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 November 15th, 2010, 02:45 PM #1 Member   Joined: Feb 2009 Posts: 54 Thanks: 0 2 questions. Find area between functions $y=x^2-6x+2$ and $y=x(2-x)+2$[/*:m:3cgn8y1c] Solve equation $x^3=x+4$ with http://en.wikipedia.org/wiki/Newton%27s_method[/*:m:3cgn8y1c]
 November 15th, 2010, 03:30 PM #2 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 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?
 November 15th, 2010, 03:35 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: 2 questions. 2.) We have $f(x)=x^3-x-4=0$ Using Newton's method, we have the recursive formula: $x_{n+1}=x_n-\frac{f(x_n)}{f#39;(x_n)}$, where $f'(x)=3x^2-1$ $x_{n+1}=x_n-\frac{x_n^3-x_n-4}{3x_n^2-1}=\frac{2(x_n^3+2)}{3x_n^2-1}$ We can easily see the root lies between 1 and 2, so choose an $x_0$ 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.
 November 15th, 2010, 03:52 PM #4 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: 2 questions. I see. I appreciate that - thank you.
 November 15th, 2010, 04:41 PM #5 Member   Joined: Feb 2009 Posts: 54 Thanks: 0 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?
 November 15th, 2010, 05:00 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: 2 questions. We found: $x_{n+1}=\frac{2(x_n^3+2)}{3x_n^2-1}$ This is what's known as a recursive formula, because given an $x_n$ we can find $x_{n+1}$ and then reiterate or reuse the formula using the previous $x_{n+1}$ as the current $x_n$ to find the next $x_{n+1}$ and so on, and we will find (with most functions at least, this one included) that $x_n$ will converge to a root of f(x). Since we know the root lies between 1 and 2, let's let $x_0=1.5$. Then: $x_1=\frac{2(1.5^3+2)}{3\cdot 1.5^2-1}\approx 1.86956521739$ $x_2=\frac{2(x_1^3+2)}{3x_1^2-1}\approx 1.79945240612$ $x_3=\frac{2(x_2^3+2)}{3x_2^2-1}\approx 1.79632797085$ $x_4=\frac{2(x_3^3+2)}{3x_3^2-1}\approx 1.79632190328$ $x_5=\frac{2(x_4^3+2)}{3x_4^2-1}\approx 1.79632190326$ $x_6=\frac{2(x_5^3+2)}{3x_5^2-1}\approx 1.79632190326$ After only 6 iterations I have reached the limit of precision of my calculator. Using Wolfram|Alpha we have $x\approx 1.796321903259441535050372726533569720184$ The exact form of the real root is: $x=\frac{1}{3}\sqrt[3]{53-3\sqrt{321}}+\frac{\sqrt[3]{18+\sqrt{321}}}{3^{\frac{2}{3}}$ As far as the first problem, the limits of integration are a and b in the expression $\int^b_a f(x)\,dx$ 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.
 November 15th, 2010, 05:44 PM #7 Member   Joined: Feb 2009 Posts: 54 Thanks: 0 Re: 2 questions. For first problem. functions meet at X=0, 4 Y=2, -6 The lowest point for $y=x^2-6x+2$ is x=3 The lowest point for $y=2x-x^2+3$ is x=1 Now what?
 November 15th, 2010, 05:58 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: 2 questions. The extremum you found for $y=2x-x^2+2$ (which is what you originally gave) is a maximum, via the 2nd derivative test. $y''(x)=-2$ 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 $y_1=-x^2+2x+2$ is the top function and $y_2=x^2-6x+2$ is the bottom function. $y_1-y_2=(-x^2+2x+2)-(x^2-6x+2)=-2x^2+8x=-2(x^2-4x)$ so to find the area A we use $A=-2\int^4_0 x^2-4x\,dx$
 November 15th, 2010, 06:06 PM #9 Member   Joined: Feb 2009 Posts: 54 Thanks: 0 Re: 2 questions. What is antiderivative of 4X?
 November 15th, 2010, 06:09 PM #10 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: 2 questions. $2x^2$ (for god's shake )

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