print barcode image c# Applications of the Integral in Software

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CHAPTER 8 Applications of the Integral
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We calculate the total work by adding all these elements together using an integral The result is W =
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624 (100 x 2 ) x dx
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= 624
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100x x 3 dx
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x4 = 624 50x 4 1 1 81 50 = 624 450 4 4 = 23,712 foot-pounds You Try It: A spring has Hooke s constant 5 How much work is performed in stretching the spring half a foot from its rest position
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Averages
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In ordinary conversation, when we average a collection p1 , , pk of k numbers, we add them together and divide by the number of items: p1 + + pk = Average = k The signi cance of the number is that if we wanted all the k numbers to be equal, but for the total to be the same, then that common value would have to be Now suppose that we want to average a continuous function f over an interval [a, b] of its domain We can partition the interval,
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P = {x0 , x1 , , xk },
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with x0 = a and xk = b as usual We assume that this is a uniform partition, with xj xj 1 = x = (b a)/k for all j Then an approximate average of f would be given by f (x1 ) + f (x2 ) + + f (xk ) k It is convenient to write this expression as app = app = 1 b a
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f (xj )
j =1
1 b a = k b a
f (xj )
j =1
Applications of the Integral
b a f (x) dx
This last is a Riemann sum for the integral (1/[b a]) the mesh of the partition go to zero, we declare average of f = = EXAMPLE 817 1 b a
Thus, letting
f (x) dx
In a tropical rain forest, the rainfall at time t is given by (t) = 01 01t + 005t 2 inches per hour, 0 t 10 What is the average rainfall for times 0 t 6
SOLUTION We need only average the function : average rainfall = = = 1 6 1 6 0
(t) dt
AM FL Y
01 01t + 005t 2 dt
005 3 1 01t 005t 2 + t = 6 3 = 01 03 + 06 = 04 inches per hour
EXAMPLE 818 Let f (x) = x/2 sin x on the interval [ 2, 5] Compare the average value
of this function on the interval with its minimum and maximum
SOLUTION Observe that 1 cos x 2 Thus the critical points occur when cos x = 1/2, or at /3, /3 We also must consider the endpoints 2, 5 The values at these points are f (x) = f ( 2) = 1 + sin 2 00907026 3 f ( /3) = + 03424266 6 2 3 03424266 f ( /3) = 6 2 5 f (5) = sin 5 3458924 2
CHAPTER 8 Applications of the Integral
Plainly, the maximum value is f (5) = 5/2 sin 5 3458924 The minimum value is f ( /3) 03424266 The average value of our function is = 1 5 ( 2)
5 2
x sin x dx 2
5 2
1 x2 = + cos x 7 4 = = 1 7
25 + cos 5 4
4 + cos 2 4
1 21 + cos 5 cos 2 7 4
084997 You can see that the average value lies between the maximum and the minimum, as it should This is an instance of a general phenomenon You Try It: On a certain tree line, the height of trees at position x is about 100 3x + sin 5x What is the average height of trees from x = 2 to x = 200 EXAMPLE 819
What is the average value of the function g(x) = sin x over the interval [0, 2 ]
SOLUTION We calculate that = 1 2 0
2 0
sin x dx =
1 [ cos x] 2
1 [ 1 ( 1)] = 0 2
We see that this answer is consistent with our intuition: the function g(x) = sin x takes positive values and negative values with equal weight over the interval [0, 2 ] The average is intuitively equal to zero And that is the actual computed value You Try It: Give an example of a function on the real line whose average over every interval of length 4 is 0
Applications of the Integral
Arc Length and Surface Area
Just as the integral may be used to calculate planar area and spatial volume, so this tool may also be used to calculate the arc length of a curve and surface area The basic idea is to approximate the length of a curve by the length of its piecewise linear approximation A similar comment applies to the surface area We begin by describing the basic rubric
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