Chapter 4 Integration

• Area
• Properties of Integrals
• Fundamental Theorem of Calculus

# Area

曲線下的面積 (只考慮$[a,b]$ 上的非負函數)

Let $S = \{(x,y) | a \le x \le b , 0 \le y \le f(x) \}.$
Divide $[a,b]$ into $n$ subintervals:
$[x_0,x_1],[x_1,x_2]...[x_{n-1},x_n], \; where \; x_0 =a, x_n = b$
Use a rectangle to approximate each $S_i$.Define that:

• $U_n = f(x_1)\Delta x_1+f(x_2)\Delta x_2+...+f(x_n)\Delta x_n = \sum^n_{i=1}f(x_i)\Delta x_i$
• $R_n = f(x_0)\Delta x_1+f(x_1)\Delta x_2+...+f(x_{n-1})\Delta x_n = \sum^n_{i=1}f(x_{i-1})\Delta x_i$
  $where \Delta x_n = x_i - x_{i-1} \forall i \in \{ 1,2,...,n \}.$

# Define

If $f: [a,b] \to \mathbb{R^+} \cup \{ 0 \}$ is continuous,
then $A = \lim_{n \to \infty}R_n = \lim_{n \to \infty} \sum^n_{i=1}f(x_i)\Delta x_i$
then $A = \lim_{n \to \infty}L_n = \lim_{n \to \infty} \sum^n_{i=1}f(x_{i-1})\Delta x_i$

# Partition

$P = \{ a = x_0, x_1,x_2,...,x_n =b \}$ is a partition of $[a,b]$.

1. 等分: $\Delta x_i = \frac{b-a}{n}$.
2. 不等分: Ex.$x_0 =a,x_1=ar,...,x_n = ar^n = b, r = \sqrt[n]{\frac{b}{a} } ab \ge 0 , a \neq 0$
   $P = \{ a,ar,ar^2,...,ar^n \},\Delta x_i =x_i - x_{i-1} = ar^i - ar^{i-1}.$

# Example

$P = \{0 = x_0,x_1 = \frac{1}{2^n}, x_2 = \frac{1}{2^{n-1}},...,x_{n-1} = \frac{1}{4}, x_n = \frac{1}{2} \}.$
$\Delta x_i = x_i - x_{i-1} = \frac{1}{2^{n+1-x}} - \frac{1}{2^{n-x}}, \left \| P \right \| = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.$

# Example

$f(x) = x^2$ on $[a,b]$. Find $A$.

1. Taking a partition of $P = \{ a = x_0, x_1,...,x_n=b \}$,where $x_i = a + \frac{b-a}{n}i,\Delta x = \frac{b-a}{n}$.
2. $L_n = f(x_0)\Delta x+f(x_1)\Delta x+...+f(x_{n-1})\Delta x = \frac{b-a}{n} \sum^{n-1}_{i=0}f(x_{i}) \\= \frac{b-a}{n} \sum^{n-1}_{i=0} \{ a^2 + 2\frac{a(b-a)}{n}i + \frac{(b-a)^2}{n^2}i^2 \} = a^2(b-a) + \frac{2a(b-a)^2}{n^2} \cdot \frac{n(n-1)}{2} + \frac{(b-a)^3}{n^3}\cdot \frac{n(n-1)(2n-1)}{6} \\= a^2(b-a) + a(b-a)^2(1-\frac{1}{n})+\frac{(b-a)^3}{6}(1-\frac{1}{n})(2-\frac{1}{n})$
3. $A = \lim_{n \to \infty}L_n = a^2(b-a)+a(b-a)^2 + \frac{1}{3}(b-a)^3$

# Define

$U_n$，上和: Taking $x_i* \in [x_{i-1},x_i]$ such that $f$ has a maximum $f(x_i*) on [x_{i-1},x_i]$,then $\sum_{i=1}^{n}f(x_i*)\Delta x_i$ is called the upper sum of $f$ on $[a,b]$ with a partition $P = \{ x_0 =a,x_1,...,x_n =b \}$.

$L_n$，下和: Taking $x_i* \in [x_{i-1},x_i]$ such that $f$ has a minimum $f(x_i*) on [x_{i-1},x_i]$,then $\sum_{i=1}^{n}f(x_i*)\Delta x_i$ is called the lower sum of $f$ on $[a,b]$ with a partition $P = \{ x_0 =a,x_1,...,x_n =b \}$.

# The Definite Integral

Divide $[a,b]$ into $n$ subintervals (子區間) of each width $\Delta x = \frac{b-a}{n}$:
$P = \{ x_0 =a,x_1,...,x_n = b \}$.
Let $x_i*$ be a sample point of subinterval $[x_{i-1},x_i]$ for all $i = 1,2,...,n$
The definite integral of $f$ from $a$ to $b$ is defined by $\int^b_af(x)dx = \lim_{n\to \infty}\sum_{i=1}^nf(x_i*)\Delta x$ if the limit exists.
Moreover, we say that $f$ is integrable on $[a,b]$.

# Example

$\int^{\frac{\pi}{2}}_0 sinxdx = ?$

Taking $x_i = 0+\frac{\frac{\pi}{2}-0}{n}\cdot i, \; i = \{ 1,2,3,...,n\}, \; \Delta x = \frac{\pi}{2n}$.

Let the right sum be $\sum_{i=1}^nf(x_i)\Delta x = \sum_{i=1}^n \frac{\pi}{2n}sin(\frac{\pi}{2n}i) = R_n$
$\because 2sin(\frac{\pi}{4n})sin(\frac{\pi}{2n}i) = cos (\frac{\pi}{4n} - \frac{\pi}{2n}i) - cos (\frac{\pi}{4n} + \frac{\pi}{2n}i) = cos(\frac{\pi(1-2i)}{4n}) - cos(\frac{\pi(1+2i)}{4n})$
$\Rightarrow sin(\frac{\pi}{2n}i) = \frac{1}{2sin(\frac{\pi}{4n})}(cos(\frac{\pi(1-2i)}{4n}) - cos(\frac{\pi(1+2i)}{4n}))$
$\therefore R_n = \frac{1}{2sin(\frac{\pi}{4n})} \cdot \frac{\pi}{2n} \sum^n_{i=1}(cos(\frac{\pi(1-2i)}{4n}) - cos(\frac{\pi(1+2i)}{4n})) = \frac{\frac{\pi}{4n}}{sin(\frac{\pi}{4n})} \{ (cos\frac{\pi}{4n} - cos\frac{3\pi}{4n})+(cos\frac{3\pi}{4n} - cos\frac{5\pi}{4n}) +...+(cos\frac{(2n-1)\pi}{4n} - cos\frac{(2n+1)\pi}{4n}) \} = \frac{\frac{\pi}{4n}}{sin(\frac{\pi}{4n})} (cos(\frac{\pi}{4n}) - cos(\frac{\pi}{4}(2+\frac{1}{n})) ).$
$\Rightarrow \lim_{n \to \infty}R_n = \{ 1-0 \} =1$.

# Example

# Example

$\int^b_a x^q dx, q \neq 1, q \in \mathbb{Q}, 0<a<b$

Let $r = (\frac{b}{a})^\frac{1}{n}$ and $x_i = ar^i, i = 0,1,...,n, \Delta x_i = ar^{i-1}(r-1)$.
$P = {x_0 = a,x_1,x_2,...x_n=b}$.
$R_n = \sum^n_{i=1}f(x_{i-1})\Delta x_i = \sum^n_{i=1}(x_{i-1})^q \cdot ar^{i-1}(r-1) = \sum^n_{i=1}(ar^{i-1})^q \cdot ar^{i-1}(r-1) \\= a^{q+1}(r-1) \sum^n_{i=1}r^{q(i-1)}r^{i-1} = a^{q+1}(r-1) \sum^n_{i=1}r^{(q+1)(i-1)} = a^{q+1}(r-1)\frac{1\cdot (r^{(q+1)n}-1)}{r^{q+1}-1} \\= \frac{r-1}{r^{q+1}-1}a^{q+1}((\frac{b}{a})^{q+1}-1) = \frac{r-1}{r^{q+1}-1} (b^{q+1} - a^{q+1})$
$\because \frac{d}{dr}|_{q=1}r^{q+1} = q+1$
$\therefore \lim_{r \to 1} \frac{r^{q+1}-1}{r-1} = \frac{d}{dr}|_{q=1}r^{q+1} = q+1$

$\Rightarrow \lim_{n \to \infty}R_n = \lim_{n \to \infty}\sum^n_{i=1}f(x_{i-1})\Delta x_i = \lim_{n \to \infty}\frac{r-1}{r^{q+1}-1} (b^{q+1} - a^{q+1}) = \lim_{r \to 1}\frac{r-1}{r^{q+1}-1} (b^{q+1} - a^{q+1}) = \frac{1}{q+1}(b^{q+1}-a^{q+1}) = \int^b_a x^q dx$.

# Properties of Integrals

1. $\int^b_af(x)dx$ is a real number.
2. $\int^b_af(x)dx = \int^b_af(y)dy = \int^b_af(t)dt$
3. $\int^a_af(x)dx = 0$
4. $\int^b_af(x)dx = -\int^a_bf(x)dx$
5. $\int^b_af(x)dx = \int^c_af(x)dx + \int^b_cf(x)dx$ ($c \in \mathbb{R}$)
6. $\int^b_a k dx = k(b-a)$
7. $\int^b_a(kf \pm hg)dx = k\int^b_af(x)dx \pm h\int^b_ag(x)dx$
8. $\int^b_af(x)dx \le |\int^b_af(x)dx| \le \int^b_a|f(x)|dx$
9. If $f$ is odd, then $\int^a_{-a}f(x)dx = 0$
   If $f$ is even, then $\int^a_{-a}f(x)dx = 2\int^a_0f(x)dx$

# Fundamental Theorem of Calculus

# Thm

If $f$ is continuous on $[a,b]$, then $g(x)= \int^x_a f(t)dt ,a\le x \le b$ is also continuous on $[a,b]$ and differentiable on $(a,b)$ and $g'(x) = f(x)$.
Moreover, $\frac{d}{dx}\int^x_af(t)dt = f(x)$.

# Example

# Thm

If $f$ is continuous on $[a,b]$,$\int^b_af(x)dx = F(b) - F(a)$,where $F(x)$ is an antiderivative of $f$.

# Example

$\int f'(x)g(x)dx = f(x)g(x) - \int f(x)g'(x)dx$
$\Rightarrow \int^b_a f'(x)g(x)dx = f(x)g(x) |^b_a - \int^b_a f(x)g'(x)dx = f(b)g(b) -f(a)g(a) - \int^b_a f(x)g'(x)dx$.

# Application

1. $\frac{d}{dx}\int^{g(x)}_a f(t)dt = f(g(x)) \cdot g'(x)$
2. $\frac{d}{dx}\int^{g(x)}_{h(x)} f(t)dt = f(g(x)) \cdot g'(x) - f(h(x))\cdot h'(x)$

# Thm

If $f$ is continuous on $[a,b]$, then

1. $\frac{d}{dx}\int^x_af(t) dt = f(x)$
2. $\int^b_af(x)dx = F(b) - F(a)$ if $F'(x) = f(x)$

# Reference

• 蘇承芳老師 - 微積分甲（一）109 學年度 - Calculus (I) Academic Year 109