Area Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x b is given by

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1 MATH 42, Fall 29 Examples from Section, Tue, 27 Oct 29 1 The First Hour Area Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x b is given by b a f(x) g(x) dx = b a f(x) dx b a f(x) dx i.e. the difference between the area under the graph of f(x) and the area under the graph of g(x): But we have to be careful if the graphs of f and g cross each other. For example, let us say we want to find the area of the shaded region below: You could say that the area is between the curves y = sin (x) and y = for x 2π, which is correct. However, it would not be correct to simply apply 1

2 the formula above and say the area should be given by 2π sin x dx = ; this would be a mistake because the graph of y = sin x does not always lie above that of y = on the interval [, 2π]. Example. Find the area of the region enclosed by the curves y = cos x, y = 2 cos x for x 2π. Solution. First, we plot the functions: Since y = 2 cos x always lies above y = cos x over the interval [, 2π], the area between these curves is given by 2π (2 cos x) cos x dx = = 2[x] 2π 2[sin x] 2π = 2(2π ) 2(sin 2π sin ) = 4π 2π 2 2 cos x dx Example (parametrised curves). Find the area enclosed by the x-axis and the parametrised curve x = 1 + e t, y = t t 2. Solution. First, we want to check if these two curves cross each other. The x-axis is just another name for the curve y =. If these two curves intersect, then the y coordinates of the intersection points satisfy both y = and y = 2

3 t t 2. So we want to sole = t t 2, i.e. = t(1 t), which has t =, 1 as the solutions. So the two curves intersect at (x, y) = (, 1 + e ) = (, 2) and (x, y) = (1, 1 + e 1 ) = (1, 1 + e). Note that y = t t 2 is positive when < t < 1 (meaning the graph of the parametrised curve is above the x-axis on this interval) and negative when t < or t > 1 (the graph of the parametrised curve is below the x-axis on this these intervals). Also, for any t, e t is always >. We can use this information to plot the curve: The region bounded by these two curves is between x = 1 and x = 1 + e. So the area is = = x=1+e x=2 t=1 t= 1 y dx (t t 2 )e t dt (substitute x = 1 + e t and use y = t t 2 ) te t t 2 e t dt Integrating by parts with u = t 2, dv = e t dt (so du = 2t dt and v = e t ), we get t 2 e t dt = t 2 e t 2te t dt. 3

4 Integrating by parts with u = t, dv = e t dt (so du = dt and v = e t ), we get te t dt = te t e t dt. Hence 1 tet dt = [te t ] 1 1 et dt = e [e t ] 1 = 1. Also, 1 t2 e t dt = [ t 2 e t ] 1 1 2tet dt = ( e) 2( 1) = 2 e. Hence 1 tet dt 1 t2 e t dt = e = 3 e. 2 The Second Hour Example. Find the area enclosed by y = ln(x), xy = 4, x = 1 and x = 3. Solution. We can plot the curves given by y = ln(x) and xy = 4: The function y = ln(x) is always increasing. On the other hand, the part of the curve xy = 4 that is in the first quadrant, which can be thought of as the graph of y = 4/x, is always decreasing. It is probably not easy to work out that these two graphs intersect somewhere on 3 < x < 4 by hand, so you may assume that they do not intersect on 1 x 3. Given this, the area we are after is ln(x) dx x = [4 ln x (x ln x x)] 3 1 = ln Example. Find the area of the region between y = cos x and y = 1 cos x, on x π. (Before you start, try to work out how many times the curves cross each other.) 4

5 Solution. To figure out what the graph of y = 1 cos x, you can start with the graph of y = cos x. To get the graph of y = cos x, you would perform a reflection (of the graph of y = cos x) with respect to the x-axis. Then, to get the graph of y = 1 cos x, you would translate the graph of y = cos x upward by 1 unit. Then it is easy to see that the two curves y = cos x, y = 1 cos x intersect each other exactly once on the interval x π: Where exactly do they intersect? To answer this question, we need to solve cos x = 1 cos x, i.e. cos x = 1/2, i.e. x = π/3 (since x π). So the area between these two curves is π/3 3 cos x (1 cos x) dx + (1 cos x) cos x dx π/3 = [2 sin x x] π/3 + [x 2 sin x] π π/3 = π/3 Example. Let M denote the bounded region between y = and y = x 4 1. Let E 1, E 2 denote the regions inside the circles of radius 1/1 and centered at (1, 1), ( 1, 1), respectively. Let F be the region enclosed by the ellipse x2 9 + y2 9 = 1, which has parametrisation x = 3 cos t, y = sin t for t 2π. Find the area of the region outside M, E 1, E 2 but inside F. Solution. First, you can plot the four curves we are dealing with: (Yes, it does look like a face. :) ) 5

6 The area of M is given by (x4 1) dx = [ x5 5 x]1 1 = 2/5. The areas of E 1, E 2 are both equal to π(1/1) 2. How do we find the area of F? Consider the part of the ellipse above the x-axis. If we can find the area under this curve (or, in other words, the area between this curve and the x-axis), then the area of F would be twice this area. It would be easier to use the parametric equations of the ellipse to compute the area of F above the x-axis. Since this part of the ellipse is given by x = 3 cos t, y = 2 sin t but with t π, its area is t= t=pi (2 sin t)( 3 sin t) dt x=3 = 6 = 6 x= 3 π π ( = 6 y(t) dx π sin 2 t dt 1 cos 2 t dt ) cos 2t dt dt π 2 = 6( π 1 2t [sin 2 2 ] π π = 3π so the area of F is 2 6π. So the area of the region inside F but outside M, E 1, E 2 is 6π 2π(1/1) 2 2/5. 6

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