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Applications of Integration I: Area and Arc Length in .NET framework
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CHAP. 32] APPLICATIONS OF INTEGRATION I: AREA AND ARC LENGTH
Fig. 321 Fig. 322 (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. 323). 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. 324, A is the area under Fig. 323 Fig. 324 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 325 shows the region R under the line y = 1 x + 2, above the parabola y = x2 , and between the yaxis 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 xaxis, as in Fig. 326. 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. 325 Fig. 326 EXAMPLE Find the area of the region bounded by the parabola y = x2 and the line y = x + 2 (see Fig. 327). The limits of integration a and b in (32.1) must be the xcoordinates 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. 327 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. 328). 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 meanvalue 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. 328

