My Math Forum Rectangle Area Approximation

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 January 10th, 2012, 05:39 PM #1 Member   Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 Rectangle Area Approximation I have to approximate the area under the curve of a graph, but I don't get how to do it at all. My teacher had to hurry, so he gave us odd problems (with the answers in the back so I'm not asking for the answer). The problem says: A particle starts at x = 0 and moves along the x-axis with velocity $v(t)\,=\,t^2\,+\,1$ for time $t\,\underline{>}\,0$. Where is the particle at t = 4[edit: I accidentally said 5 before]? Approximate the area under the curve using four rectangles of equal widths and heights determined by the midpoints of the intervals, as in Example 1. The answer says: Each rectangle has base 1. The area under the curve is approximately $1(\,\frac{5}{4}\,+\,\frac{13}{4}\,+\,\frac{29}{4}\ ,+\,\frac{53}{4}\,)\,=\,25$, so the particle is close to x=25. If you can just explain how to work it in the form my book does, I would be very appreciative.
 January 10th, 2012, 06:06 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rectangle Area Approximation If each rectangle has a base of 1, then there must be 5 rectangles. To have 4 rectangles, the base of each would be 5/4. And the sum of the areas A would be: $A=\frac{5}{4}\sum_{k=0}^4v$$k\cdot\frac{5}{4}+\fra c{5}{8}$$=\frac{5}{4}$$v\(\frac{5}{8}$$+v$$\frac{1 5}{8}$$+v$$\frac{25}{8}$$+v$$\frac{35}{8}$$\)=$ $\frac{5}{4}$$\frac{5^2+15^2+25^2+35^2+256}{64}$$=\ frac{2945}{64}=46.015625$ We can now compare this to the actual value: $\int_0\,^{x(5)}\,dx=\int_0\,^5 t^2+1\,dx$ $x(5)=\frac{1}{3}5^3+5=\frac{140}{3}=46.\bar{6}$ I assume you meant to approximate x(4)?
January 10th, 2012, 06:18 PM   #3
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Re: Rectangle Area Approximation

Quote:
 Originally Posted by MarkFL If each rectangle has a base of 1, then there must be 5 rectangles. To have 4 rectangles, the base of each would be 5/4. And the sum of the areas A would be: $A=\frac{5}{4}\sum_{k=0}^4v$$k\cdot\frac{5}{4}+\fra c{5}{8}$$=\frac{5}{4}$$v\(\frac{5}{8}$$+v$$\frac{1 5}{8}$$+v$$\frac{25}{8}$$+v$$\frac{35}{8}$$\)=$ $\frac{5}{4}$$\frac{5^2+15^2+25^2+35^2+256}{64}$$=\ frac{2945}{64}=46.015625$ We can now compare this to the actual value: $\int_0\,^{x(5)}\,dx=\int_0\,^5 t^2+1\,dx$ $x(5)=\frac{1}{3}5^3+5=\frac{140}{3}=46.\bar{6}$ I assume you meant to approximate x(4)?
Crap! I meant to say at point t=4! Could you explain it that way?

 January 10th, 2012, 06:28 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Rectangle Area Approximation I should have caught by just looking at the answer given by your book! But I blindly plodded forward. Anyway, we have 4 rectangles of base 1. The mid-point of the first rectangle is $t=\frac{1}{2}$ and we find by evaluating v(t) at t = 1/2 to get the height of the rectangle at the mid-point of the interval: $v$$\frac{1}{2}$$=$$\frac{1}{2}$$^2+1=\frac{1}{4}+1 =\frac{5}{4}$ We add 1 to the previous mid-point to get $t=\frac{3}{2}$ and we find: $v$$\frac{3}{2}$$=$$\frac{3}{2}$$^2+1=\frac{9}{4}+1 =\frac{13}{4}$ We add 1 to the previous mid-point to get $t=\frac{5}{2}$ and we find: $v$$\frac{5}{2}$$=$$\frac{5}{2}$$^2+1=\frac{25}{4}+ 1=\frac{29}{4}$ We add 1 to the previous mid-point to get $t=\frac{7}{2}$ and we find: $v$$\frac{7}{2}$$=$$\frac{7}{2}$$^2+1=\frac{49}{4}+ 1=\frac{53}{4}$ So, factoring out the base of 1 and adding the heights, we get: $A=1$$\frac{5+13+29+53}{4}$$=\frac{100}{4}=25$ Compare this to the actual value of $25.\bar{3}$.
January 10th, 2012, 06:47 PM   #5
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Re: Rectangle Area Approximation

Thank you! That's much more simple than I had hoped!
Quote:
 Originally Posted by MarkFL I should have caught by just looking at the answer given by your book! But I blindly plodded forward. Anyway, we have 4 rectangles of base 1. The mid-point of the first rectangle is $t=\frac{1}{2}$ and we find by evaluating v(t) at t = 1/2 to get the height of the rectangle at the mid-point of the interval: $v$$\frac{1}{2}$$=$$\frac{1}{2}$$^2+1=\frac{1}{4}+1 =\frac{5}{4}$ We add 1 to the previous mid-point to get $t=\frac{3}{2}$ and we find: $v$$\frac{3}{2}$$=$$\frac{3}{2}$$^2+1=\frac{9}{4}+1 =\frac{13}{4}$ We add 1 to the previous mid-point to get $t=\frac{5}{2}$ and we find: $v$$\frac{5}{2}$$=$$\frac{5}{2}$$^2+1=\frac{25}{4}+ 1=\frac{29}{4}$ We add 1 to the previous mid-point to get $t=\frac{7}{2}$ and we find: $v$$\frac{7}{2}$$=$$\frac{7}{2}$$^2+1=\frac{49}{4}+ 1=\frac{53}{4}$ So, factoring out the base of 1 and adding the heights, we get: $A=1$$\frac{5+13+29+53}{4}$$=\frac{100}{4}=25$ Compare this to the actual value of $25.\bar{3}$.

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# a particle starts at x=0 and moves along the x-axis with velocity t^2 1

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