MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 25

abf(x)dx really means limΔx0 ( f(a)Δx+ f(a+Δx)Δx++ f(b-Δx)Δx ) =limΔx0 ( add up the areas of little boxes } f(a+kΔx) Δx ) a b x y Δx The first box } f(a) Δx has area f(a)Δx.

The second box } f(a+Δx) Δx has area f(a+Δx)Δx.

So think of abf(x)dx as adding up areas from a to b of infinitesimally small boxes } f(x) dx with area f(x)dx.

02exdx 1 2 x y Suppose Δx=13 e0Δx+eΔx Δx+e2ΔxΔx +e3ΔxΔx+ e4ΔxΔx++ e2-ΔxΔx = e013+ e1313+ e2313+ e3313+ e4313+ e5313 = 13 ( 1+e13+ (e13)2+ (e13)3+ (e13)4+ (e13)5 ) = 13 ((e13)6-1e13-1) = 13(e63-1e13-1) = (e2-1) (13e13-1). Suppose Δx=15 e0Δx+ eΔxΔx+ e2ΔxΔx++ e2-ΔxΔx = e015+e15 15+e2515+ e3515++ e9515 = 15 ( e0+ e15+ e25+ e35+ e45++ e95 ) = 15 ( e0+ e15+ (e15)2+ (e15)3++ (e15)9 ) = 15 ( (e15)10-1 e15-1 ) = (e2-1) (15e15-1). Suppose Δx=1N e0Δx+ eΔxΔx+ e2ΔxΔx+ e3ΔxΔx++ e2-ΔxΔx = e01N+ e1N1N+ e2N1N+ e3N1N+ e4N1N++ e2-1N1N = (e2-1) (1Ne1N-1). So limΔx0 ( e0Δx+ eΔxΔx++ e2-ΔxΔx ) =limΔx0 (e2-1) Δx(eΔx-1) =(e2-1)·1= e2-1. Note: 02exdx = ex+c|x=0x=2 = (e2+c)- (e0+c) = e2+c-e0-c = e2-e0 = e2-1.

-111x2dx -1 1 x 1 y By adding up little boxes: -111x2dx = limΔx0 ( 1(-1)2Δx+ 1(-1+Δx)2Δx+ 1(-1+2Δx)2Δx++ 1(1-Δx)2Δx ) = limΔx0 ( 1·Δx+ 1(-1+Δx)2Δx++ 102OOPS!!Δx++ 1(1-Δx)2Δx ) So -111x2dx is UNDEFINED.

Note: -111x2dx = -11x-2dx = x-1-1+c |x=-1x=1 = (1-1-1+c)- ((-1)-1-1+c) = -1+c-1-c = -2. So this is a case where abdfdxdx f(b)-f(a). i.e. adding up areas of little boxes and doing the indefinite integral and plugging in give different answers.

The fundamental theorem of calculus says abdfdxdx =f(b)-f(a) (is not a lie) provided f(x) doesn't do anything bad between a and b. It should be

(a) defined everywhere between a and b,
(b) continuous everywhere between a and b,
(c) differentiable everywhere between a and b.

The fundamental theorem of calculus says Area underg(x)from atob=A(b)- A(a) where g(x)dx=A(x)+c.

Why does this work?

Let A(x)=area under g(x) from a to x. Then dAdx = limΔx0 A(x+Δx)-A(x) Δx a x x y y=g(x) = limΔx0 (area of last little box)Δx = limΔx0 g(x)ΔxΔx = limΔx0g(x) =g(x). a x Δ x + x x y y=g(x) So g(x)dx= A(x)+c. So A(b)-A(a) = (area underg(x)fromatob)- (area underg(x)fromatoa) = area underg(x)fromatob.

Notes and References

These are a typed copy of Lecture 25 from a series of handwritten lecture notes for the class MATH 221 given on November 6, 2000.

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