Chapter 2 Proofs: Theorems on Definite Integrals

Proof.

Let $P = \{a = x_0 < x_1 < \cdots < x_n = b\}$ be an arbitrary partition of $[a,b]$. By the telescoping sum we have

$$F(b) - F(a) = \sum_{i=1}^{n} \bigl[F(x_i) - F(x_{i-1})\bigr]$$

Applying the Mean Value Theorem to each subinterval $[x_{i-1}, x_i]$, there exists $c_i \in (x_{i-1}, x_i)$ such that

$$F(x_i) - F(x_{i-1}) = F'(c_i)(x_i - x_{i-1}) = f(c_i)\,\Delta x_i$$

Therefore

$$F(b) - F(a) = \sum_{i=1}^{n} f(c_i)\,\Delta x_i$$

The right-hand side is a Riemann sum. Since $f$ is continuous on $[a,b]$, it is integrable, and as the mesh $\|P\| = \max_i \Delta x_i \to 0$, every Riemann sum converges to $\displaystyle\int_a^b f(x)\,dx$. Since the left-hand side $F(b) - F(a)$ is a constant independent of the partition,

$$F(b) - F(a) = \int_a^b f(x)\,dx$$

as required. $\blacksquare$

Proof.

Fix $x \in [a,b]$ and consider the difference quotient for $h \neq 0$ (with $x+h \in [a,b]$).

$$\frac{G(x+h) - G(x)}{h} = \frac{1}{h}\left[\int_a^{x+h} f(t)\,dt - \int_a^x f(t)\,dt\right] = \frac{1}{h}\int_x^{x+h} f(t)\,dt$$

Since $f$ is continuous on a closed interval, by the Mean Value Theorem for integrals there exists $c_h$ between $x$ and $x+h$ such that

$$\frac{1}{h}\int_x^{x+h} f(t)\,dt = f(c_h)$$

As $h \to 0$, we have $c_h \to x$, and by the continuity of $f$, $f(c_h) \to f(x)$. Therefore

$$G'(x) = \lim_{h \to 0} \frac{G(x+h) - G(x)}{h} = \lim_{h \to 0} f(c_h) = f(x)$$

as required. $\blacksquare$

Proof.

For any partition $P = \{x_0, x_1, \dots, x_n\}$ of $[a,b]$ and sample points $\xi_i \in [x_{i-1}, x_i]$, consider the Riemann sum.

$$\sum_{i=1}^{n} \bigl[\alpha\,f(\xi_i) + \beta\,g(\xi_i)\bigr]\,\Delta x_i = \alpha \sum_{i=1}^{n} f(\xi_i)\,\Delta x_i + \beta \sum_{i=1}^{n} g(\xi_i)\,\Delta x_i$$

This holds trivially by the linearity of finite sums. Taking the limit as $\|P\| \to 0$, since both $f$ and $g$ are integrable, each Riemann sum converges to the respective definite integral. Therefore

$$\int_a^b \bigl[\alpha\,f(x) + \beta\,g(x)\bigr]\,dx = \alpha\int_a^b f(x)\,dx + \beta\int_a^b g(x)\,dx$$

as required. $\blacksquare$

Proof.

Consider a partition $P$ of $[a,b]$ that includes $c$ as a partition point (say $c = x_k$). Then

$$P_1 = \{x_0, x_1, \dots, x_k\}, \quad P_2 = \{x_k, x_{k+1}, \dots, x_n\}$$

are partitions of $[a,c]$ and $[c,b]$ respectively, and the Riemann sum with sample points $\xi_i$ decomposes as

$$\sum_{i=1}^{n} f(\xi_i)\,\Delta x_i = \sum_{i=1}^{k} f(\xi_i)\,\Delta x_i + \sum_{i=k+1}^{n} f(\xi_i)\,\Delta x_i$$

For a partition that does not include $c$, we can consider the refinement obtained by adding $c$. The Riemann sum changes by at most $2M\|P\|$ (where $M = \sup|f|$), which vanishes as $\|P\| \to 0$.

Therefore, taking the limit as $\|P\| \to 0$,

$$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$$

as required. $\blacksquare$

Proof.

For $a < b$, the definite integral is defined as the limit of Riemann sums:

$$\int_a^b f(x)\,dx = \lim_{\|P\| \to 0} \sum_{i=1}^{n} f(\xi_i)\,(x_i - x_{i-1})$$

When $a > b$, the integral is defined by taking a partition $b = x_0 < x_1 < \cdots < x_n = a$ as a Riemann sum. Here $\Delta x_i = x_i - x_{i-1} > 0$, but the direction from $a$ to $b$ is reversed, so by convention

$$\int_b^a f(x)\,dx := -\int_a^b f(x)\,dx$$

This can also be derived from the interval splitting property. Setting $c = a$ in the interval splitting theorem gives

$$\int_a^a f(x)\,dx = \int_a^b f(x)\,dx + \int_b^a f(x)\,dx$$

Since the left-hand side is $0$ (proved in the next section),

$$\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$$

as required. $\blacksquare$

Proof.

Since the length of the integration interval is $0$, for any partition $P$ the interval $[a,a]$ consists only of $\Delta x_i = 0$. Therefore the Riemann sum is

$$\sum_{i=1}^{n} f(\xi_i)\,\Delta x_i = 0$$

Since every Riemann sum equals $0$, the limit is also $0$.

$$\int_a^a f(x)\,dx = 0$$

$\blacksquare$

Proof.

Define $h(x) = g(x) - f(x)$. By hypothesis, $h(x) \ge 0$ for all $x \in [a,b]$. Since $h$ is the difference of two integrable functions, it is integrable.

For any partition $P$ and sample points $\xi_i$, since $h(\xi_i) \ge 0$ and $\Delta x_i > 0$,

$$\sum_{i=1}^{n} h(\xi_i)\,\Delta x_i \ge 0$$

Taking the limit as $\|P\| \to 0$,

$$\int_a^b h(x)\,dx \ge 0$$

By linearity,

$$\int_a^b g(x)\,dx - \int_a^b f(x)\,dx \ge 0$$

That is, $\displaystyle\int_a^b f(x)\,dx \le \int_a^b g(x)\,dx$. $\blacksquare$

Proof.

By the interval splitting property,

$$\int_{-a}^{a} f(x)\,dx = \int_{-a}^{0} f(x)\,dx + \int_0^a f(x)\,dx$$

Apply the substitution $x = -t$ ($dx = -dt$) to the left integral. When $x = -a$ we have $t = a$, and when $x = 0$ we have $t = 0$, so

$$\int_{-a}^{0} f(x)\,dx = \int_a^0 f(-t)\,(-dt) = \int_0^a f(-t)\,dt$$

Therefore

$$\int_{-a}^{a} f(x)\,dx = \int_0^a f(-t)\,dt + \int_0^a f(x)\,dx = \int_0^a \bigl[f(-x) + f(x)\bigr]\,dx$$

(1) Even function case: Since $f(-x) = f(x)$,

$$\int_{-a}^{a} f(x)\,dx = \int_0^a \bigl[f(x) + f(x)\bigr]\,dx = 2\int_0^a f(x)\,dx$$

(2) Odd function case: Since $f(-x) = -f(x)$,

$$\int_{-a}^{a} f(x)\,dx = \int_0^a \bigl[-f(x) + f(x)\bigr]\,dx = \int_0^a 0\,dx = 0$$

$\blacksquare$