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Math Help - Using properties of integrals

  1. #1
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    Using properties of integrals

    Use the properties of integrals to verify that

    2 <= ∫ -1 to 1 (1 + x^2) ^ 1/2 dx <= 2(2) ^ 1/2
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    Quote Originally Posted by SyNtHeSiS View Post
    Use the properties of integrals to verify that

    2 <= ∫ -1 to 1 (1 + x^2) ^ 1/2 dx <= 2(2) ^ 1/2
    consider the areas of the upper and lower rectangles as shown in the graph
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    where are the lines y = 2 and y = 2(2) ^ 1/2? I know that the parabola is between the graph, but I dont know how to verify it by the properties of integrals
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    Quote Originally Posted by SyNtHeSiS View Post
    where are the lines y = 2 and y = 2(2) ^ 1/2? I know that the parabola is between the graph, but I dont know how to verify it by the properties of integrals
    not the lines ... the areas.
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    I am talking about the line y = 1.4 (which should be 2(2)^1/2) and and y = 1 (which should be 2) as the parabola is squeezed between them.
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    Quote Originally Posted by SyNtHeSiS View Post
    I am talking about the line y = 1.4 (which should be 2(2)^1/2) and and y = 1 (which should be 2) as the parabola is squeezed between them.

    f(0) = 1 = minimum value of y = \sqrt{x^2+1} on [-1,1]

    f(1) = f(-1) = \sqrt{2} = maximum value of y = \sqrt{x^2+1} on [-1,1]

    f(0)[1 - (-1)] < \int_{-1}^{1} \sqrt{x^2+1} \, dx < f(1)[1 - (-1]<br />

    2 < \int_{-1}^{1} \sqrt{x^2+1} \, dx < 2\sqrt{2}<br />
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    Quote Originally Posted by skeeter View Post
    f(0) = 1 = minimum value of y = \sqrt{x^2+1} on [-1,1]

    f(1) = f(-1) = \sqrt{2} = maximum value of y = \sqrt{x^2+1} on [-1,1]
    Sorry but why did you sub in 0 and 1 into f ?
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    Quote Originally Posted by SyNtHeSiS View Post
    Sorry but why did you sub in 0 and 1 into f ?
    Upper and lower bounds. An integrable function f on [a, b], is necessarily bounded on that interval. Thus there are real numbers m and M so that mf (x) ≤ M for all x in [a, b]. Since the lower and upper sums of f over [a, b] are therefore bounded by, respectively, m(ba) and M(ba), it follows that

    m(b-a) \le \int_a^b f(x) \, dx \le M(b-a)

    if you require further explanation, see your instructor.
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