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Math Help - graphs of antiderivatives!

  1. #1
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    graphs of antiderivatives!



    can someone show me the antiderivative of this graph...thanks...

    mathaction
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  2. #2
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    Start drawing vertical lines.

    At x = 0, G(x) = 0

    As x increases from x = 0, you get trapeziods. One parallel length obviously is 10. The other is obtained from the equation of the line between (0,-10) and (10,-20). Calculate the area of ALL such trapezoids, depending on the value of x.

    Of course, this piece will be negative.

    Do the same with the next line segment, starting where the last piece left off.
    Last edited by TKHunny; November 20th 2007 at 06:31 AM.
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  3. #3
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    Hello, mathaction!

    Edit: I'll re-do this solution with the new information: g(0) = 50




    Can someone show me the antiderivative of this graph?

    For reference, label the points: . A(0,\text{-}10),\;B(10,\text{-}20),\;C(15,0),\;D(20,10),\;E(40,0)


    Segment AB has slope: . m \:=\:\frac{(\text{-}20) - (\text{-}10)}{10 - 0} \:=\:-1 . and contains A(0,\text{-}10)

    . . Its equation is: .  y -(\text{-}10) \:=\:\text{-}1(x-0)\quad\Rightarrow\quad y \:=\:\text{- }x-10

    We have: . g'(x) \:=\:\text{- }x-10

    . . Integrate: . g(x)\:=\;\text{- }\frac{1}{2}x^2-10x + C_1

    Since g(0) = 50, we have: . 50 \:=\:-\frac{1}{2}(0^2) - 10(0) + C_1\quad\Rightarrow\quad C_1 \:=\:50

    . . Therefore: {\color{blue}\text{on }[0,10]\!:\;\;g(x) \:=\;\text{- }\frac{1}{2}x^2 - 10x + 50}

    Note that: . g(10) \:=\:\text{- }\frac{1}{2}(10^2) - 10(10) + 50 \:=\:\text{- }100



    Segment BC has slope: . m \:=\:\frac{0 - (\text{-}20)}{15-10} \:=\:4 . and contains C(15,0)

    . . Its equation is: .  y - 0  \:=\:4(x-15)\quad\Rightarrow\quad y \:=\:4x-60

    We have: . g'(x) \:=\:4x-60

    . . Integrate: . g(x)\:=\;2x^2-60x + C_2

    Since g(10) = \text{- }100, we have: . \text{- }100 \:=\:2(10^2) - 60(10) + C_2\quad\Rightarrow\quad C_2 = 300

    . . Therefore: . {\color{blue}\text{on }[10,\,20]\!:\;\;g(x) \:=\:2x^2 - 60x - 100}

    Note that: . g(20) \:=\:2(20^2) - 60(20) + 300 \:=\:\text{- }100



    Segment CD has slope: . m \:=\:\frac{10 - 0}{20 - 15} \:=\:2 . and contains C(15,0)

    . . Its equation is: .  y - 0 \:=\:21(x-15)\quad\Rightarrow\quad y \:=\:2x-30

    We have: . g'(x) \:=\:2x-30

    . . Integrate: . g(x)\:=\;x^2-30x + C_3

    Since g(20) = \text{- }100\!:\;\text{- }100 \:=\:20^2 - 30(20) + C_3\quad\Rightarrow\quad C_3 = 100

    . . Therefore: . {\color{blue}\text{on }[15,20],\;g(x) \;=\;x^2-30x + 100}

    Note that: . g(20) \:=\:20^2 - 30(20) + 100 \:=\:\text{- }100



    Segment DE has slope: . m \:=\:\frac{0-10}{40 - 20} \:=\:\text{- }\frac{1}{2} . and contains E(40,0)

    . . Its equation is: .  y -0 \:=\:\text{- }\frac{1}{2}(x-40)\quad\Rightarrow\quad y \:=\:\text{- }\frac{1}{2}x+10

    We have: . g'(x) \:=\:\text{- }\frac{1}{2}x+10

    . . Integrate: . g(x)\:=\;\text{- }\frac{1}{4}x^2+10x + C_4

    Since g(20) = \text{- }100, we have: . \text{- }100 \:=\:-\frac{1}{4}(20^2) + 10(20) + C_4\quad\Rightarrow\quad C_4 = \text{- }200

    . . Therefore: . {\color{blue}\text{on }[20,\,40]\!:\;\;g(x) \:=\:\text{- }\frac{1}{4}x^2 + 10x - 200}



    We have a piecewise function: . g(x) \;=\;\begin{Bmatrix}\text{- }\frac{1}{2}x^2 - 10x + 50 & 0 \leq x \leq 10 \\<br />
2x^2-60x - 100 & 10 \leq x \leq 15 \\<br />
x^2-30x + 100 & 15 \leq x \leq 20 \\<br />
\text{- }\frac{1}{4}x^2+10x - 200 & 20 \leq x \leq 40<br />
\end{Bmatrix}

    Last edited by Soroban; November 20th 2007 at 03:57 PM.
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  4. #4
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    yeah sorry it does give g(0) it equals 50
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