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Math Help - Using Midpoint rule to appromiate area bounded by curvves

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    Using Midpoint rule to appromiate area bounded by curvves

    I thought I had the process but I was off apparently. The equations are y=sin^2(pi x/4), y=cos^2(pi x/4) 0, greater or equal to x, greater or equal to 1)
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    Quote Originally Posted by sfgiants13 View Post
    I thought I had the process but I was off apparently. The equations are y=sin^2(pi x/4), y=cos^2(pi x/4) 0, greater or equal to x, greater or equal to 1)
    The directions of the inequalities are wrong, but assume you mean for 0 \le x \le 1. Then area is:

    A=\int_{x=0}^1 \cos^2(\pi x/4)-\sin^2(\pi x/4)\ dx

    now what is the difficulty?

    CB
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    Quote Originally Posted by CaptainBlack View Post
    The directions of the inequalities are wrong, but assume you mean for 0 \le x \le 1. Then area is:

    A=\int_{x=0}^1 \cos^2(\pi x/4)-\sin^2(\pi x/4)\ dx

    now what is the difficulty?

    CB
    Yeah sorry...it was use the midpoint rule and n=4 to approximate the area between the curves.
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  4. #4
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    Quote Originally Posted by sfgiants13 View Post
    Yeah sorry...it was use the midpoint rule and n=4 to approximate the area between the curves.
    You want four subintervals of the interval interval [0,1], so they are each of length 0.25, and have mid points 0.125, 0.375, 0.625, 0.875, or:

    x_i=\frac{b-a}{n} \times (k+0.5), \ \ k=0,1,.., n-1

    are the interval mid points, where a, b are the limits of integration, and n is the number of subintervals.

    Then:

    \int_a^b f(x)\ dx \approx \frac{b-a}{n}\sum_{k=0}^{n-1}f(x_i)

    or writing h=(b-a)/n

    x_i=h \times (k+0.5), \ \ k=0,1,.., n-1

    \int_a^b f(x)\ dx \approx h\sum_{k=0}^{n-1}f(x_i)

    which is the sum of the function at the mid points of the intervals times the interval length

    CB
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