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Math Help - Piecewise Function problem

  1. #1
    Junior Member
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    Piecewise Function problem

    Let f(x) =

    (0) if x< 0
    (x) if 0 <= x <= 1
    (2 - x) if 1 < x <= 2
    (0) if x > 2

    and g(x) = [integrate] f(t)dt on [0, x]

    i) Find an expression for g(x) similar to the one for f(x).

    ii) Where is f differentiable? Where is g differentiable?


    I'm pretty much stuck, and don't know how to start this question.
    I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).

    But from that, how do I figure out an expression for g(x) from the piecewise function?

    Any and all help is appreciated here.
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  2. #2
    Senior Member
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    Quote Originally Posted by Tulki View Post
    Let f(x) =

    (0) if x< 0
    (x) if 0 <= x <= 1
    (2 - x) if 1 < x <= 2
    (0) if x > 2

    and g(x) = [integrate] f(t)dt on [0, x]

    i) Find an expression for g(x) similar to the one for f(x).

    ii) Where is f differentiable? Where is g differentiable?


    I'm pretty much stuck, and don't know how to start this question.
    I know first of all that by the fundamental theorem of Calculus, g'(x) = f(x).

    But from that, how do I figure out an expression for g(x) from the piecewise function?

    Any and all help is appreciated here.
    i)

    for x < 0, f(x) = 0 and g(x) = \int f(t) ~dt = 0

    for 0 \leq x \leq 1, f(x) = x and g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{x} t ~dt = \frac{1}{2}x^2

    for 1 < x \leq 2, f(x) = 2 - x and g(x) = \int_{0}^{x} f(t) ~dt = \int_{0}^{1} f(t) ~dt + \int_{1}^{x} f(t) ~dt= \int_{0}^{1} t ~dt + \int_{1}^{x} (2-t) ~dt= \frac{1}{2} + (2x - 2) - (\frac{1}{2}x^2 - \frac{1}{2})

    for x > 2
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  3. #3
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    Hello, Tulki!

    Here's part (a).
    Did you make a sketch?


    \text{Let }\:f(x) \;=\;\left\{\begin{array}{cccc}0 && x < 0 \\ \\[-4mm]<br />
x && 0 \leq x \leq 1 \\ \\[-4mm]<br />
2 - x && 1 < x \leq 2 \\ \\[-4mm]<br />
0 && x > 2 \end{array}\right\}

    and: . g(x) \:=\: \int_0^x f(t)\,dt

    a) Find an expression for g(x) similar to the one for f(x).

    g(x) gives the area under the graph from 0\text{ to }x

    The graph looks like this:
    Code:
              |
              |
             1+       *
              |     *:::*
              |   *:::::::*
              | *:::::::::| *
        - * * * - - - * - + - * * * - -
              0       1   x   2

    Then: . \displaystyle g(x) \;=\;\left\{ \begin{array}{cccc}<br />
0 && x < 0 \\ \\[-3mm]<br />
\int^x_0t\,dt && 0 \leq x \leq 1 \\ \\[-2mm]<br />
\frac{1}{2} + \int^x_1 (2-t)\,dt && 1 < x \leq 2 \\ \\[-3mm]<br />
1 && x > 2<br />
\end{array} \right\}

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  4. #4
    MHF Contributor

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    Because all of the graphs are straight lines, you don't really need to integrate, just find the areas of the polygons involved.

    For x< 0, f(x)= 0 so its integral is 0 (and its graph is just the x-axis so there is 0 area).

    For x between 0 and 1, y= x so the "area under the curve" is the area of a right triangle with base x and height x: its area is \frac{1}{2}x^2.

    For x between 1 and 2, y= 2- x and we can break the "area under the currve" from 0 to x into two parts: The area of the right triangle with base 1 and height 1 is \frac{1}{2}(1)(1)= \frac{1}{2}. The area from 1 to x is a trapezoid with bases 1 and 2-x and height x. It's area is \frac{2-x+ 1}{2}x= \frac{(3-x)x)}{2}= \frac{3x- x^2}{2}.
    The total area is \frac{1}{2}+ \frac{3x- x^2}{2}= \frac{1+ 3x- x^2}{2}. AT x= 2, the integral is the entire area of a triangle with base 2 and height 1 which is 1.

    For x> 2, the graph is again the x-axis so no new area is added. For all x> 2, the function is the constant 1.
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  5. #5
    Junior Member
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    Thank you very much, all of you!

    I suppose it was only the piecewise function that scared me. Graphing it DOES indeed help greatly. The pieces of the function are pretty darn simple, in retrospect.
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