Week 11-12 - Antidifferentiation & Integration #

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.
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• the sum of antiderivatives
  • $\int(f(x) +g(x)) dx = \int(f(x)) dx + \int(g(x) dx$
• The antiderivative of $f(mx+b)$
  • $\int f(mx+b) dx = \dfrac{F(mx+b)}{m} + C$

Integration #

Summation Notation #

• Sum of n, n², or n³(Brilliant.org)

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

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

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• $\displaystyle\sum_{i=0}^{n}i^2 = \frac{(2n+1)(n+1)n}{6}$

  • Animation: http://www.takayaiwamoto.com/Sums_and_Series/sumsqr_1_anim.gif
  • khanacademy.org

• $\displaystyle\sum_{i=0}^{n}i^3 = (\sum_{i=0}^{n}i)^2$

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  • From Math can be fun with CAD

• What about $\displaystyle\sum_{i=0}^{n}i^m$?

  • Sum of 1 to N^m,

The Area Problem #

• Find the area of the region S that lies under the curve $y = f(x)$ from a to b.
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• 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: $x_i^* = x_{i}$
    • $\displaystyle\sum_{i=1}^nf(x_{i}) \cdot (x_i-x_{i-1})$
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  • PS: $x_i^*$ can be any number in the ith subinterval $[x_{i-1}, x_i]$.

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 $[a, b]$.

Compute Integrals #

• What is the integral of x^2 from x=0 to 1? ($\int_0^1 x^2 dx$)
  • $\displaystyle \int_0^1 x^2 dx = \sum_{i=1}^n f(x_i^*) \cdot (x_i - x_{i-1})$
  • Each rectangle has width $\frac{1}{n}$ and the heights are the values of the function $f(x) = x^2$ at the points $\frac{1}{n}, \frac{2}{n}, \ldots, \frac{n}{n}$; that is, the heights are $(\frac{1}{n})^2, (\frac{2}{n})^2, \ldots, (\frac{n}{n})^2$.
  • $\begin{aligned}\int_0^1 x^2 dx &= \lim_{n \to \infty} \sum_{i=1}^n (\frac{i}{n})^2 \cdot \frac{1}{n} \\ &= \lim_{n \to \infty} \frac{1}{n^2} \sum_{i=1}^n i^2 \cdot \frac{1}{n} = \lim_{n \to \infty} \frac{1}{n^3} \cdot \sum_{i=1}^n i^2 \\ &= \lim_{n \to \infty}(\frac{1}{n^3} \cdot \frac{(n)(n+1)(2n+1)}{6}) \\ &= \lim_{n \to \infty}(\frac{1}{n^3} \cdot \frac{2n^3+3n^2+n}{6}) \\ &= \frac{1}{3} \end{aligned}$
  • So, $3\int_0^1 x^2 dx = \int_0^1 3x^2 dx = 1$: like stretching the area vertically, then we get $\int_0^1 3x^2 dx$
• 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
• $\displaystyle \int_a^b [f(x) + g(x)] dx = \int_a^b f(x) dx + \int_a^b g(x) dx$
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• $\displaystyle \int_a^b [f(x) - g(x)] dx = \int_a^b f(x) dx - \int_a^b g(x) dx$
• $\displaystyle \int_a^b c f(x) dx = c \int_a^b f(x) dx$
• $\displaystyle \int_a^c f(x) dx + \int_c^b f(x) dx = \int_a^b f(x) dx$
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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).
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• 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.

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.

Refers #

• http://www.takayaiwamoto.com/Sums_and_Series/Sums_and_Series.html