# Week 11-12 - Antidifferentiation & Integration

## 1. Antiderivatives

• Anti-differentiation is a bridge between the differentiation and the integration.
• Theorem If $f$ is an antiderivative of $f$ on an interval I, then the most general antiderivative of $f$ on I is $F(x) + C$ where C is an arbitrary constant.
• the sum of antiderivatives
• $\int(f(x) +g(x)) dx = \int(f(x)) dx + \int(g(x) dx$
• The antiderivative of $f(mx+b)$
• $\int f(mx+b) dx = \dfrac{F(mx+b)}{m} + C$

## 2. Integration

### 2.1. Summation Notation

• Sum of n, n², or n³(Brilliant.org)

• $\displaystyle\sum_{i=0}^{n}i = \frac{n(n+1)}{2}$

• $\displaystyle\sum_{i=0}^{n}(2i-1) = n^2$
• $\displaystyle\sum_{i=0}^{n}i^2 = \frac{(2n+1)(n+1)n}{6}$
• $\displaystyle\sum_{i=0}^{n}i^3 = (\sum_{i=0}^{n}i)^2$
• What about $\displaystyle\sum_{i=0}^{n}i^m$?

### 2.2. The Area Problem

• Find the area of the region S that lies under the curve $y = f(x)$ from a to b.
• Riemann Sum:
• Patition the intervel $[a, b]$ and $x_0 = a < x_1 < x_2 < \ldots < x_n = b$. Choose sample points $x_i$, then $\text{Area}\ \approx \sum_{i=1}^nf(x_i) \cdot (x_i - x_{i-1})$
• Left Riemann Sum: $x_i^* = x_{i-1}$
• $\displaystyle\sum_{i=1}^nf(x_{i-1}) \cdot (x_i-x_{i-1})$
• Right Riemann Sum: $x_i^* = x_{i}$
• $\displaystyle\sum_{i=1}^nf(x_{i}) \cdot (x_i-x_{i-1})$
• PS: $x_i^*$ can be any number in the ith subinterval $[x_{i-1}, x_i]$.

### 2.3. Definition of a Definite Integral

• If $f$ is a function defined for $a \le x \le b$, we divide the interval $[a, b]$ into n subintervals of equal width $\Delta x = (b - a)/n$. We let $x_0(=a), x_1, x_2, \ldots, x_n(=b)$ be the endpoints of these subintervals and we let $x_1^*, x_2^*, \ldots, x_n^*$ be any samples points in these subintervals, so $x^*$ lies in the ith subinterval $[x_{i-1}, x_i]$. Then the definite integral of f from a to b is $\int_a^bf(x)dx = \lim_{n \to \infty}^n f(x_i^*) \Delta x$
• provided that this limit exists and gives the same value for all possible choices of sample points. If it does exist, we say that f is integrable on $[a, b]$.

### 2.4. Compute Integrals

• What is the integral of x^2 from x=0 to 1? ($\int_0^1 x^2 dx$)

• $f(x) = x^2$ is continuous so it is integrable $x_i = \frac{1}{n}$, check the figure above.
• $\displaystyle\sum_{i=1}^n f(x_i^*) \cdot (x_i - x_{i-1}) = \sum_{i=1}^n (\frac{1}{n})^2 \cdot \frac{1}{n}$
• \begin{aligned}\int_0^1 x^2 dx &= \sum_{i=1}^n (\frac{1}{n})^2 \cdot \frac{1}{n} \\ &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n^3} \cdot \frac{1}{n} = \lim_{n \to \infty} \frac{1}{n^3} \cdot \sum_{i=1}^n i^2 \\ &= \lim_{n \to \infty}(\frac{1}{n^3} \cdot \frac{(n)(n+1)(2n+1)}{6}) \\ &= \lim_{n \to \infty}(\frac{1}{n^3} \cdot \frac{2n^3+3n^2+n}{6}) \\ &= \frac{1}{3} \end{aligned}
• So, $3\int_0^1 x^2 dx = \int_0^1 3x^2 dx = 1$: like stretching the area vertically, then we get $\int_0^1 3x^2 dx$
• This deduction doesn't say that the area of x^2 from x=0 to 1 is 1/3. It use limit theorem to point out that, if we removed a little piece of area 1/3 then we can fit it in area $\int_0^1 3x^2 dx$. And the same, if we add a little piece of area 1/3 then we can cover it up. It is all about limits.
• What is the integral of x^3 from x = 1 to 2?

• \begin{aligned} \sum_{i=1}^n f(x_i^*) \cdot (x_i - x_{i-1}) &= \sum_{i=1}^n f(\frac{2}{n} \cdot i) \cdot \frac{2}{n} \\ &= \sum_{i=1}^n (\frac{2}{n} \cdot i)^3 \cdot \frac{2}{n} \\ &= \sum_{i=1}^n\frac{16}{n^4} \cdot i^3 \end{aligned}
• \begin{aligned}\int_0^2 x^3 dx &= \lim_{n \to \infty} \sum_{i=1}^n \frac{16}{n^4} \cdot i^3 \\ &= \lim_{n \to \infty} \sum_{i=1}^n \frac{16}{n^4} \cdot \sum_{i=1}^n i^3 = \lim_{n \to \infty} \sum_{i=1}^n \frac{16}{n^4} \cdot (\sum_{i=1}^n i)^2 \\ &= \lim_{n \to \infty} \sum_{i=1}^n \frac{16}{n^4} \cdot (\frac{(n)(n+1)}{2})^2 = \lim_{n \to \infty} \frac{4n^2(n+1)^2}{n^4} \\ &= 4 \end{aligned}
• Use same deduction, we get $\displaystyle\int_0^1 x^3 dx = \frac{1}{4}$
• Then $\displaystyle\int_0^2 x^3 dx = \displaystyle\int_0^1 x^3 dx + \displaystyle\int_1^2 x^3 dx = 1/4 + 4$

#### Properties of the Integral

• $\displaystyle \int_a^b c dx = c(b-a)$, where c is any constant
• $\displaystyle \int_a^b [f(x) + g(x)] dx = \int_a^b f(x) dx + \int_a^b g(x) dx$
• $\displaystyle \int_a^b [f(x) - g(x)] dx = \int_a^b f(x) dx - \int_a^b g(x) dx$
• $\displaystyle \int_a^b c f(x) dx = c \int_a^b f(x) dx$
• $\displaystyle \int_a^c f(x) dx + \int_c^b f(x) dx = \int_a^b f(x) dx$

#### The Accumulation Function Increasing/Decreasing

• If f takes on both positive and negative values, as in the figure below, then the Riemann sum is the sum of the areas of the rectangles that lie above the x-axis and the negatives of the areas of the rectangles that lie below the x-axis (the areas of the blue rectangles minus the areas of the gold rectangles).
• So when f is positive, A(x) increasing, and A'(x) > 0. In other way, when f is negative, A(x) decreasing, and A'(x) < 0.

### 2.5. The Integral of Sin x From -1 To 1

• \begin{aligned} \int_{-1}^1 \sin x dx &= \int_{-1}^0 \sin x dx + \int_0^1 \sin x dx \\ &= \int_0^1 \sin (-x) dx + \int_0^1 \sin x dx \\ &= \int_0^1 - \sin x dx + \int_0^1 \sin x dx \\ &= - \int_0^1 \sin x dx + \int_0^1 \sin x dx \\ &= 0 \end{aligned}
• So, symmetry can be exploited to calculate some integrals.