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Math Help - Area Approx. with Rectangles

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    Area Approx. with Rectangles

    Area Approx. with Rectangles-2x-3-4.jpgI'm trying to get an area approximation under the graph,  f(x) = 2x^{3} + 4 from  x = -1 to  x = 5 by using 6 and 12 left, right, and midpoint rectangles. I've computed 6 LE and 6 & 12 for both the RE and MP rectangles but I can't figure out how to correctly do 12 rectangles for LE and MP rectangles. I already know I got the right answer for the ones I mentioned, but I can't figure out what I'm doing wrong for 12 LE and 12 MP rectangles.

     \Delta x = 1 for 6 rectangles, and  \Delta x = \frac{1}{2} for 12 rectangles.

    I'll write out all my summations that I used my calculator for. I marked the incorrect calculations with **

    6 LE  \Sigma^{4}_{i=-1} (0.5(f(-1)+f(0)+f(1)+f(2)+f(3)+f(4))) is 222

    **12 LE  \Sigma^{4}_{i=-1}  {0.5(f(-1)+f(-0.5)+f(0)+ f(0.5)+f(1)+f(1.5)+f(2)+f(2.5)+f(3)+f(3.5)+f(4))} ... I got 182.875

    6 RE  \Sigma^{5}_{i=0} 0.5(f(0)+f(1)+f(2)+f(3)+f(4)+f(5)) is 474

    12 RE  \Sigma^{5}_{i=0} {0.5(f(0)+ f(0.5)+f(1)+f(1.5)+f(2)+f(2.5)+f(3)+f(3.5)+f(4)+f(  4.5)+f(5))} is 402

    6 MP  \Sigma^{4.5}_{i=-0.5} 1(f(-0.5)+f(0.5)+f(1.5)+f(2.5)+f(3.5)+f(4.5)) is 330

    **12 MP  \Sigma^{4.25}_{i=-0.75} {0.5(f(-0.75)+f(-0.25)+f(0.25)+f(0.75)+f(1.25)+f(1.75)+f(2.25)+f(2.  75)+f(3.25)+f(3.75)+f(4.25))} ... I got 225.328125
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    Re: Area Approx. with Rectangles

    You have forgotten about f(-0.5) in your 12RE.
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    Re: Area Approx. with Rectangles

    Quote Originally Posted by Prove It View Post
    You have forgotten about f(-0.5) in your 12RE.
    I thought for a right end approximation that the counter starts at 1+ whatever 'a' is. So, in this case if it goes from -1 to 5, the counter would start at 0?
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    Re: Area Approx. with Rectangles

    If you had 6 points, yes, because the change in x is 1.

    If you have 12 points, you start 1/2 a unit from a, because the change in x is 1/2.
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    Re: Area Approx. with Rectangles

    Oh that makes a lot more sense. Also, in the graph that I attached, does it look like there are any points on it where f(x) is 0? What confuses me is I've seen some videos on YouTube doing this approximations and certain x values would be skipped because the rectangle had no height (no area) at that point.
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    Re: Area Approx. with Rectangles

    Just a \LaTeX tip:

    For summations, use the \sum command. Look at the difference:

    \Sigma_{k=1}^nk=\frac{n(n+1)}{2}: \Sigma_{k=1}^nk=\frac{n(n+1)}{2}

    \sum_{k=1}^nk=\frac{n(n+1)}{2}: \sum_{k=1}^nk=\frac{n(n+1)}{2}
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    Re: Area Approx. with Rectangles

    Quote Originally Posted by AZach View Post
    Oh that makes a lot more sense. Also, in the graph that I attached, does it look like there are any points on it where f(x) is 0? What confuses me is I've seen some videos on YouTube doing this approximations and certain x values would be skipped because the rectangle had no height (no area) at that point.
    If the x intercept is one of the endpoints of your interval of integration, you can't disregard it. Draw the graph and the rectangles if you don't believe me...
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    Re: Area Approx. with Rectangles

    I tried to do this to compute the left hand sum with 12 rectangles.  \sum_{i=-1}^{11} f(xi)dx where  f(xi) = 2(-1+\frac{1i}{2})^{3})+4 since  xi = -1 + \frac{1}{2}i

    And then I basically expanded the summation out
     [2(-1 - \frac{1}{2})^{3} + 4] + [2(-1 - 0)^{3} + 4] + [2(-1 + \frac{1}{2})^{3} + 4] + [2(-1 + 1)^{3} + 4] + [2(-1 + \frac{3}{2})^{3} + 4] + [2(-1 + 2)^{3} + 4] + [2(-1 + \frac{5}{2})^{3} + 4] + [2(-1 + 3)^{3} + 4] + [2(-1 + \frac{7}{2})^{3} + 4] + [2(-1 + 4)^{3} + 4] + [2(-1 + \frac{9}{2})^{3} + 4] + [2(-1 + 5)^{3} + 4] + [2(-1 + \frac{11}{2})^{3} + 4]


    I know that's a lot to look at, but I basically just plug  i = -1 , i = 0, i = 1 ... i = 11 into f(xi) ... Does this look like the proper method to find the area under the curve of the function I listed originally  f(x) = 2x^{3} + 4
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    Re: Area Approx. with Rectangles

    Quote Originally Posted by AZach View Post
    I tried to do this to compute the left hand sum with 12 rectangles.  \sum_{i=-1}^{11} f(xi)dx where  f(xi) = 2(-1+\frac{1i}{2})^{3})+4 since  xi = -1 + \frac{1}{2}i

    And then I basically expanded the summation out
     [2(-1 - \frac{1}{2})^{3} + 4] + [2(-1 - 0)^{3} + 4] + [2(-1 + \frac{1}{2})^{3} + 4] + [2(-1 + 1)^{3} + 4] + [2(-1 + \frac{3}{2})^{3} + 4] + [2(-1 + 2)^{3} + 4] + [2(-1 + \frac{5}{2})^{3} + 4] + [2(-1 + 3)^{3} + 4] + [2(-1 + \frac{7}{2})^{3} + 4] + [2(-1 + 4)^{3} + 4] + [2(-1 + \frac{9}{2})^{3} + 4] + [2(-1 + 5)^{3} + 4] + [2(-1 + \frac{11}{2})^{3} + 4]


    I know that's a lot to look at, but I basically just plug  i = -1 , i = 0, i = 1 ... i = 11 into f(xi) ... Does this look like the proper method to find the area under the curve of the function I listed originally  f(x) = 2x^{3} + 4
    I think you're relying too much on the formula. Actually think about what you are doing. You are approximating the area under a curve by subdividing it into rectangles and adding up the areas of all the rectangles. The length of each rectangle is the same as the length of each subdivision, and the width of each rectangle is the value of the function at the left-hand endpoint. So that means the area of each rectangle is \displaystyle \begin{align*} f(x_i) \Delta x \end{align*}, where \displaystyle \begin{align*} x_i \end{align*} is the left-hand endpoint. Remembering that your interval is \displaystyle \begin{align*} \frac{1}{2} \end{align*} a unit in length, that means your approximation is

    \displaystyle \begin{align*} A \approx \frac{1}{2}f(-1) + \frac{1}{2}f\left(-\frac{1}{2}\right) + \frac{1}{2}f(0) + \frac{1}{2}f\left(\frac{1}{2}\right) + \dots + \frac{1}{2}f\left(\frac{9}{2}\right) \end{align*}
    Thanks from AZach
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    Re: Area Approx. with Rectangles

    I see what's going on now. I had been mostly just rearranging the numbers without really using the graph.

    Thanks for pointing that out, Plato.
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    Re: Area Approx. with Rectangles

    *Prove It :P
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    Re: Area Approx. with Rectangles

    My mistake. I'm not sure why I thought you were Plato...

    The 12 MP approximation came out to be  0.5 ((f(-0.75)+ .... + f(4.75)) ... It's so much easier to find a starting point and ending point on a graph and then just add the 'change in x'.
    Last edited by AZach; November 29th 2012 at 11:21 AM.
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