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Math Help - Integrals and Derivatives

  1. #1
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    Integrals and Derivatives

    Use the values to estimate integral from 0 to 6 using six equal subintervals with left endpoints.
    (0, 9.7) (1, 8.9) (2, 8.4) (3, 6.1) (4, 4.5) (5, -6.9) (6, -10.3)

    Find the interval on which the curve f(x)=integral of 1/(6+t+t^2)dt from 0 to x is concave downward.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by sw3etazngyrl View Post
    Use the values to estimate integral from 0 to 6 using six equal subintervals with left endpoints.
    (0, 9.7) (1, 8.9) (2, 8.4) (3, 6.1) (4, 4.5) (5, -6.9) (6, -10.3)
    You are expected to use a numerical integration rule to do this. The use of
    the term "left end points" seems to imply that you are supposed to something
    like:

    <br />
I = \int_0^6 f(x) dx = \sum_{x=0}^5 f(x) \delta x = (9.7+8.9+8.4+6.1+4.5-6.9)\times 1 = 30.7<br />

    (note the last data point (6,-10.3) does not contribute to this as it is not the
    left end point of one of the sub-intervals of (0,1). Simpsons rule would be a better
    way of doing this if you know that).

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by sw3etazngyrl View Post

    Find the interval on which the curve f(x)=integral of 1/(6+t+t^2)dt from 0 to x is concave downward.
    Concave downwards means that the derivative is decreasing.

    Now if:

    <br />
f(x)=\int_0^x 1/(6+t+t^2)dt <br />

    by the fundamental theorem of calculus:

    <br />
\frac{d}{dx} f(x)= 1/(6+x+x^2) <br />

    So can you take it from there?

    RonL
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