose an in that interval and use the iterative or recursive formula found to approximate the root to the desired accuracy.

edit: didn't see your post, jason.spade, so I removed my information regarding problem 1.
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November 15th, 2010, 02:52 PM   #4
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Re: 2 questions.

I see. I appreciate that - thank you.
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November 15th, 2010, 03:41 PM   #5
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Re: 2 questions.

And how do I use method of iteration? I need more detailed explanation.
jason.spade What do you mean limits of integration?
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November 15th, 2010, 04:00 PM   #6
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Re: 2 questions.

We found:



This is what's known as a recursive formula, because given an we can find and then reiterate or reuse the formula using the previous as the current to find the next and so on, and we will find (with most functions at least, this one included) that will converge to a root of f(x).

Since we know the root lies between 1 and 2, let's let . Then:








After only 6 iterations I have reached the limit of precision of my calculator.

Using Wolfram|Alpha we have



The exact form of the real root is:



As far as the first problem, the limits of integration are a and b in the expression

You need to find where the two given functions intersect to determine these. You also need to determine which function is the upper function and which is the lower and integrate their difference (i.e. upper - lower) on the interval you find.
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November 15th, 2010, 04:44 PM   #7
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Re: 2 questions.

For first problem.
functions meet at X=0, 4 Y=2, -6
The lowest point for is x=3
The lowest point for is x=1
Now what?
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November 15th, 2010, 04:58 PM   #8
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Re: 2 questions.

The extremum you found for (which is what you originally gave) is a maximum, via the 2nd derivative test.

meaning the concavity is downward, meaning the extremum is a maximum. This function is a parabola opening downward, as indicated by the negative sign in front of the squared term.

Thus is the top function and is the bottom function.

so to find the area A we use

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November 15th, 2010, 05:06 PM   #9
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Re: 2 questions.

What is antiderivative of 4X?
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November 15th, 2010, 05:09 PM   #10
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Re: 2 questions.

(for god's shake )
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