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Applications of Integration I: Area and Arc Length
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32.1 AREA BETWEEN A CURVE AND THE y-AXIS We have learned how to nd the area of a region like that shown in Fig. 32-1. Now let us consider what happens when x and y are interchanged. EXAMPLES
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(a) The graph of x = y2 + 1 is a parabola, with its nose at (1, 0) and the positive x-axis as its axis of symmetry (see Fig. 32-2). Consider the region R consisting of all points to the left of this graph, to the right of the y-axis, and between y = 1 and y = 2. If we apply the reasoning used to calculate the area of a region like that shown in Fig. 32-1, but with x and y interchanged, we must integrate along the y-axis. Thus, the area of R is given by the de nite integral
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(y2 + 1) dy
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The fundamental theorem gives
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(y2 + 1) dy =
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y3 +y 3
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( 1)3 + ( 1) 3
1 9 8 +2 1 = +3=3+3=6 3 3 3
CHAP. 32]
APPLICATIONS OF INTEGRATION I: AREA AND ARC LENGTH
Fig. 32-1
Fig. 32-2
(b) Find the area of the region above the line y = x 3 in the rst quadrant and below the line y = 4 (the shaded region of Fig. 32-3). Thinking of x as a function of y, namely, x = y + 3, we can express the area as
(y + 3) dy =
y2 + 3y 2
42 + 3(4) 2
16 02 + 3(0) = + 12 = 20 2 2
Check this result by computing the area of trapezoid OBCD by the geometrical formula given in Problem 31.9.
32.2 AREA BETWEEN TWO CURVES Assume that 0 g(x) f (x) for x in [a, b]. Let us nd the area A of the region R consisting of all points between the graphs of y = g(x) and y = f (x), and between x = a and x = b. As may be seen from Fig. 32-4, A is the area under
Fig. 32-3
Fig. 32-4
APPLICATIONS OF INTEGRATION I: AREA AND ARC LENGTH
[CHAP. 32
the upper curve y = f (x) minus the area under the lower curve y = g(x); that is, A=
a b b
f (x) dx
g(x) dx =
(f (x) g(x)) dx
(32.1)
EXAMPLE Figure 32-5 shows the region R under the line y = 1 x + 2, above the parabola y = x2 , and between the y-axis and 2 x = 1. Its area is
1 x + 2 x 2 dx = 2 = =
x3 x2 + 2x 4 3 13 12 + 2(1) 4 3
03 02 + 2(0) 4 3
1 9 1 27 4 23 1 +2 = = = 4 3 4 3 12 12
Formula (32.1) is still valid when the condition on the two functions is relaxed to g(x) f (x) that is, when the curves are allowed to lie partly or totally below the x-axis, as in Fig. 32-6. See Problem 32.3 for a proof of this statement. Another application of (32.1) is in nding the area of a region enclosed by two curves.
Fig. 32-5
Fig. 32-6
EXAMPLE Find the area of the region bounded by the parabola y = x2 and the line y = x + 2 (see Fig. 32-7).
The limits of integration a and b in (32.1) must be the x-coordinates of the intersection points P and Q, respectively. These are found by solving simultaneously the equations of the curves y = x 2 and y = x + 2. Thus, x2 = x + 2 whence, x = a = 1 and x = b = 2. Thus, A= = =
2 1
x2 x 2 = 0
(x 2)(x + 1) = 0
(x + 2) x 2 dx =
2 x3 x2 + 2x 2 3
23 22 + 2(2) 2 3 8 4 +4 2 3
1 2
+ 2( 1)
( 1)3 3
1 3 9 1 2+ = +6 2 3 2 3 3 3+6 9 3 = = +6 3= +3= 2 2 2 2
CHAP. 32]
APPLICATIONS OF INTEGRATION I: AREA AND ARC LENGTH
Fig. 32-7 32.3 ARC LENGTH Consider a differentiable (not just continuous) function f on a closed interval [a, b]. The graph of f is a curve running from (a, f (a)) to (b, f (b)). We shall nd a formula for the length L of this curve. Divide [a, b] into n equal parts, each of length x. To each xi in this subdivision corresponds the point Pi (xi , f (xi )) on the curve (see Fig. 32-8). For large n, the sum P0 P1 + P1 P2 + + Pn 1 Pn n Pi 1 Pi of the lengths of the i=1 line segments Pi 1 Pi is an approximation to the length of the curve. Now, by the distance formula (2.1), Pi 1 Pi = But xi xi 1 = (xi xi 1 )2 + ( f (xi ) f (xi 1 ))2
x; also, by the mean-value theorem (Theorem 17.2), f (xi ) f (xi 1 ) = (xi xi 1 )f (xi ) = ( x)f (xi )
for some xi in (xi 1 , xi ). Hence, Pi 1 Pi = = ( x)2 + ( x)2 ( f (xi ))2 = 1 + ( f (xi ))2 ( x)2 = 1 + ( f (xi ))2 ( x)2 1 + ( f (xi ))2 x
Fig. 32-8