Riemann Integration fX a b By Joey Stehno

Riemann Integration f(X) a b By: Joey Stehno

f(X) How can we find the area under this curve? a b

a b] in Lets divide [ , half (meaning two intervals) [ X 0 , X 1 ] & [ X 1 , X 2 ] a= X 0 X 1 b= X 2

Note Δ means “change in” so ΔX 1 = X 1 – X 0 ΔX 2 = X 2 – X 1 In this case Δ can be thought as the width of the interval a= X b= X X 1 0 ΔX 1 ΔX 2 2

f(X 0) Now label f(X 0) & f(X 1) on the function f(X 1) a= X b= X X 1 0 ΔX 1 ΔX 2 2

f(X 0) Do you remember how to find the area of a rectangle? f(X 1) a= X b= X X 1 0 ΔX 1 ΔX 2 2

f(X 0) Rectangle 1 f(X 1) Rectangle 2 ΔX 1 ∙ f(X 0) a= X ΔX 2 ∙ f(X 1) b= X X 1 0 ΔX 1 ΔX 2 2

Area of the rectangles= (of Left Sided Integral) f f ΔX 1 ∙ (X 0) + ΔX 2 ∙ (X 1) f(X 0) f(X 1) ΔX 1 ∙ f(X 0) a= X ΔX 2 ∙ f(X 1) b= X X 1 0 ΔX 1 ΔX 2 2

Area of the rectangles= (of Left Sided Integral) f f ΔX 1 ∙ (X 0) + ΔX 2 ∙ (X 1) Note: Using only two intervals is a bad estimate for finding the area under a curve. f(X 0) So lets try splitting the function up into four intervals f(X 1) ΔX 1 ∙ f(X 0) a= X ΔX 2 ∙ f(X 1) b= X X 1 0 ΔX 1 ΔX 2 2

a= X 0 X 1 X 2 X 3 b= X 4

f(X 3) f(X 0) f(X 2) f(X 1) a= X 0 ΔX 1 X 2 ΔX 2 b= X X 3 ΔX 4 4

Area of the rectangles= f f f ΔX 1 ∙ (X 0) + ΔX 2 ∙ (X 1) + ΔX 3 ∙ (X 2) + (of Left Sided Integral) f ΔX 4 ∙ (X 3) f(X 0) f(X 2) f(X 1) a= X 0 ΔX 1 Ehh…it’s a better estimate but lets try splitting it up into more intervals! X 2 ΔX 2 b= X X 3 ΔX 4 4

Better but still needs more intervals! a= X X 0 1 X 2 X 3 X 4 X 5 X 6 X 7 b= X 8

Hmm…are you seeing a pattern here? The more intervals added, the better the estimate. a b= X 16

If we were to keep adding an infinite amount of rectangles, then we would be guaranteed to find the area under the curve a = Xn b

R= So are you ready for the formula for Riemann’s Sum?

R= It looks INTENSE but it’s really not. We’ll break it down piece by piece

R= a= X 0 b= Xn

R= For this example we will use a= X 0 X 1 n=5 X 2 X 3 X 4 b= X 5

R= We will refer to every interval by writing Xi – Xi-1 from ≤ 1 ≤i 5 So by saying i=1 we are referring to ΔX 1 a= X 0 X 1 X 2 X 3 X 4 b= X 5

R= Let i * be an arbitrary point in the interval [Xi – Xi-1 ] X 1 * a= X 0 X 3 * X 2 * X 1 X 2 X 4 * X 3 X 5 * X 4 b= X 5

R= If a= X 0 X 1 i *= Xi-1 for all i then this is known as a left Riemann Sum X 2 X 3 X 4 b= X 5

R= If i *= Xi for all i then this is known as a right Riemann Sum a= X 0 X 1 X 2 X 3 X 4 b= X 5

R= Since n approaches infinity, there is an infinite amount of intervals To be precise there are b-a n intervals Because the length of the entire interval is b -a. Then we divide that by n intervals a= X 0 b= Xn