Tag: Step Functions

Prove an inequality relating sums and integrals

by RoRi

March 4, 2016

Prove that

$\sum_{k=1}^{n-1} f(k) \leq \int_1^n f(x) \, dx \leq \sum_{k=2}^n f(k)$

where $f$ is a nonnegative, increasing function defined for all $x \geq 1$ .

Use this to deduce the inequalities

$e n^n e^{-n} < n! < e n^{n+1} e^{-n}$

by taking $f(x) = \log x$ .

Proof. Since $f(x)$ is increasing we know $f([x]) \leq f(x) \leq f([x+1])$ (where $[x]$ denotes the greatest integer less than or equal to $x$ ) for all $x \geq 1$ . Define step functions $s$ and $g$ by

$s(x) = f([x]), \qquad t(x) = f([x+1]).$

Then $s$ and $t$ are constant on the open subintervals of the partition

$P = \{ 0,1,2, \ldots, n \}.$

So,

$\begin{align*} \int_1^n s(x) \, dx = \sum_{k=1}^{n-1} s_k (x_k - x_{k-1}) = \sum_{k=1}^{n-1} f(k) \\[9pt] \int_1^n t(x) \, dx = \sum_{k=1}^{n-1} t_k (x_k - x_{k-1}) = \sum_{k=1}^{n-1} f(k+1) = \sum_{k=2}^n f(k). \end{align*}$

Since $f$ is integrable we must have

$\int_1^n s(x) \, dx \leq \int_1^n f(x) \, dx \leq \int_1^n t(x) \, dx$

for every pair of step functions $s \leq f \leq t$ . Hence,

$\sum_{k=1}^{n-1} f(k) \leq \int_1^n f(x) \, dx \leq \sum_{k=2}^n f(k). \qquad \blacksquare$

Next, if we take $f(x) = \log x$ (which is nonnegative and increasing on $x \geq 1$ ) we have

$\begin{align*} &&\sum_{k=1}^{n-1} \log k \leq \int_1^n \log x \, dx \leq \sum_{k=2}^n \log k \\[9pt] \implies && \log(n-1)! \leq n \log n - n + 1 \leq \log n! \\[9pt] \implies && (n-1)! \leq e n^n e^{-n} \leq n!. \end{align*}$

From this we then get two inequalities

$n! \leq en^{n+1} e^{-n} \quad \text{and} \quad e n^n e^{-n} \leq n!.$

Therefore,

$en^n e^{-n} \leq n! \leq e n^{n+1} e^{-n}.$

Prove some statements about integrals of bounded monotonic increasing functions

by RoRi

February 29, 2016

Consider a bounded, monotonic, real-valued function $f$ on the interval $[0,1]$ . The define sequences

$s_n = \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{n}{k} \right), \qquad t_n = \frac{1}{n} \sum_{k=1}^n f \left( \frac{k}{n} \right).$

Prove that
$s_n \leq \int_0^1 f(x) \, dx \leq t_n$

and that

$0 \leq \int_0^1 f(x) \,dx - s_n \leq \frac{f(1) - f(0)}{n}.$
Prove that the two sequences $\{ s_n \}$ and $\{ t_n \}$ converge to $\int_0^1 f(x) \, dx$ .
State and prove a generalization of the above to interval $[a,b]$ .

Proof.First, we define two step functions,
$s(x) = f \left( \frac{[nx]}{n} \right), \qquad t(x) = f \left( \frac{[nx+1]}{n} \right)$

where $[x]$ denotes the greatest integer less than or equal to $x$ . Then we define a partition of $[0,1]$ ,

$P = \left\{ 0, \frac{1}{n}, \frac{2}{n}, \ldots, 1 \right\}.$

For any $x_{k_1} \leq x < x_k$ we have

$s(x) = f \left( \frac{k-1}{n} \right), \qquad t(x) = f \left( \frac{k}{n} \right).$

So, $s(x)$ and $t(x)$ are constant on the open subintervals of the partition $P$ .
Since $f(x)$ is monotonically increasing and $s(x) \leq f(x) \leq t(x)$ for all $x$ (by the definition of $s$ and $t$ ) we have

$\begin{align*} &&\int_0^1 s(x) \, dx \leq \int_0^1 f(x) \, dx \leq \int_0^1 t(x) \, dx \\[9pt] \implies && \sum_{k=0}^{n-1} s_k (x_k - x_{k-1} \leq \int_0^1 f(x) \, dx \leq \sum_{k=0}^{n-1} t_k (x_k - x_{k-1}) && (\text{Def step function integral}) \\[9pt] \implies && \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \leq \int_0^1 f(x) \, dx \leq \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k+1}{n} \right) &&(x_k - x_{k-1} = \frac{1}{n} \forall k) \\[9pt] \implies && \fract{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \leq \int_0^1 f(x) \, dx \leq \frac{1}{n} \sum_{k=1}^nf \left( \frac{k}{n} \right) \\[9pt] \implies && s_n \leq \int_0^1 f(x) \, dx \leq t_n && \text{for all } n \\[9pt] \implies && 0 \leq \int_0^1 f(x) \, dx - s_n \leq t_n - s_n \\[9pt] \implies && 0 \leq \int_0^1 f(x) \, dx - s_n \leq \frac{f(1) - f(0)}{n}. \qquad \blacksquare \end{align*}$
Proof. From part (a) we have
$0 \leq \int_0^1 f(x) \, dx - s_n \leq \frac{f(1) - f(0)}{n} \quad \implies \quad \lim_{n \to \infty} \left( \int_0^1 f(x) \, dx - s_n \right)= 0$

since $\lim_{n \to \infty} \frac{f(1)- f(0)}{n} = 0$ . Since $\int_0^1 f(x) \, dx$ does not depend on $n$ we have

$\int_0^1 f(x) \, dx - \lim_{n \to \infty} s_n = 0 \quad \implies \quad \lim_{n \to \infty} s_n = \int_0^1 f(x) \, dx.$

Therefore,

$\begin{align*} && s_n \leq \int_0^1 f(x) \, dx \leq t_n \\[9pt] \implies && s_n - t_n \leq \int_0^1 f(x) \, dx - t_n \leq 0 \\[9pt] \implies && \frac{f(0)-f(1)}{n} \leq \int_0^1 f(x) \, dx - t_n \leq 0 \\[9pt] \implies && \lim_{n \to \infty} \left( \int_0^1 f(x) \, dx - t_n \right) = 0 \\[9pt] \implies && \lim_{n \to \infty} t_n = \int_0^1 f(x) \, dx. \qquad \blacksquare \end{align*}$
Claim: If $f$ is a real-valued function that is monotonic increasing and bounded on the interval $[a,b]$ , then
$\lim_{n \to \infty} s_n = \int_a^b f(x) \, dx, \qquad \text{and} \qquad \lim_{n \to \infty} t_n = \int_a^b f(x) \, dx$

for $s_n$ and $t_n$ defined as follows:

$s_n = \frac{b-a}{n} \cdot sum_{k=0}^{n-1} f \left( a + k \frac{b-a}{n} \right), \qquad t_n = \frac{b-a}{n} \cdot \sum_{k=1}^n f \left( a + k \frac{b-a}{n} \right).$

Proof. Let

$P = \left\{ a, a + \frac{b-a}{n}, a + \frac{2(b-a)}{n}, \ldots, a + \frac{n(b-a)}{n} = b \right\}$

be a partition of the interval $[a,b]$ . Then, define step functions $s(x)$ and $t(x)$ with $s(x) = f(x_{k-1})$ and $t(x) = f(x_k)$ for $x_{k-1} \leq x < x_k$ . By these definitions we have $s(x) \leq f(x) \leq t(x)$ for all $x \in [a,b]$ (since $f$ is monotonic increasing). Since $f$ is bounded and monotonic increasing it is integrable, and

$\begin{align*} && \int_a^b s(x) \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b t(x) \, dx \\[9pt] \implies && \sum_{k=1}^n s_k (x_k - x_{k-1}) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n t_k (x_k - x_{k-1}) \\[9pt] \implies && \sum_{k=1}^n s_k \left( \frac{b-a}{n} \right) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n t_k \left( \frac{b-a}{n} \right). \end{align*}$

And, since $s_k = f(x_{k-1} = f \left( a + \frac{(k-1)(b-a)}{n} \right)$ , and $t_k = f(x_k) = f \left( a + \frac{k(b-a)}{n} \right)$ we have

$\begin{align*} \implies && \sum_{k=1}^n f \left( a + \frac{(k-1)(b-a)}{n} \right) \left( \frac{b-a}{n} \right) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n f \left( a + \frac{k(b-a)}{n} \right) \left( \frac{b-a}{n} \right) \\[9pt] \implies && \left( \frac{b-a}{n} \right) \sum_{k=0}^{n-1} f \left( a+ \frac{k(b-a)}{n} \right) \leq \int_a^b f(x) \, dx \leq \left( \frac{b-a}{n} \right) \sum_{k=1}^n f \left( a+ \frac{k(b-a)}{n} \right). \qquad \blacksquare \end{align*}$

Prove inequalities of the logarithm with respect to some series

by RoRi

November 6, 2015

Consider a partition $P = \{ a_0, a_1, a_2, \ldots, a_n \}$ of the interval $[1,x]$ for some $x > 1$ .

Find step functions that are constant on the open subintervals of $P$ and integrate to derive the inequalities:
$\sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) < \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right).$
Give a geometric interpretation of the inequalities in part (a).
Find a particular partition $P$ (i.e., choose particular values for $a_0, a_1, \ldots, a_n$ ) to establish the following inequalities for $n > 1$ ,
$\sum_{k=2}^n \frac{1}{k} < \log n < \sum_{k=1}^{n-1} \frac{1}{k}.$

Proof. We define step function $s$ and $t$ by
$\begin{align*} s(x) &= \frac{1}{a_k} & \text{for } x &\in [a_{k-1}, a_k ) \\ t(x) &= \frac{1}{a_{k-1}} & \text{for } x &\in [a_{k-1}, a_k ). \end{align*}$

Since $\frac{1}{x}$ is strictly decreasing on $\mathbb{R}_{>0}$ , we have

$s(x) < \frac{1}{x} < t(x) \qquad \text{for all } x \in \mathbb{R}_{>0}.$

Therefore, using the definition of the integral of a step function as a sum,

$\begin{align*} && \int_1^x s(u) \, du &< \int_1^x \frac{1}{u} \, du < \int_1^x t(u) \, du \\ \implies && \sum_{k=1} s_k (a_k - a_{k-1} ) &< \log x < \sum_{k=1}^n t_k (a_k - a_{k-1}) \\ \implies && \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) &< \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right). \qquad \blacksquare \end{align*}$
Geometrically, these inequalities say that the area under the curve $\frac{1}{x}$ lies between the step functions that take on the values $\frac{1}{a_k}$ and $\frac{1}{a_{k-1}}$ for each $x \in [ a_{k-1}, a_k )$ .
Proof. To establish these inequalities we pick the partition,
$P = \{ 1, 2, 3, \ldots, n \} \qquad \text{for } n > 1.$

Then, applying part (a) we have

$\begin{align*} && \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_k} \right) &< \log x < \sum_{k=1}^n \left( \frac{a_k - a_{k-1}}{a_{k-1}} \right) \\ \implies && \sum_{k=1}^n \frac{1}{a_k} &< \log n < \sum_{k=1}^n \frac{1}{a_{k-1}} &(\text{since } a_k - a_{k-1} = 1) \\ \implies && \sum_{k=2}^n \frac{1}{k} &< \log n < \sum_{k=1}^{n-1} \frac{1}{k}. \end{align*}$

The final line follows since $a_0 = 1, a_1 = 2, \ldots$ so the sum on the left starts with $\frac{1}{2}$ and the sum on the right only runs to $\frac{1}{n-1}$ . These were the inequalities requested $. \qquad \blacksquare$

Prove the integral of a step function is invariant under translation

by RoRi

July 20, 2015

For a step function $s(x)$ on an interval $[a,b]$ prove

$\int_a^b s(x) \, dx = \int_{a+c}^{b+c} s(x-c) \, dx.$

Proof. Let $P = \{ x_0, \ldots, x_n \}$ be a partition of $[a,b]$ such that $s(x) = s_k$ is constant on the open k-th subinterval of the partition.
Then $P = \{ x_0 + c, x_1 + c, \ldots, x_n + c \}$ is a partition of $[a+c, b+c]$ and $s(x-c) = s_k$ for $x_{k-1} + c < x < x_k + c$ . So,

$\int_a^b s(x) \, dx = \sum_{k=1}^n s_k (x_k - x_{k-1}),$

and,

$\int_{a+c}^{b+c} s(x-c) \, dx = \sum_{k=1}^n s_k (x_k - x_{k-1}).$

Thus, $\int_a^b s(x) \, dx = \int_{a+c}^{b+c} s(x-c) \, dx. \qquad \blacksquare$

Prove integrals of step functions are additive wrt interval of integration

by RoRi

July 20, 2015

For a step function $s(x)$ prove that

$\int_a^c s(x) \, dx + \int_c^b s(x) \, dx = \int_a^b s(x) \, dx \qquad \text{if } a < c < b.$

Proof. Let $P_1 = \{ x_0, \ldots, x_m \}$ be a partition of $[a,c]$ and $P_2 = \{ y_0, \ldots, y_n \}$ be a partition of $[c,b]$ such that $s(x)$ is constant on the open subintervals of $P_1$ and $P_2$ . Then, let $P = P_1 \cup P_2$ so $P$ is a partition of $[a,b]$ and $s(x)$ is constant on the open subintervals of $P$ . Then, let $y_0 = x_m, y_1 = x_{m+1}, \ldots, y_n = x_{m+n}$ . Then,

$\begin{align*} \int_a^c s(x) \, dx + \int_c^b s(x) \, dx &= \sum_{k=1}^m s_k (x_k - x_{k-1}) + \sum_{k=1}^n s_k (y_k - y_{k-1}) \\ &= \sum_{k=1}^m s_k (x_k - x_{k-1}) + \sum_{k=m+1}^{m+n} s_k (x_k - x_{k-1}) \\ &= \sum_{k=1}^{m+n} s_k (x_k - x_{k-1}) \\ &= \int_a^b s(x) \, dx. \qquad \blacksquare \end{align*}$

Prove the comparison theorem for integrals of step functions

by RoRi

July 20, 2015

For step functions $s(x) < t(x)$ for every $x \in [a,b]$ prove that

$\int_a^b s(x) \, dx < \int_a^b t(x) \, dx.$

Proof. Let $P = \{ x_0, \ldots, x_n \}$ be a partition of $[a,b]$ such that $s(x)$ and $t(x)$ are constant on the open subintervals of $P$ (such a partition exists since we can take the common refinement of partitions on which $s$ and $t$ are constant on open subintervals). Then, assume $s(x) = s_k, \ t(x) = t_k$ if $x_{k-1} < x < x_k$ . From the definition of integrals of step functions, we have

$\int_a^b s(x) \, dx = \sum_{k=1}^n s_k (x_k - x_{k-1}) \qquad \text{and} \qquad \int_a^b t(x) \, dx = \sum_{k=1}^n t_k (x_k - x_{k-1}).$

Then, $s_k < t_k$ (since $s(x) < t(x)$ for every $x \in [a,b]$ ), which implies,

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

Prove the linearity property of integrals of step functions

by RoRi

July 20, 2015

Prove that for step function $s(x), t(x)$ defined on an interval $[a,b]$ and for constants $c_1, c_2 \in \mathbb{R}$ , we have

$\int_a^b \left(c_1 s(x) + c_2 t(x) \right) \, dx = c_1 \int_a^b s(x) \, dx + c_2 \int_a^b t(x) \, dx.$

In Apostol, this is Theorem 1.4.

Proof. By the additive property of the integral of step functions we know

$\int_a^b ( c_1 s(x) + c_2 t(x) ) \, dx = \int_a^b c_1 s(x) \, dx + \int_a^b c_2 t(x) \, dx.$

Then, by the homogeneous property (Theorem 1.3 in Apostol) we know

$\int_a^b c_1 s(x) \, dx + \int_a^b c_2 t(x) \, dx = c_1 \int_a^b s(x) \, dx + c_2 \int_a^b t(x) \, dx. \qquad \blacksquare$

Prove the additive property of integrals of step functions

by RoRi

July 20, 2015

For $s(x), t(x)$ step functions defined on an interval $[a,b]$ , prove

$\int_a^b (s(x) + t(x)) \, dx = \int_a^b s(x) \, dx + \int_a^b t(x) \, dx.$

Proof. Let $P = \{ x_0, x_1, \ldots, x_n \}$ be a partition of $[a,b]$ such that $s(x)+t(x)$ is constant on the open subintervals of $P$ (we know such a $P$ exists since we can take the common refinement of the partition of $[a,b]$ on which $s(x)$ and $t(x)$ individually are constant on open subintervals). Let $s(x) + t(x) = s_k + t_k$ if $x_{k-1} < x < x_k$ for $k=1,2, \ldots, n$ . Then, working from the definition of the integral of a step function on an interval, we have

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

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

by RoRi

July 20, 2015

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:

$\displaystyle{\int_a^b s + \int_b^c s = \int_a^c s}$ .
$\displaystyle{\int_a^b (s+t) = \int_a^b s + \int_a^b t}$ .
$\displaystyle{\int_a^b c \cdot s = c \cdot \int_a^b s}$ .
$\displaystyle{\int_{a+c}^{b+c} s(x) \, dx = \int_a^b s(x+c) \, dx}$ .
If $s(x) < t(x)$ for all $x \in [a,b]$ , then $\displaystyle{ \int_a^b s < \int_a^b < t}$ .

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,

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

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

$\begin{align*} \int_a^b s(x) \, dx &= \sum_{k=1}^n c \cdot s_k \cdot (x_k^2 - x_{k-1}^2) \\ &= c \cdot \sum_{k=1}^n s_k (x_k^2 - x_{k-1}^2) \\ &= c \cdot \int_a^b s(x) \, dx. \qquad \blacksquare \end{align*}$
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,
$\begin{align*} \int_{0+1}^{1+1} s(x) \, dx &= 1 \cdot (2^2 - 1^2) = 3 \\ \int_0^1 s(x+1) \, dx &= 1 \cdot (1^2 - 0^2) = 1. \end{align*}$
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

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

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

by RoRi

July 19, 2015

Consider if we chose the following definition for the definite integral of a step function:

$\int_a^b s(x) \, dx = \sum_{k=1}^n s_k^3 \cdot (x_k - x_{k-1}).$

Which of the following properties (all valid for the actual definition) would remain valid under this new definition:

$\displaystyle{\int_a^b s + \int_b^c s = \int_a^c s}$ .
$\displaystyle{\int_a^b (s+t) = \int_a^b s + \int_a^b t}$ .
$\displaystyle{\int_a^b c \cdot s = c \cdot \int_a^b s}$ .
$\displaystyle{\int_{a+c}^{b+c} s(x) \, dx = \int_a^b s(x+c) \, dx}$ .
If $s(x) < t(x)$ for all $x \in [a,b]$ , then $\displaystyle{ \int_a^b s < \int_a^b < t}$ .

True.
Proof. Let $P_1 = \{ x_0, \ldots, x_m \}$ be a partition of $[a,c]$ and $P_2 = \{ y_0, \ldots, y_n \}$ be a partition of $[c,b]$ 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 $x_m = y_0, x_{m+1} = x_0, \ldots, x_{m+n} = y_n$ , so $P$ is a partition of $[a,b]$ and $s(x)$ is constant on the open subintervals of $P$ . Then,

$\begin{align*} \int_a^c s(x) \, dx + \int_c^b s(x) \, dx &= \sum_{k=1}^m s_k^3 (x_k - x_{k-1}) + \sum_{k=1}^n s_k^3 (y_k - y_{k-1}) & (\text{``def." of } \int_a^b s)\\ &= \sum_{k=1}^m s_k^3 (x_k - x_{k-1}) + \sum_{k=m+1}^{n+m} s_k^3 (x_k - x_{k-1}) \\ &= \sum_{k=1}^{m+n} s_k^3 (x_k - x_{k-1}) \\ &= \int_a^b s(x) \, dx. \qquad \blacksquare \end{align*}$
False.
Counterexample: Let $s(x) = t(x) = 1$ for all $x \in [0,1]$ . Then,

$\int_0^1 (s(x) + t(x)) \, dx = 8$

while

$\int_0^1 s(x) \, dx + \int_0^1 t(x) \, dx = 1 + 1 = 2.$

Hence, this property does not hold. (More generally, since $(s+t)^3 \neq s^3 + t^3$ .)
False. Again, this is because $(cs)^3 \neq c(s^3)$ . A specific counterexample is given.
Counterexample: Let $s(x) = 1$ for all $x \in [0,1]$ and let $c = 2$ . Then,

$\int_0^1 c \cdot s(x) \, dx = (2)^3 = 8,$

while,

$c \cdot \int_0^1 s(x) \, dx = 2 \cdot 1 = 2.$
True.
Proof. Let $P =\{ x_0, \ldots, x_n \}$ be a partition of $[a,b]$ such that $s(x) = s_k$ on the $k$ th open subinterval of $P$ . Then,

$P = \{ x_0 + c, x_1 + c, \ldots, x_n + c \}$

is a partition of $[a+c, b+c]$ and $s(x-c) = s_k$ on $x_{k-1} + c < x < x_k + c$ . So,

$\int_a^b s(x) \, dx = \sum_{k=1}^n s_k^3 (x_k - x_{k-1}) \qquad \text{and} \qquad \int_{a+c}^{b+c} s(x-c) \, dx = \sum_{k=1}^n s_k^3 (x_k - x_{k-1}).$

Thus, indeed we have

$\int_a^b s(x) \, dx = \int_{a+c}^{b+c} s(x-c) \, dx. \qquad \blacksquare$
True.
Proof. Since $s(x) < t(x)$ implies $s(x)^3 < t(x)^3$ , the result follows immediately $. \qquad \blacksquare$