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Properties of another alternate definition of the integral of a step function

by RoRi

Suppose we defined the integral of a step function as:

    \[ \int_a^b s(x) \, dx = \sum_{k=1}^n s_k \cdot (x_k^2 - x_{k-1}^2). \]

Which of the following properties of the integral would still hold:

  1. \displaystyle{\int_a^b s + \int_b^c s = \int_a^c s}.
  2. \displaystyle{\int_a^b (s+t) = \int_a^b s + \int_a^b t}.
  3. \displaystyle{\int_a^b c \cdot s = c \cdot \int_a^b s}.
  4. \displaystyle{\int_{a+c}^{b+c} s(x) \, dx = \int_a^b s(x+c) \, dx}.
  5. If s(x) < t(x) for all x \in [a,b], then \displaystyle{ \int_a^b s < \int_a^b < t}.
  1. True.
    Proof. Let P = \{ x_0, \ldots, x_m \} be a partition of [a,c] and P_2 = \{ y_0, \ldots, y_n \} be partition of [b,c] such that s(x) is constant on the open subintervals of P_1 and P_2. Then, let P = P_1 \cup P_2 = \{ x_0, \ldots, x_m, x_{m+1}, \ldots, x_{m+n} \}, where x_m = y_0, x_{m+1} = y_1, \ldots, x_{m+n} = y_n, so P is a partition of [a,c] and s(x) is constant on the open subintervals of P. Then,

        \begin{align*} \int_a^b s(x) \, dx + \int_b^c s(x) \, dx &= \sum_{k=1}^m s_k (x_k^2 - x_{k-1}^2) + \sum_{k=1}^n s_k (y_k^2 - y_{k-1}^2) & (\text{``def." of } \int_a^b s)\\ &= \sum_{k=1}^m s_k (x_k^2 - x_{k-1}^2) + \sum_{k=m+1}^{m+n} s_k (x_k^2 - x_{k-1}^2) \\ &= \sum_{k=1}^{m+n} s_k (x_k^2 - x_{k-1}^2) \\ &= \int_a^c s(x) \, dx. \qquad \blacksquare \end{align*}

  2. True.
    Proof. Let P = \{ x_0, \ldots, x_n \} be partition of [a,b] such that s(x) + t(x) is constant on the open subintervals of P. (We know such a partition exists since we can take the common refinement of the partitions of [a,b] on which s and t individually are constant.) Assume s(x) + t(x) = s_k + t_k if x_{k-1} < x < x_k for k = 1,2, \ldots, n. Then,

        \begin{align*} \int_a^b (s(x) + t(x)) \, dx &= \sum_{k=1}^n (s_k + t_k)(x_k^2 - x_{k-1}^2) \\ &= \sum_{k=1}^n \left( s_k (x_k^2 - x_{k-1}^2) + t_k (x_k^2 - x_{k-1}^2) \right) \\ &= \sum_{k=1}^n s_k (x_k^2 - x_{k-1}^2) + \sum_{k=1}^n t_k (x_k^2 - x_{k-1}^2) \\ &= \int_a^b s(x) \, dx + \int_a^b t(x) \, dx. \qquad \blacksquare \end{align*}

  3. True.
    Proof. Let P = \{ x_0, \ldots, x_n \} be a partition of [a,b] such that s is constant on the open subintervals of P. Assume s(x) = s_k if x_{k-1} < x < x_k for k = 1, \ldots, n. Then c \cdot s(x) = c \cdot s_k if x_{k-1} < x < x_k so using our alternative definition of the integral we have,

        \begin{align*} \int_a^b s(x) \, dx &= \sum_{k=1}^n c \cdot s_k \cdot (x_k^2 - x_{k-1}^2) \\ &= c \cdot \sum_{k=1}^n s_k (x_k^2 - x_{k-1}^2) \\ &= c \cdot \int_a^b s(x) \, dx. \qquad \blacksquare \end{align*}

  4. False. Since s((x_k + c)^2 - (x_{k-1} +c)^2) \neq (s+c)(x_k^2 - x_{k-1}^2). In particular, a counterexample is given by letting s(x) = 1 for all x \in [1,2] and let c = 1. Then,

        \begin{align*} \int_{0+1}^{1+1} s(x) \, dx &= 1 \cdot (2^2 - 1^2) = 3 \\ \int_0^1 s(x+1) \, dx &= 1 \cdot (1^2 - 0^2) = 1. \end{align*}

  5. False.
    A counterexample is given by considering s(x) = 0 and t(x) = 1 on the interval [-1,0]. Then, s < t on the interval, but

        \[ \int_{-1}^0 s(x) \, dx = 0 \not< \int_{-1}^0 t(x) \, dx = 1 \cdot (0^2 - (-1)^2) = -1. \]

3 comments

  1. Hteica says:

    Hi! I appreciate your efforts. But (b) doesn’t seem convincing enough, I would suggest to use induction to prove that the s+t is constant on every open subinterval of union of Pt and Ps (partition of [a,b] according to t and s respectively). I don’t think it is a good choice to assume this is true by default. By the way, thanks a lot!

    • fitnesstendency says:

      Yeah, it’s negative, but keep in mind that we aren’t using the original definition of integral, but a modified one, so we can’t say we are dealing with “real” integrals here.

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