Definite Integral

class: center, middle, inverse, title-slide

.title[
# Definite Integral
]
.subtitle[
## Session 02
]
.author[
### Alejandro Ucan
]
.date[
### 2023-10-29
]

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# Goals:

* Prove that Riemann sums approximate the area under the curve. 
 * Define the definite integral. 
 * Compute definite integrals using the definition. 
 * State the Foundamental Theorem of Calculus. 
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# Recall Riemann sums

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# Definite Integral

> __Definition:__ Let `$f$` be a function defined on the interval `$[a,b]$`. The definite integral of `$f$` from `$a$` to `$b$` is defined as the limit of a Riemann sum when we increase the number of subintervals in a partition. In symbols `$$\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x_i$$` provided that this limit exists and gives the same value for all possible choices of sample points `$x_i^*$` in the subintervals `$[x_{i-1}, x_i]$`.

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## Some conventions:

* We will only use regular partitions but the definition is valid for any kind of partition. 
* We will only consider left Riemann sums, but this computation is also valid for right and middle sums. 
* We will use the notation `$\Delta x = \frac{b-a}{n}$` and `$x_i = a + i \Delta x$`.

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#### Example 1:

1. Compute the definite integral `$\int_0^1 x dx$` using the definition.

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Let's start by computing the Riemann sum for `$n$` subintervals. The length of the subintervals is given by `$\Delta x= \frac{1-0}{n}$` and the left points are `$$x_j=0 + j \Delta x = \frac{j}{n}$$` for `$j=0,1,2,\ldots,n$`. The Riemann sum is given by `$$\sum_{j=1}^n f(x_j) \Delta x = \sum_{j=1}^n \frac{j}{n} \frac{1}{n} = \frac{1}{n^2} \sum_{j=1}^n j = \frac{1}{n^2} \frac{n(n+1)}{2} = \frac{n+1}{2n}$$`

Let's take the limit in the number of intervals: `$$\lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2}.$$`

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#### Example 2:

1. Compute the definite integral `$\int_1^4 4x+3 dx$` using the definition.

--
Let's start by computing the Riemann sum for `$n$` subintervals. The length of the subintervals is given by `$\Delta x= \frac{1-{-1}}{n}$` and the left points are `$$x_j=-1 + j \Delta x = -1 + \frac{j}{n}$$` for `$j=0,1,2,\ldots,n$`. The Riemann sum is given by `$$\sum_{j=1}^n f(x_j) \Delta x = \sum_{j=1}^n (4(-1 + \frac{j}{n})+3) \frac{2}{n} = \frac{2}{n} \sum_{j=1}^n (4(-1 + \frac{j}{n})+3) = \frac{2}{n} \sum_{j=1}^n (4-4\frac{j}{n}+3) = \frac{2}{n} \sum_{j=1}^n (7-4\frac{j}{n}) = \frac{2}{n} (7n-4\frac{n(n+1)}{2}) = 7-2\frac{n+1}{n}$$`

Let's take the limit in the number of intervals: `$$\lim_{n \to \infty} 7-2\frac{n+1}{n} = 7-2 = 5.$$`

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#### Example 3:

1. Compute the definite integral `$\int_0^1 x^2 dx$` using the definition.

--
Let's start by computing the Riemann sum for `$n$` subintervals. The length of the subintervals is given by `$\Delta x= \frac{1-0}{n}$` and the left points are `$$x_j=0 + j \Delta x = \frac{j}{n}$$` for `$j=0,1,2,\ldots,n$`. The Riemann sum is given by `$$\sum_{j=1}^n f(x_j) \Delta x = \sum_{j=1}^n \frac{j^2}{n^2} \frac{1}{n} = \frac{1}{n^3} \sum_{j=1}^n j^2 = \frac{1}{n^3} \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}$$`

Let's take the limit in the number of intervals: `$$\lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \frac{2}{3}.$$`

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# The Fundamental Theorem of Calculus

> __Theorem:__ Let `$f$` be a continuous function on the interval `$[a,b]$`. If `$F$` is any antiderivative of `$f$`, then `$$\int_a^b f(x) dx = F(b) - F(a).$$`

#### Example 4:

1. Compute the definite integral `$\int_0^1 x dx$` using the Fundamental Theorem of Calculus.

The antiderivative of `$x$` is `$\frac{x^2}{2}$` and the definite integral is given by `$$\int_0^1 x dx = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}.$$`

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#### Example 5:

Compute the following definite integrals using the Fundamental Theorem of Calculus.

1. `$\int_-1^4 4x+3 dx$` 
1. `$\int_0^1 x^2 dx$` 
1. `$\int_0^3 x^2-3x+5 dx$` 
1. `$\int_1^e \frac{1}{x} dx$`