Calculus 220 section 6 4 Areas in the x y Plane notes by Tim Pilachowski Example A Find the area of the geometric figure pictured to the right Answer 6 2 2 3 In Lectures 6 2 and 6 3 we were able to equate the area under a curve on an interval a x b with the definite integral from a to b b a f x dx Combining this concept with geometric ideas such as those used in Example A supports the assertion of some of the properties of integrals already used in Lecture 6 1 b b a f x g x dx a f x dx g x dx a f x g x dx a f x dx g x dx b b b b b a k f x dx k a f x a b a dx We ll apply these properties to some new scenarios Example B Find the area between the curves f x x 1 and g x 1 on the interval 1 x 3 x As long as f x g x for all x in the interval we can use the second property above and subtract the areas Answer 6 ln 3 1 2 on the interval 1 x 3 x The shape remains the same because the curves from Example B were simply shifted down by 2 The area should be the same However in this case some of the area lies below the x axis and would thus be a negative area as noted in Lecture 6 3 Does this affect the calculation of area between the two curves Example C Find the area between the curves f x x 1 and g x So the integral calculation gives us the same result as in Example B area 6 ln3 It doesn t matter whether the two curves lie above or below the x axis only that we subtract the higher curve minus the lower Example D Find the area between the curves y x 2 and y x 2 4 x 12 on the interval 0 x 4 Looking at the graph window 1 5 by 1 17 y x 2 4 x 12 lies above y x 2 for a portion of the interval but is below for the rest of the interval We need to determine where the two intersect then set up two integrals one for each portion Answer 20 15 0 0 5 Example E Find the area between the curves y 2x and y x 2 4 x 8 In this case although no interval is specified by looking at the graph window 1 5 by 1 10 we can see that there are two points of intersection Between these two points the area is bounded and lies between the two curves We need to solve for the intersections to find the boundaries or limits of integration Answer 4 3 10 0 0 5 Example F Find the area bounded by y x 2 y 6 x and y 8 2 x on the interval 0 x 2 Looking at the graph window 1 5 by 1 10 one of the corners of the bounded area is clearly 0 0 We ll need to solve for the others Answer 16 3 10 0 0 5 The key to all of these is looking at the graph to determine which function is above the other finding intersections as needed then setting up and evaluating the appropriate integrals In the next section we ll delve into applications which use all of the integration concepts developed so far If a function represents a rate of change we ll use an integral area under the curve to determine an amount For example from a velocity function which is the rate of change of position with respect to time we can integrate to determine the distance traveled Another example Knowing how costs are changing with respect to the number of units produced Marginal Cost we can integrate to find the amount of Cost