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Math Help - Another Area Estimate Question

  1. #1
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    Another Area Estimate Question

    The area between the graph of and the x-axis for 1 ≤ x ≤ 4 s estimated using 20 rectangles and x* k is taken to be the right-hand boundary of each interval.

    The result is rounded off to three decimal places.

    The answer is 4.741

    Could someone help explain this? Thanks.
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  2. #2
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    Well maybe a little bit more than that. Okay here's the deal:
    We devided b-a=4-1=3 into 20 rectangles each of width \frac{3}{20}=0.15 then the approximation could be done by adding the rectangles r_k of width \Delta\,x_k\cdot = 0.15 and length/hight f(x_k) that is \sum_{k=1}^{20}\,f(x_k)\Delta x_k=4.8912 whuc will turn out to be very soon to be \int_1^4\,\sqrt{x}\,dx=\frac{14}{3}=4.66666...
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  3. #3
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    I was wondering if I could get a little more explanation with this one. I dont really understand.
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  4. #4
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    Well if you have reached Rieman sums you should be getting that, but I'm assuming you've never heard about the guy. So, when approximating areas under a curve of y=f(x), above the x-axis and between 2 vertical lines lets say x=a and x= b we divide the segment b-a into partitions of width \Delta x_k each... It's like cutting a piece of bread into slices of equal width (the \Delta x_ks don't have to be of equal width though, later to that...). So when you finish dividing the segment into smaller subsegments, and then drawing a recatangle over each subsegment of height = the (average) value of f on that segment (the avergae value of f on [1;1.15] is around 1 because  \sqrt{1} \ \mbox{ almost equal to}  \sqrt{1.15} \ \mbox{ almost equal to}  1)
    Then, you add the areas of these rectangles, each has an area of  \ A= \frac{b-a}{20} \cdot f(c) where c is any point in the interval [x_k;x_{k+1}] , I would go for the edge points(  \ [x_k;x_{k+1}], \ [1;1.15], \ [1.15;1.30],\  [1.30;1.45] and so on...) So when you sum up the whole lot you're doing \frac{b-a}{20} \cdot ( f(x_0)+f(x_1)+...+f(x_{19}) ), note that 20 is your choice, it could ran up to 100, a 1000 or infinity.
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