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First Statement:

Let $s$ be an arbitrary step function defined on the closed interval $[a,b]$. Then we have $$\int_{a}^{b} s(x) \ dx = \int_{a+c}^{b+c} s(x-c) \ dx.$$

Second Statement:

Let $s$ be an arbitrary step function defined on the closed interval $[a+c, b+c]$. Then we have $$\int_{a+c}^{b+c} s(x) \ dx = \int_{a}^{b} s(x+c) \ ds.$$

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

We are not supposed to use substitution techniques for integration, and are only to use the following definition:

If there is a partition $P = \{ a = x_0, x_1, \ldots , x_{n-1}, x_n = b \}$ of the closed interval $[a,b]$ such that the function $s$ defined on $[a,b]$ is constant on the open sub-intervals of $P$, then $s$ is called a step function.

Furthermore, if $s(x) = s_k$ for $x_{k-1} < x < x_k $, where $k = 1, 2, \ldots, n$, then we define the definite integral $\int_{a}^{b} s(x) \ dx$ as follows: $$ \int_{a}^{b} s(x) \ dx = \sum_{k=1}^n s_k \cdot (x_k - x_{k-1}).$$

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  • $\begingroup$ are you sure that you want to show that these statements can be deduced from each other, or that each statement is true? $\endgroup$ – sranthrop Jul 13 '14 at 19:57
  • $\begingroup$ sranthrop, yes, I am. $\endgroup$ – Saaqib Mahmood Jul 13 '14 at 22:46
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(2) -> (1): Define $t(x):=s(x-c)$. Then $t$ is a step-function on $[a+c,b+c]$. From (2) we obtain $$ \int_{a+c}^{b+c}s(x-c)dx=\int_{a+c}^{b+c}t(x)dx=\int_a^bt(x+c)dx=\int_a^bs(x)dx, $$ and this is (1). The reverse implication goes analogously.

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  • $\begingroup$ I'am still not sure if this is what you wanted, but if so, I hope it helped ;) $\endgroup$ – sranthrop Jul 14 '14 at 18:13

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