anonymous
  • anonymous
Use the method of quadrature to estimate the area under the curve 1102-07-06-00-00_v2_files/i0040000.jpg that is above the x-axis from x = 0 to x = 3. a. 2.3 c. 27.6 b. 55.3 d. 1.1
Geometry
schrodinger
  • schrodinger
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anonymous
  • anonymous
Try attaching the image.
anonymous
  • anonymous
y =-\frac{ 1 }{ 4 }x ^{2}+10\]
anonymous
  • anonymous
Great. Is your method of quadrature the trapezoidal rule, or Simpson's rule? (or other?)

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anonymous
  • anonymous
@SithsAndGiggles im not sure
anonymous
  • anonymous
Left to our own devices, are we? Okay, not a problem. If we end up not using the expected method, it won't matter much; our answer should be approximately equal. I'll go ahead and use trapezoidal partitions. |dw:1440869313600:dw|
anonymous
  • anonymous
Since a trapezoidal approximation converges pretty well, I won't use more than 3 partitions, that should do just fine. (Plus the interval can be split very easily.) |dw:1440869436882:dw|
anonymous
  • anonymous
So, \[\int_0^3\left(-\frac{1}{4}x^2+10\right)\,dx\approx \sum(\text{areas of trapezoids})\] Recall the area of a trapezoid: |dw:1440869582894:dw| so you have \[\int_0^3\left(-\frac{1}{4}x^2+10\right)\,dx\approx \frac{f(0)+f(1)}{2}+\frac{f(1)+f(2)}{2}+\frac{f(2)+f(3)}{2}\] since \(C=1\) for all trapezoids. Here, \(f(x)=-\dfrac{1}{4}x^2+10\).
anonymous
  • anonymous
What does that mean?
anonymous
  • anonymous
What does *what* mean?

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