anonymous
  • anonymous
Use the Trapezoidal Rule with n = 4 to approximate the integral from negative 1 to 0 of 1 over the square of the quantity x minus 1, dx.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
https://gyazo.com/00654619772734b85d9d1d0e0bf61444
anonymous
  • anonymous
@jim_thompson5910 @Kenshin
jim_thompson5910
  • jim_thompson5910
Trapezoidal Rule: \[\int_{a}^{b}f(x)dx = \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+\cdots+2*f(x_{n-2})+2*f(x_{n-1})+f(x_n)\right]\] where \[\Large \Delta x = \frac{b-a}{n}\]

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jim_thompson5910
  • jim_thompson5910
in this case a = -1 b = 0 n = 4 \[\Large \Delta x = \frac{b-a}{n}\] \[\Large \Delta x = \frac{0-(-1)}{4}\] \[\Large \Delta x = \frac{1}{4}\] \[\Large \Delta x = 0.25\]
jim_thompson5910
  • jim_thompson5910
If n = 4, then \[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+2*f(x_3)+f(x_4)\right]\]
anonymous
  • anonymous
I'm quite puzzled, especially with that last step
jim_thompson5910
  • jim_thompson5910
this? \[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+2*f(x_3)+f(x_4)\right]\]
anonymous
  • anonymous
Yes
jim_thompson5910
  • jim_thompson5910
do you recall from your lesson this formula? \[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+\cdots+2*f(x_{n-2})+2*f(x_{n-1})+f(x_n)\right]\] btw I'm referencing this page http://www.mathwords.com/t/trapezoid_rule.htm
jim_thompson5910
  • jim_thompson5910
sorry the formula is getting cut off, let me retry
anonymous
  • anonymous
Yea, I've been taking a look at that page too
jim_thompson5910
  • jim_thompson5910
\[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\cdots+2*f(x_{n-2})+2*f(x_{n-1})+f(x_n)\right]\]
jim_thompson5910
  • jim_thompson5910
that's the general formula for any integer n if n were 4, then we have 5 terms we deal with, which are f(x0), f(x1), f(x2), f(x3), f(x4)
anonymous
  • anonymous
Ok, I think I see where you're going with this
jim_thompson5910
  • jim_thompson5910
we'll need the values of x0 through x4 they are... \[\large x_0 = a = -1\] \[\large x_1 = x_0 + \Delta x = -1+0.25 = -0.75\] \[\large x_2 = x_1 + \Delta x = -0.75+0.25 = -0.5\] \[\large x_3 = x_2 + \Delta x = -0.5+0.25 = -0.25\] \[\large x_4 = x_3 + \Delta x = -0.25+0.25 = 0\] ------------------------------------------------------- so in summary \[\large x_0= -1\] \[\large x_1= -0.75\] \[\large x_2 = -0.5\] \[\large x_3= -0.25\] \[\large x_4 = 0\]
jim_thompson5910
  • jim_thompson5910
then you plug those values into \[\int_{a}^{b}f(x)dx \approx \frac{\Delta x}{2}*\left[f(x_0)+2*f(x_1)+2*f(x_2)+2*f(x_3)+f(x_4)\right]\] and you evaluate
anonymous
  • anonymous
I arrived at 1.0178, is that correct?
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
what are f(x0), f(x1), f(x2), f(x3), f(x4) equal to?
anonymous
  • anonymous
Just a moment
anonymous
  • anonymous
@BarfBag \[f(x)=\frac{ 1 }{ (x-1)^2 }\] You would plug in each value given by @jim_thompson5910 to get what f(x0), f(x1), f(x2), f(x3), f(x4) are equal to.
anonymous
  • anonymous
f(x0) = 1/4, f(x1) = 0.327, f(x2) = 0.444, f(x3) = 0.64, f(x4) = 1
jim_thompson5910
  • jim_thompson5910
looks good @BarfBag
jim_thompson5910
  • jim_thompson5910
now compute `f(x0) + 2*f(x1) + 2*f(x2) + 2*f(x3) + f(x4)` and tell me what you get
anonymous
  • anonymous
I got 4.072
anonymous
  • anonymous
Multiply that by dx/2, correct?
jim_thompson5910
  • jim_thompson5910
correct
anonymous
  • anonymous
Multiply it by (1/4)/2
anonymous
  • anonymous
I ended up with B as my final answer
anonymous
  • anonymous
0.5089
anonymous
  • anonymous
B should be correct.
jim_thompson5910
  • jim_thompson5910
yep it's B) 0.5089
anonymous
  • anonymous
Thank you! That took over an hour to solve so I'm very appreciative of your help. XD

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