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Math Help - Trapezoid Rule is really ruling, need to be free, help!

  1. #1
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    Trapezoid Rule is really ruling, need to be free, help!

    --------------------------------------------------------------------------------

    I am sitting here stuck on these three problems. Please help me!

    1. Find an approximation for the area in the first quadrant between the x-axis and the curve y=4-(x-2)^2 using 4 equally spaced intervals and a Left Hand Rieman Sum.

    Im not sure about the left hand part. But I've tried and I come up with 10 and 10.666.



    2. I have graph the function of y=1/x, where x goes from 0 to 5. Using the trapexoid rule with 4 equal intervals, I must approximate the area under the curve between x=1 and x=3.
    Please help.


    Okay, this is what I have for the second one so far. But for some reason I keep missing it from here. I know its just a matter of calculating but I just cant get it right.

    f(x) = 1/x
    a = 1
    b = 3
    n = 4

    T = [(b - a)/(2n)]*[f(x0) + 2f(x1) + 2f(x2) + ... + 2f(xn-1) + f(xn)]

    T = [(3 - 1)/(2*4)]*[f(1) + 2f(3/2) + 2f(2) + 2f(5/2) + f(3)]

    T = (1/4)[1 + 2(2/3) + 2(1/2) + 2(2/5) + (1/3)]

    Now for the next trapezoid problem(ruler).

    If the trapezoid rule is used with 5 interval,, what is the integral dx/1 + x^2 ,with limits 1,0.

    Please, please help!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by roseh
    --------------------------------------------------------------------------------

    I am sitting here stuck on these three problems. Please help me!

    1. Find an approximation for the area in the first quadrant between the x-axis and the curve y=4-(x-2)^2 using 4 equally spaced intervals and a Left Hand Rieman Sum.

    Im not sure about the left hand part. But I've tried and I come up with 10 and 10.666.
    You are being asked to estimate:

    <br />
\int_0^4 y(x) dx=\int_0^4 4-(x-2)^2\ dx<br />

    using 4 equal sub-intervals of [0,4]. These sub-intervals are:

    [0,1], [1,2], [2,3] and [3,4].

    As we are instructed to use left hand Riemann sums we will approximate the
    function on each of these subintervals by its value at the left hand point of
    the sub-interval, so:

    <br />
\int_0^4 y(x) dx\approx (y(0)\times \delta x) +(y(1) \times \delta x)  + (y(2) \times \delta x) + (y(3) \times \delta x)<br />
,

    where \delta x is the width of the intervals in this case \delta x=1

    Hence:

    <br />
\int_0^4 y(x) dx\approx (0\times \delta x) +(3 \times \delta x)  + (4 \times \delta x) + (3 \times \delta x)=10<br />
.

    This may be compared with the result of doing the integral analytically, which
    gives an area of 32/3 \approx 10.667.

    RonL
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  3. #3
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    Quote Originally Posted by roseh
    -----------------------------------------------2. I have graph the function of y=1/x, where x goes from 0 to 5. Using the trapexoid rule with 4 equal intervals, I must approximate the area under the curve between x=1 and x=3.
    You need to approximate,
    \int^3_1\frac{dx}{x}
    You are using 4 equal intervals thus, n=4.
    Next, you need equal widths which is \Delta x=\frac{b-a}{n}=\frac{2}{4}=.5.
    Now, by the trapezoidal rule,
    \frac{1}{2}[f(a)+2f(a+\Delta x)+2f(a+2\Delta x)+ 2f(a+3\Delta x)+f(a+4\Delta x)] \Delta x
    But in your values,
    \frac{1}{2}[f(1)+2f(1.5)+2f(2)+2f(2.5)+f(3)].5
    Now, find the value f(x)=\frac{1}{x},
    \frac{1}{2}[1+1.33+1+.8+.33].5\approx 1.11
    -----
    Now, compare to the actual vale. This is what I am afraid of I do not know if you studied natrual logarithms. In that case, skip this part.
    The actual value of,
    \int^3_1\frac{dx}{x}=\ln 3
    because this is the very definition of the natural logarithm. But \ln 3\approx 1.09

    Notice that you error by using only 4 subintervals is only .02
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