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anonymous
 one year ago
Use the method of quadrature to estimate the area under the curve 110207060000_v2_files/i0040000.jpg that is above the xaxis from x = 0 to x = 3.
a.
2.3
c.
27.6
b.
55.3
d.
1.1
anonymous
 one year ago
Use the method of quadrature to estimate the area under the curve 110207060000_v2_files/i0040000.jpg that is above the xaxis from x = 0 to x = 3. a. 2.3 c. 27.6 b. 55.3 d. 1.1

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Try attaching the image.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y =\frac{ 1 }{ 4 }x ^{2}+10\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great. Is your method of quadrature the trapezoidal rule, or Simpson's rule? (or other?)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles im not sure

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Left 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
 one year ago
Best ResponseYou've already chosen the best response.0Since 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
 one year ago
Best ResponseYou've already chosen the best response.0So, \[\int_0^3\left(\frac{1}{4}x^2+10\right)\,dx\approx \sum(\text{areas of trapezoids})\] 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
 one year ago
Best ResponseYou've already chosen the best response.0What does that mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What does *what* mean?
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