Calculus II MAT 146 Integration Applications: Area Between Curves

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1 Calculus II MAT 46 Integration Applications: Area Between Curves A fundamental application of integration involves determining the area under a curve for some interval on the x- or y-axis. In a previous chapter, we reviewed or learned for the first time various methods for evaluating integrals, including approximation techniques based on areas. When we study improper integrals, we will determine how to calculate areas over infinite-length x-axis or y-axis intervals and areas involving discontinuities within limits of integration. Here, as we consider applications of integration, we extend the idea of area under a curve to area between two curves.! g( x) for If f x all x on some closed interval [a,b], then the region bounded by f and g has area A, where A = b " ( f ( x)! g( x) ) dx a x = a x = b Example Calculate the area between y = x + and y =! x for! x!. y = f (x) y = g(x) 4.5 y = x + y = x a typical height: x + width:!x!! x a = b =

2 Step : Graph the two functions and identify the x-axis interval. This is useful in determining that x +! " x on! x!. Step : Sketch a typical within the bounded region. Determine the height and width of this typical. Label the dimensions. Step : Set up an appropriate integral to show the area of the typical. (!! x ) A = " x + dx The integral serves to sum the areas of an infinite number of the typical s packed into the x-axis interval from a = to b =. This is the length (height) of the typical. This is the width of the typical. Step 4: Evaluate the integral. Height x Width represents area calculation. A = " (( x + )! (! x) ) dx = x x " ( + x +) dx = + x # + x % % = 6 You can, of course, use a CAS, such as your TI-89 or an online app, to complete this calculation. Example Determine the area between the curves x!= y and x!= y on the interval, [ ]. a typical! x! height: x! width:!x

3 " x! ) Look at the graph to see that x! ) Sketch a typical and determine its area. Area = height! width " x " = # x " ) Build an integral that sums these areas. on! x!. (! x! ) % & 'x A = " x! dx 4) Evaluate the integral. A = " (( x!)! ( x!)) dx = x x " (! x) 4 dx = + x # % % = 6 Example Determine the area bounded by curves f (x) = x! 4.5 and g(x) =!x +. (,4) f( x) = ( x ).5 a typical (,) g( x) = x a =.5 b = The area of a typical is: Area = height! width = #("x + ) " ( x ") % & 'x To determine limits of integration, we need to determine the x-coordinates of the points of intersection f (x) = g(x) :!x + = ( x!) " x =!, x = (using TI-89 solve command)

4 Integrate: A = " (!x + )! ( x!) dx = "!x + x + dx = 9!! Example 4 Determine the area bounded by curves y = x! x + and y = x +. y = x x (,4).5 y = x + Area Rectangle! x + = " # x! x + one typical.5 (,) another typical.5 % &x Area Rectangle (,).5 = " # x +! x! x + % &x 4 a =.5 b = a = b =.5 Here, the graph shows two bounded regions. We also needed to determine two x-axis intervals, for integration. We solved x! x + = x + to determine the x-coordinates of the ordered-pair points of intersection noted in the figure. This leads to the following area calculation:! x + A = " # x! x + & % dx + " & # x +! = 4 + 4! x! x + % dx = 8 Notice, too, the symmetry of the graphs. The two bounded regions have equal area. Of course, we d need to formally justify that indeed there is such symmetry. Example 5 Calculate the area of the region bounded by x = y! and x =. Here, it may be more efficient to consider a typical oriented with respect to the y-axis. If so, we need integration limits on the y-axis. To

5 determine these (potential) y values, we set the two functions equal to each other and solve for y: y!= " y = ± These are the values for a and b shown in the plot accompanying this example. This generates the following area calculation, with y as the variable of integration: A = " #! y! & % dy =! Notice, too, that we could have transformed the functions into functions with x as the independent variable ( x = y! " y = ± x + ). The integral for area calculation with x as the independent variable is: A = " # x +! (! x + & )% dx = & x + dx = ' 6 * ) (, = +.!! Example 6 Determine the area of region R, between the curves y = cos (! x) and y =, where R is the first region to the right of the origin (i.e., x > ) bounded by the curves. We solve cos! x = to determine the smallest positive values of x that satisfy the equation: x = / and x = 5 /. This leads to

6 5 " A =! cos (! x % # & ' dx = + ).57 These examples help illustrate important steps you can carry out when calculating area between two curves: Graph the functions in question and identify the number of bounded regions as well as which function is greater than the other for each region. Determine the x-axis intervals (or y-axis intervals) for the bounded regions. The interval endpoints may be explicitly stated or can be determined using algebraic techniques, most typically by setting the two functions equation to each other. Draw in a typical and determine its area. This provides essential information for the area integral you need to create. For each bounded region, create a definite integral to represent the sum of the areas of an infinite number of typical s. Evaluate this integral to determine the area of each bounded region. Note that your TI-89 or other CAS can be a useful tool for several components of your solution process.

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