Integrals

Intuition

The derivative is taking a distance function (position) and calculating it's velocity over time. Integration is the process of taking a velocity function and deducing the distance traveled. Hence the nickname "anti-derivative"

1. Given a position of an object, the derivative represents its velocity.
2. Given the velocity of an object, the area under the curve to the x axis represents it's position.

Distance vs displacement

1. Distance is given by the area above the x-axis plus the area below the x-axis.
2. Displacement is given by the area above the x-axis minus the area below the x-axis

Displacement is also sometimes called the "signed area between a graph and the horizontal axis".

Integration is typically interested in displacement.

When the velocity function dips below the x axis, the area under that negative portion of the curve should typically be subtracted from your final answer because it is negative velocity.

Intuitively if you run 3 miles and turn around and run 2 miles your displacement is only 1 mile.

Approximation

Let's say we have a velocity function f(x) = x(8-x)

Endpoint Approximation

Remember distance = velocity * time and the area of a rectangle is area = x * y

We can approximate the distance traveled over an interval by breaking the interval into rectangle and calculating the distance of each rectangle.

For each rectangle you can pick an approximation of the velocity at any point in the rectangles width f(x) where x is inside rectangle:

left side = left-endpoint approximation right side = right-endpoint approximation any point in the middle would also work.

If you calculate the distance (area) of each rectangle at left-endpoint and at right endpoint, you know that the real answer is somewhere between those two answers.

Improving the Approximation

If you take endpoint approximation and imagine approaching infinite rectangles or the width of each rectangle approaching 0 you can calculate the actual area under the curve.

Infinite Rectangles

Define a rectangle as width: deltax, height: f(n)

Suppose f(x) is a continuous function on [a,b]. Divide the interval into n subintervals. The width of each subinterval is deltax = (b-a)/n (b-a is length of full interval and you divide by number of rectangles)

The total area would be f(x)deltax for n-1 of n rectangles using left endpoint approximation. Rewritten as deltax(...f(n - 1) for each n)

As an example, estimate the area of f(x) = x^2 on [0, 2] using four rectangles and left-endpoint approximation:

area = deltax(f(x0) + f(x1) + f(x2) + f(x3))

deltax = (b-a)/n = 2-0/4 = 1/2

x0 = 0, x1 = 1/2, x2 = 1, x3 = 3/2

area = 1/2(f(0) + f(1/2) + f(1) + f(3/2))
     = 1/2(0^2 + (1/2)^2 + 1^2 + (3/2)^2)
     = 1.75

If we let n approach infinity, we get a formula called a Riemann Sum:

limn->inf [(b-a)/n][...f(n - 1) for each n]

This exact concept can be rewritten with integral notation (I am going to just write int but google to see real symbol like ∫):

The definite integral of f from a to b is inta-b f(x)dx

dx tells you what the independent variable is, in this case the width of each rectangle. It is implied in this notation that dx approaches 0.

This is the exact area under the graph f(x) on [a, b].

f(x) is called the integrand

a and b are the limits of integration

Types

Definite Integral

A definite integral has given limits on the function like [a, b]

The definite integral represents the signed area under the graph of f(x) on [a, b]

Properties of definite integrals

∫ac fx dx + ∫cb fx dx = ∫ab fd dx
∫aa fx dx = 0
∫ab fx dx = -(∫ba fx dx)

linearity property
∫[a to b] (f(x) + g(x)) dx = ∫[a to b] f(x) dx + ∫[a to b] g(x) dx

Units

Definite integrals take rate of change and give you the quantity

If f(x) is miles/hr then F(x) = miles

If f(x) is liters/min then F(x) = liters

Indefinite Integral

Indefinite integrals have no limits on the input of the function (-inf, inf)

The indefinite integral represents the family of anti derivatives of f(x)

F(x) + C (any constant)

Improper Integrals

Improper integrals involve infinity.

type 1: ∫[1,inf] 1/x^2 dx

type 2: ∫[0,2] 1/sqrt(x) dx

You can solve this by writing it as a limit and then solving the limit:

lim[m->inf] ∫[1,m] 1/x^2 dx

lim[m->inf] -1/x |1,m

lim[m->inf] -1/m - -1/1

-1/inf + 1 = 1

Because the limit is a number, we say this integral converges.

The integral diverges if the limit goes to infinity or is undefined.

You are not allowed to send both bounds to infinity at the same time, so if your integral is bounded above and below with inf/-inf you need to split it into two limits and bound 1 side with a number for each.

Remember infinity does not cancel out. So inf - inf just diverges.

In type 2, the limit is set up so m-> the bound that results in undefined. You might have to split it up if the undefined behavior is in the middle, like ∫[-1,1] 1/x dx

You can have combo of part1 and part2. Split it into however many you need and add them if they result in numbers. If any is inf, the whole thing diverges.

Factorials

You can write factorials as an integral with this formula:

n! = ∫[0,inf] x^n e^(-x) dx

(1/2)! = ∫[0,inf] x^(1/2) e^(-x) dx

You can prove this with induction using integration by parts. Start at 1!, then 2! works out to 2 _ 1! and 3! works out to 3 _ 2! etc…

∫[0, inf] x^3 e^-x dx

integration by parts:

u = x^3
du = 3x^2
dv = e^-x
v = -e^-x

(-x^3e^-x |[0,inf]) + 3∫[0,inf] x^2 e^-x dx

0 + 3∫[0,inf] x^2 e^-x dx

= 3 * 2!