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First Statement:

Let $s$ be an arbitrary step function defined on the closed interval $[a, b]$. Then we have $$ \int_{ka}^{kb} s\left(\frac{x}{k}\right) \ dx = k \int_a^b s(x) \ dx $$ for every $k > 0$.

Second Statement:

Let $s$ be an arbitrary step function defined on the closed interval $[ka, kb]$. Then we have $$ \int_{ka}^{kb} s(x) \ dx = k \int_a^b s(kx) \ dx $$ for every $k > 0$.

Then how to establish the equivalence of the above two statements?

We are not supposed to use the substitution technique of integration, and are only to use the following definition:

If there is a partition $P = \{ a = x_0, x_1, \ldots, x_n = b \}$ of the closed interval $[a, b]$ such that a function $s$ defined on $[a,b]$ is constant on the open sub-intervals of $P$, then $s$ is said to be a step function.

Moreover, if $s(x) = s_i $ for $x_{i-1} < x < x_i$, then we define the definite integral $\int_a^b s(x) \ dx$ as follows: $$ \int_a^b s(x) \ dx = \sum_{i=1}^n s_i \cdot (x_i - x_{i-1}).$$

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HINT: Let us partition two intervals $I = [a,b]$ and $K = [ka,kb]$ into $n$ sub–intervals:

\begin{alignat}{2} P_n(I)&:=\{&a&=x_0,\, x_1,\, x_2,\, &\dots,\, x_{n-1},\, x_n,\, x_{n+1} &= b\} \\ P_n(K)&:=\{&ak&=y_0,\, y_1,\, y_2,\, &\dots,\, y_{n-1},\, y_n,\, x_{n+1} &= kb\} \end{alignat}

Since $\displaystyle\int_a^b\!\!\! s(x) \, dx = \lim_{n\to\infty}\sum_{i=0}^n s_i\!\cdot\!(x_{i+1} - x_{i})\,$ and $\displaystyle\int_{ka}^{kb}\!\!\! s(x) \, dx = \int_{ka}^{kb}\!\!\! s(y) \, dy = \lim_{n\to\infty}\sum_{i=0}^n s_i\!\cdot\!(y_{i+1} - y_{i})$ where $s_{i} = s(x)$ for some $x\in[x_{i},\,x_{i+1}]$ and $s_j= s(y_{j})$ for $y\in[y_{i},\,y_{i+1}]$, it is sufficient to show that

$\displaystyle\lim_{n\to\infty}\sum_{i=0}^n s_i\!\cdot\!(x_{i+1} - x_{i}) = \lim_{n\to\infty}\sum_{j=0}^n s_j\!\cdot\!(y_{j+1} - y_{j})$

Obviously $\,y_{j+1} - y_{j} = k\,(x_{i+1} - x_{i}),\,$ and we can se that $s_j = s(y),\; y\in[y_{j}, y_{j+1}]$ is proportional to $s_i = s(x),\; x\in[x_{i}, x_{i+1}]$ with scaling factor $\dfrac{1}{k}$ by expanding function $s\,(\cdot)$ into Taylor Series:

\begin{align} s_i = s(x) &\approx s(x_{i}) \cdot(x - x_{i}) + \mathcal{O}(x-x_i) &&\approx s(x_{i}) \cdot(x_{i+1} - x_{i}) + \mathcal{O}(x_{i+1}-x_i) \\ s_j = s(x) &\approx s(y_{j}) \cdot(y - y_{j}) + \mathcal{O}(y-y_j) &&\approx s(y_{j}) \cdot(y_{j+1} - y_{j}) + \mathcal{O}(y_{j+1}-y_j) \\ &&&= s(y_{j}) \cdot(x_{j+1} - x_{j})\cdot k + \mathcal{O}(x_{j+1}-x_j) \cdot k \end{align}

Thus for $i=j$ we get

$$ s(x_{i}) \cdot(x_{i+1} - x_{i}) \approx s(y_{i}) \cdot(x_{i+1} - x_{i})\cdot k \implies \bbox[1.25ex, border:solid 2pt #e10000]{s(x_{i}) \approx s(y_{i})\cdot k } $$

Hope you can pick it from here.

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