November 15th, 2010, 02:45 PM  #1 
Member Joined: Feb 2009 Posts: 54 Thanks: 0  2 questions.

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: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  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. 
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: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  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 WolframAlpha 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. 
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 is x=3 The lowest point for 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: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  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 
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. (for god's shake )


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