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    Demonstrating behaviors of calculus

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    Numerical approximations to integrals typically get better -- i.e., their error goes down -- proportional to a power of N, the number of subintervals in the interval of integration. For the upper and lower sums, the error typically goes down like 1/N as N increases. For the midpoint and trapezoidal rules, the error typically goes down like 1/N^2. For Simpson's rule, the error typically goes down like 1/N^4.

    (a) Demonstrate this behavior numerically, using the integral of x^3 on [0,1] as a typical integral. (b) Demonstrate that this normal behavior is NOT seen in the integral of sqrt(x) on [0,1]! Apparently the slightly bad behavior of sqrt(x) at 0 (it is not differentiable there) is to blame.

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    Hi, here I have attached a file, which includes answer for your question.

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    Hi, here is the solution.

    (a) Demonstrate this behavior numerically, using the integral of x^3 on [0,1] as a typical integral.

    Typical integral
    1 Use
    1 Power rule x^n = x^n+1
    x^3 . dx = [ x^4/4] --------
    0 n+1

    0

    [ ¼ - 0] => ¼ => 0.25

    Now, we have to use the Trapezoidal rule and simpson's rule, then we have to compare the result with typical integral and show that the error goes down when n increase (ie) If the subinterval increases the error will be reduce.

    Trapezoidal rule:
    b

    f(x) dx = h/2 [ f 1+ 2 f2 + 2 f3+......+2 fn+ fn+1]

    a

    Where h= (b-a)/n n- is the subinterval (we can take any number for n, when n-increases the error will be reduce)

    Here the interval [ 0,1], we take n=4

    h= (1-0)/4 = 0.25

    So, we take subinterval 0, 0.25, 0.5, 0.75,1
    1

    x^3 dx = ...

    Solution Summary

    This solution is comprised of a detailed explanation to demonstrate this behavior numerically, using the integral of x^3 on [0,1] as a typical integral.

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