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Thread: Midpoint Rule

  1. #1
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    Midpoint Rule

    midpoint rule to approximate the integral where n=3
    upper limit=11 ; lower limit=5 (5x+2x^2)dx

    so far i've got delta x and just cant get the numbers to plug into the original function
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  2. #2
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    Mid-point rule

    Hello sjara
    Quote Originally Posted by sjara View Post
    midpoint rule to approximate the integral where n=3
    upper limit=11 ; lower limit=5 (5x+2x^2)dx

    so far i've got delta x and just cant get the numbers to plug into the original function
    $\displaystyle \Delta = \frac{11-5}{3}=2$. Do you agree?

    So we want three rectangles, each of width $\displaystyle 2$ units, starting at $\displaystyle x=5$, ending at $\displaystyle x=11$; total width $\displaystyle = 3\times2 = 6$

    The mid-point of the first rectangle, then, is at $\displaystyle x = 5 + \tfrac12\Delta = 5 + 1=6$. So this is the first value of $\displaystyle x$ to plug into the function; i.e.
    $\displaystyle f(6) = 5\times6 + 2\times 6^2 = 30 +72 = 102$
    This is the height of the first rectangle, then. Its area is therefore $\displaystyle 102 \times 2$.

    The next value of $\displaystyle x$ is at the mid-point of the second rectangle; i.e. at $\displaystyle x = 6 + 2 = 8$; and the final value will be at $\displaystyle x = 8+2 = 10$.

    Work out $\displaystyle f(8)$ and $\displaystyle f(10)$, and then the total area.

    I make the answer $\displaystyle 1040$. Do you?

    Grandad
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