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Math Help - Estimate the area

  1. #1
    Junior Member
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    Estimate the area

    The problem asks: Estimate the area under the graph of f(x0 = 4x^2 + 4x + 4 from x=5 to x= 8.
    using 3 approximating rectangles and right endpoints.
    R3=?
    Using 3 approximating rectangles and left endpoints.
    L3=?
    I under stand up to here:
    4x^3/3-2x^2-4x
    I know I am supposed to evaluate from 5 to 8 but that is as far as it goes.
    How about a push.
    Thank You,
    Keith
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  2. #2
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    Well, first find the co-ordinatdes where the graph crosses the x-axis.
    y = 4x^2 + 4x + 4
    4x^2 + 4x + 4 = 0 (0 is the y co-ordinate where the graph crosses the x-axis)
    Using the discriminant b-4ac = 4-(4*4*4)
    = -48 < 0 therefore the graph doesn't cross the x-axis at all. So to do this i would suggest drawing the graph of 4x^2 + 4x + 4. But it is still possible to do it without drawing the graph it just takes a little bit more thought of which ill divulge into tomorrow. It's late now .
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    See here.
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  4. #4
    MHF Contributor
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    Quote Originally Posted by keith View Post
    The problem asks: Estimate the area under the graph of f(x0 = 4x^2 + 4x + 4 from x=5 to x= 8.
    using 3 approximating rectangles and right endpoints.
    R3=?
    Using 3 approximating rectangles and left endpoints.
    L3=?
    I under stand up to here:
    4x^3/3-2x^2-4x
    I know I am supposed to evaluate from 5 to 8 but that is as far as it goes.
    How about a push.
    Thank You,
    Keith
    I don't know what the problem means but I can guess. It is about finding the approximate area under the parabola from x=5 to x=8 using 3 approximating rectangles. It is not through the usual Integral, from 5 to 8, of (4x^2 +4x +4)dx.

    The approximating "dx" here is actually 1 unit wide. The 3 approximating rectangles are:
    a) For R3 or right endpoints, f(6)high by 1 wide, f(7)high by 1 wide, and f(8) high by 1 wide.
    b) For L3 or left endpoints, f(5)high by 1 wide, f(6)high by 1 wide, and f(7)high by 1 wide.

    So,
    f(x) = 4x^2 +4x +4
    f(x) = 4(x^2 +x +1) --------***

    R3 = 4[6^2 +6 +1]*1 +4[7^2 +7 +1]*1 +4[8^2 +8 +1]*1 = 692 sq.units. ----answer

    L3 = 4[5^2 +5 +1]*1 +4[6^2 +6 +1]*1 +4[7^2 +7 +1]*1 = 524 sq.units. ----answer.

    --------------------------
    So, the approximate area under the curve from x=5 to x=8 is the average of R3 and L3,
    = (R3 +L3)/2 = (692 +524)/2 = 608 sq.units ------***

    If we use integration to find the exact area,
    A = INT.(5-->8)[4x^2 +4x +4]dx
    A = 4*INT.(5-->8)[x^2 +x +1] dx
    A = 4[(x^3)/3 +(x^2)/2 +x]|(5-->8)
    A = 4[{(8^3)/3 +(8^2)/2 +8} -{(5^3)/3 +(5^2)/2 +5}]
    A = 4[151.5]
    A = 606 sq.units exactly. ------------***

    So the approximate 608 sq.units above is not bad.
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