Introduction of Integration

作者: AnonTokyo

简介: Explaining principle of integration using high school mathematics.

最后修改: 2025-04-08 09:07:13.060416

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Introduction to Integ ration Ra y Zhu December 24, 2024 1 What is Integration Imagine y ou ha v e a cur v e, and y ou w ant to nd the Ar ea under it. This is where integ ration comes in. It helps calculate that area precisel y . But bef ore w e jump to integ ration, let s rs t look at a simpler idea: breaking the area into smaller pieces. 2 Riemann Sum: The Building Bloc k A q uadratic function ( ) = 100 2 is giv en. Y ou w ant to nd the area under the function from = 0 to = 10. This smooth shape cannot be computed b y area f or mula from regular g eometr ic shapes. Thus, w e es timate this area. 2.1 Step 1: Divide the Ar ea into R ect angles Divide the inter v al [ 0 , 10 ] into n eq ual par ts. Eac h par t has a width: F igure 1: f(x) divided into 10 rectangles 1
F or this e x ample w e let = 10 as sho wn in F igure 1 . Ov er eac h sub-inter v al, dra w a rectangle. The height of the rectangle is deter mined b y the v alue of ( ) at some point in that sub-inter v al. 2.2 Step 2: Calculate the Ar ea of Eac h R ect angle The area of a rectangle is: Here, height is ( ) , and the width is Δ . N otice that since Δ is cons tant, the v alue of f ollo w s a ar ithmetic seq uence = + Δ . Thus denotes the inde x in ar ithmetic seq uence and denotes the rs t v alue in seq uence. 2.3 Step 3: A dd U p All the R ect angles The total area is appro ximatel y the sum of all rectangle areas: This sum is called a Riemann Sum . 3 Mor e Riemann Sum Appr o ximations N otice that the rectangle in F igure 1 ha v e its left v er te x f all on ( ) . W e can obser v e a clear l y o v eres timated total area. This is called a Left Riemann Sum Thus, if w e c hang e the intersection point of rectangle and ( ) , w e could g et dierent Riemann Sums: F igure 2: f(x) divided into 10 rectangles with intersection in the r ight and middle 2
F igure 2 demons trates Midpoint Riemann Sum and Right Riemann Sum of ( ) . The es timation v alue e videntl y decreases from the Left Riemann Sum of 715 to Midpoint Riemann Sum of 667.5 then to the Right Riemann Sum of 615. 4 Fr om Appr o ximated Sum to Pr ecise Ar ea As y ou can obser v e, the total sum of area could v ar y b y taking dierent Riemann Sum. T o acq uire more accurate v alue of area under ( ) , consider increasing , the number of rectangles used to appro ximate area. Imagine a process of cutting ( ) approac hes innity . When , the width of eac h rectangle Δ 0. The area of ( ) could be appro ximated more precisel y . F igure 3: f(x) divided into 100 rectangles F igure 3 demons trates the Left, Midpoint and Right Riemann Sum with 100 rectangles. The y ha v e a es timated area of 671.65, 666.675, and 661.65. The dierence betw een 3 dierent approac h decreases as increases. Bac k to the imagination where , 3 v alue will e v entuall y be the same. T ak e Right Riemann Sum as e x ample. T o nd the limit v alue, rs t e v aluate the sum of this eq uation: (note that Δ = 10 0 and = + Δ ) The sum of the sq uares 1 2 + 2 2 + 3 2 + 4 2 . . . + 2 could be computed b y ( 1 ) ( 2 1 ) 6 . This f or mula could be p ro v ed b y high sc hool alg ebra thus will not be discussed here. 3
Imagine as , the fraction of 500 and 500 3 2 will approac h 0. Thus this lea v es us to a precised area under ( ) as 2000 3 . A t this point, y ou ha v e acq uired the basic idea of Integ ration: Ar ea = The limit of Riemann Sum as n appr oac h innity . The eq uation of Í = 1 ( ) Δ as could be denoted in another w a y : 4.1 Further Extension N o w y ou ha v e lear ned ho w to nd the integ ral of ( ) = 100 2 o v er x rang e of [ 0 , 10 ] , tr y the pro v e that result is same when computing limit of Midpoint Riemann Sum and Left Riemann Sum . 4
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