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Fig 42
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We take it for granted that the area of a rectangle of length and width w is w Now our strategy is to divide the base interval [a, b] into equal subintervals Fix an integer k > 0 We designate the points
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P = {x0 , x1 , x2 , , xk },
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with x0 = a and xk = b We require that |xj xj 1 | = |b a|/k x for j = 1, , k In other words, the points x0 , x1 , , xk are equally spaced We call
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The Integral
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the set P a partition Sometimes, to be more speci c, we call it a uniform partition (to indicate that all the subintervals have the same length) Refer to Fig 43
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b_a k x0 = a xj xk = b
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Fig 43
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The idea is to build an approximation to the area A by erecting rectangles over the segments determined by the partition The rst rectangle R1 will have as base the interval [x0 , x1 ] and height chosen so that the rectangle touches the curve at its upper right hand corner; this means that the height of the rectangle is f (x1 ) The second rectangle R2 has as base the interval [x1 , x2 ] and height f (x2 ) Refer to Fig 44
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y = f (x)
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x0 = a x1 x2
xk = b
Fig 44
Continuing in this manner, we construct precisely k rectangles, R1 , R2 , , Rk , as shown in Fig 45 Now the sum of the areas of these rectangles is not exactly equal to the area A that we seek But it is close The error is the sum of the little semi-triangular pieces that are shaded in Fig 46 We can make that error as small as we please by making the partition ner Figure 47 illustrates this idea Let us denote by R(f, P ) the sum of the areas of the rectangles that we created from the partition P This is called a Riemann sum Thus
R(f, P ) =
j =1
f (xj )
x f (x1 )
x + f (x2 )
x + + f (xk )
Here the symbol k =1 denotes the sum of the expression to its right for each of j the instances j = 1 to j = k
CHAPTER 4 The Integral
y = f (x)
x0 = a x1 x2
xk = b
Fig 45
y = f (x)
x0 = a x1 x2
xk = b
Fig 46
y = f (x)
Fig 47
The reasoning just presented suggests that the true area A is given by
lim R(f, P )
We call this limit the integral of f from x = a to x = b and we write it as
f (x) dx
Thus we have learned that area A =
The Integral
f (x) dx
It is well to take a moment and comment on the integral notation First, the integral sign
is an elongated S, coming from summation The dx is an historical artifact, coming partly from traditional methods of developing the integral, and partly from a need to know explicitly what the variable is The numbers a and b are called the limits of integration the number a is the lower limit and b is the upper limit The function f is called the integrand Before we can present a detailed example, we need to record some important information about sums: I We need to calculate the sum S = 1 + 2 + + N = goal, we write
N j =1 j To
achieve this
S =1+2 + + (N 1) + N S = N + (N 1) + + 2 +1 Adding each column, we obtain 2S = (N + 1) + (N + 1) + + (N + 1) + (N + 1)
N times
Thus 2S = N (N + 1) or N (N + 1) 2 This is a famous formula that was discovered by Carl Friedrich Gauss (1777 1855) when he was a child S= II The sum S = 12 + 22 + + N 2 = S=
n 2 j =1 j
is given by
2N 3 + 3N 2 + N 6 We shall not provide the details of the proof of this formula, but refer the interested reader to [SCH2]
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