The Area bounded by Two Functions

The Area bounded by Two Functions The graph below shows 2 functions f(x) and g(x) that are continuous between x = a and x = b and f(x) g(x). The area shaded in green is the area between the 2 curves. We refer to this area as the area bounded by f(x) and g(x). The area between the 2 functions f(x) and g(x) that are both continuous between x = a and x = b and where f(x) g(x) is found by the following integral x=b x=a [ f(x) g(x) ] where a x b This formula requires the first function to be the upper of the two functions and the second function be the lower function. The following word formula can help you remember that the area is always the larger function minus the smaller function. x=b x=a [ upper function lower function] where a x b Note: You must be be sure that f(x) and g(x) are continuous between x = a and x = b and f(x) g(x). It is normally easy to see that the 2 functions are continuous between x = a and x = b. It will normally be require that you graph the 2 functions to be sure that f(x) g(x) between x = a and x = b. Note: The requirement that f(x) g(x) is only on the interval from x = a and x = b. In many cases f(x) g(x) on other intervals. The integral slight change in the integral can be used in intervals where f(x) g(x). x=b x=a [ f(x) g(x) ] cannot be used in this case but a Section 5 9" Page 1 of 14" 2018 Eitel

Example: Find the area enclosed by the functions y = x 2 and y = We must first graph both functions on one axis. x What do we mean by area enclosed by and which function is the upper function. In this case most would probably say that y = x 2 is the upper function and they would be right for the vast majority of the x s. y = x 2 is not above y = x for all values of x. y = x 2 is under y = x from x = 0 to 1 and then y = x 2 is above y = x from that point on. Because of this you should always sketch of a graph of the region. Without a sketch it s often easy to mistake which of the two functions is the larger. The limits of integration for this will be the intersection points of the two curves. Set the functions equal to each other and solve for x. This will give you the left and right In this case it s pretty easy to see that they will intersect at x = 0 and x = 1 so these are the limits of integration. x 2 = x 4 = x x x 4 1 = 0 ( x 2 +1) x +1 x = 0,1 and 1 ( ) ( x 1) = 0 the x values for the points of intersection are x = 0 and x = 1 1 x x 2 0 1 = 2 x/2 1 x ] 0 = 2 1 0 1 1 [ ] Note: In most cases we do not include units. The area is in square units and the unit depends on the application. Section 5 9" Page 2 of 14" 2018 Eitel

Example: Find the area enclosed by the functions y = 4x +16and y = 2x 2 +10 2x 2 +10 = 4x +16 2x 2 4x 6 = 0 ( ) = 0 2 x 2 2x ( ) ( x ) = 0 2 x +1 x = 1 and the x values for the points of intersection are x = 1 and x = 4x +16 1 ( ) ( ) 2x 2 +10 2x 2 + 4x + 6 1 = 2 x + 2x 2 + 6x ] 1 = [ 18 +18 +18] 2 2 6 18 2 + 2 + 6 76 Section 5 9" Page of 14" 2018 Eitel

Example: Find the area enclosed by the functions y = + 2x x 2 and the x axis y = 0 + 2x x 2 = 0 ( ) = 0 x 2 2x ( ) ( x ) = 0 x +1 x = 1 and the x values for the points of intersection are x = 1 and x = x 2 + 2x + 1 = 1 x + x 2 + x ] 1 = [ 9 + 9 + 9] 1 1 1 9 1 +1+1 2 Section 5 9" Page 4 of 14" 2018 Eitel

Example: Find the area enclosed by the functions y = 8 and y = 2x and x = 4 x 2x = 8 x 2x 2 = 8 2x 2 8 = 0 ( ) = 0 2 x 2 4 ( )( x 2) = 0 2 x + 2 x = 2 and 2 the only x values for the point of intersection that work is x =2 The other point of intersection where the verticle line x = 4 intersects the graphs this occurs when x = 4 2 2x 8 x = x 2 e 8ln(x) ] 2 [ ] = e 2 8ln(e) 4 8ln 4 e 2 8 4 + ln 4 e 2 12 + 8ln 4 Section 5 9" Page 5 of 14" 2018 Eitel

Example: Find the area enclosed by the functions y = x +1 and y = xe x2 and x = 0 and x = 2 The 2 points of intersection where the verticle lines x = 0 x = 4 intersect the 2 graphs this occurs when x = 0 and x = 4 2 0 x +1 xe x2 = 1 2 x2 + x 1 2 e x2 ] 0 2 = 2 + 2 + 1 2 e 4 0 + 0 + 1 2 e0 = 4 + 1 2e 4 1 2 7 2 1 2e 4 Section 5 9" Page 6 of 14" 2018 Eitel

Example: Find the area enclosed by the functions y = cos(x) and y = sin(x) and x = 0 and x = π / 2 cos(x) = sin(x) the x value for the only point of intersection is x = π / 4 and at the verticle lines x= 0 and x = π / 2 π /4 cos(x) sin(x) plus sin(x) cos(x) 0 π /4 = sin(x) + cos(x) ] 0 π /2 π /4 π /2 plus cos(x) sin(x) ] π /4 = [ sin(π / 4) + cos(π / 4) ] [ sin(0) + cos(0) ] plus [ sin(π / 2) cos(π / 2) ] sin(π / 4) cos(π / 4) [ ] = 2 2 + 2 2 0 +1 = 2 2 2 1+ 2 2 2 1 2 2 2 [ ] plus 1 0 [ ] 2 2 2 2 Section 5 9" Page 7 of 14" 2018 Eitel

Example: Set up but do not integrate the integral that would be used to find the area enclosed by the functions y = 2x +16 and 2x 2 +10 from x = 2 and x = 5 2x 2 +10 = 4x +16 2x 2 4x 6 = 0 ( ) = 0 2 x 2 2x ( ) ( x ) = 0 2 x +1 x = 1 and the x values for the points of intersection are x = 1 and x = There are separate enclosed regions. The graph in red is the upper function in the first and third region and it is the lower region in the middle region. 1 Area left = 2x 2 +10 2 Area middle = 4x +16 5 ( ) ( 4x +16) ( 1 ) ( 2x 2 +10) ( ) ( 4x +16) Area right = 2x 2 +10 if you evaluated the sum of the definite Integrals the area would be 142 Section 5 9" Page 8 of 14" 2018 Eitel

Area enclosed by functions in terms of y The formula we used in the example above required that we have 2 Continuous Functions f(x) and g(x) on the interval a x b and that f(x) g(x). The enclosed area was bounded by the upper function and the lower function. We refer to this area as the area bounded by f(x) and g(x). What if we have a graph that encloses a region but the graphs do NOT represent a function in terms of x. If we have 2 relations are functions in terms of y expressed as f(y) and g(y) and one relation forms a right bound and the other forms a left bound with f(y) g(y) in the interval c,d following integral to find the enclosed region. [ ] then we can use the The area between the 2 relations in terms of y f(y) and g(y) that are both continuous between y = c and y = d and where f(y) g(y) is found by the following integral y=d y=c [ f(y) g(y) ] where c y bd Note: You must be be sure that f(y) and g(y) are continuous between y = c and y = d and f(y) g(y). It is normally easy to see that the 2 functions are continuous between y = c and y = d It will normally be require that you graph the 2 functions to be sure that f(y) g(y) between y = c and y = d. Note: The requirement that f(x) g(x) is only on the interval from x = a and x = b. In many cases f(x) g(x) on other intervals. The integral slight change in the integral can be used in intervals where f(y) g(y). y=d y=c [ f(y) g(y) ] cannot be used in this case but a Section 5 9" Page 9 of 14" 2018 Eitel

Example: Find the area enclosed by x = y 2 +10 and x = ( y 2) 2 y 2 +10 = ( y 2) y 2 +10 = y 2 4y + 4 2y 2 4y 6 = ( ) ( y ) = 0 2 y +1 y = 1 and y = If y = 1 then x = If y = then x = 1 The points of intersection are (, 1) and (1,) The integral to find the enclosed region will be in terms of y. The relation x = y +1 is always to the left of the right most equationx = 1 2 y2 so the integral will be It will involve the right most relation minus the left most equation y=d y=c y= [ right relation left relation] ( ) 2 ( y 2 +10) y 2 y= 2 y= y 2 +10 y 2 + 4y 4 y= 2 y= 2y 2 + 4y + 6 y= 1 2 y + 2y 2 + 6y ] 1 = [ 18 +18 +18] 2 2 6 76 Section 5 9" Page 10 of 14" 2018 Eitel

Example: Find the area enclosed by x = y 2 and x = 1 Note that y = x y 2 and x = 1 is not a function in term of x but it is a function in terms of so so we can write the g(y) = y 2 Find the y values for the intersection of y = y 2 and x = 1 y 2 = 1 y 2 4 = 0 ( y + 2) y 2 y = 2 and y = 2 ( ) = 0 The points of intersection are ( 1, 2) and ( 1, 2) y=d y=c y=2 y= 2 y=2 [ right relation left relation] ( ) ( y 2 ) 1 y 2 + 4 y= 2 1 2 y + 4y ] 2 = 8 + 8 8 8 2 Section 5 9" Page 11 of 14" 2018 Eitel

Example: Find the area enclosed by x = 1 2 y2 and y = x 1 which can be written in term of x = y +1 1 2 y2 = y +1 y 2 6 = 2y + 2 y 2 2y 8 = 0 ( y 4) y + 2 y = 4 and y= 2 ( ) = 0 If y = 4 then x = 5 The points of intersection are (5, 4) and ( 1, 2) If y = 2 then x = 1 The integral to find the enclosed region will be in terms of y. The relation x = y +1 is always to the left of the right most equationx = 1 2 y2 so the integral will be It will involve the right most relation minus the left most equation y=b y=a [ right relation left relation] y=4 (y+1) 1 2 y2 y= 2 y=4 1 2 y2 + y + 4 y= 2 1 6 y + 1 4 2 y2 + 4y ] 2 = 64 6 + 8 +16 8 6 + 2 8 18 Section 5 9" Page 12 of 14" 2018 Eitel

A second look at the last problem. The graph in red is not a function in terms of x so you cannot use the first integral method x=b x=a [ upper function lower function] Both the graph in red and the graph in blue are functions in terms of y so we can use the second integral method. This was the example on the previous page. y=d y=c [ right relation left relation] Section 5 9" Page 1 of 14" 2018 Eitel

There is a way to use the first integral method x=b x=a [ upper function lower function] It involves solving the relation x = 1 2 y2 for y to get 2 separate functions. One function is y = + 2x + 6 and graphed in red and the second function y = 2x + 6 is graphed in green. 1 2 y2 = y +1 y 2 6 = 2y + 2 y 2 2y 8 = 0 ( y 4) y + 2 y = 4 and y= 2 ( ) = 0 The points of intersection are (5, 4) and ( 1, 2) The area in yellow has an upper bound of y = + 2x + 6 and a lower bound of y = x 1 so we can find that area by the following integral. Area in yellow = x=5 x= 1 2x + 6 (x-1) The area in pink has an upper bound of y = + 2x + 6 and a lower bound of y = 2x + 6 so we can find that area by the following integral. x= ( ) Area in pink = 2x + 6 2x + 6 x= 1 x= Area in pink = 2 2x + 6 x= 1 x= Area in pink = 2 2x + 6 x= 1 The total of these 2 areas is the total area. Completing the definite integrals yields an area of 18 Section 5 9" Page 14 of 14" 2018 Eitel