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Math Help - Question about area under a curve using midpoint rule

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    Question about area under a curve using midpoint rule

    okay, the question says to use the midpoint rule with n=4 to approximate the area of the region bounded by the graph of the function and the x axis over the indicated interval

    function is f(x)= x^2 + 4x interval is [0,4]

    midpoint rule

    sigma from i=1 to n of f((xi-xi-1)/2) times delta x

    thanks very much
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by doctorgk View Post
    okay, the question says to use the midpoint rule with n=4 to approximate the area of the region bounded by the graph of the function and the x axis over the indicated interval

    function is f(x)= x^2 + 4x interval is [0,4]

    midpoint rule

    sigma from i=1 to n of f((xi-xi-1)/2) times delta x

    thanks very much
    why is this in the pre-calc section?

    divide [0,4] into 4 equal sub-intervals, we get the intervals: [0,1],~[1,2],~[2,3],~[3,4]

    obviously (i hope) we have \Delta x = 1 here. if you're not sure, remember \Delta x = \frac {b - a}{n} where [a,b] is the interval we are considering, and n, of course, is the number of sub-intervals

    the midpoints of the subintervals respectively are: x_1 = \frac 12,~x_2 = \frac 32,~ x_3 = \frac 52,~x_4 = \frac 72

    so, by the midpoint rule, our estimate for the integral is given by:

    \int_0^4f(x)~dx = \sum_{i = 1}^{4} f(x_i) \Delta x = \Delta x \left( f \left( \frac 12 \right) + f \left( \frac 32 \right) + f \left( \frac 52 \right) + f \left( \frac 72 \right) \right)

    = f \left( \frac 12 \right) + f \left( \frac 32 \right) + f \left( \frac 52 \right) + f \left( \frac 72 \right)

    i leave the rest to you
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