Area Between Two Curves
The area between two curves refers to the region enclosed by the graphs of two functions on a given interval. To find this area, one typically computes the definite integral of the absolute difference between the two functions over the specified interval. This concept is commonly used in calculus to solve problems involving finding the area of a region bounded by two curves.
4 Key excerpts on "Area Between Two Curves"
eBook - ePub- Gregory Hill, Mel Friedman(Authors)
- 2013(Publication Date)
- Research & Education Association(Publisher)
...There are times, though, where several functions exist, and the area between those functions is desired. If your understanding of area from Riemann sums is strong, then the concepts presented here should go smoothly. To determine the Area Between Two Curves on an interval [ a, b ], instead of drawing rectangles from each curve to the x -axis, rectangles are simply drawn from function to function. The differences between the two function values, larger minus smaller, will then determine the height of each rectangle. Additionally, if the smaller function value is always subtracted from the larger value, it does not matter whether the functions lie above or below the x -axis. Again, as the width of each rectangle approaches 0, it also does not matter which x i in each subinterval is used to determine function values being subtracted. Figure 7.4 illustrates the idea with a few rectangles. Figure 7.4 Area Between Two Functions If functions f (x) and g (x) are each continuous on an interval [ a, b ], and f (x) ≥ g (x) over the entire interval, the area between f and g is If the functions happen to cross each other inside the interval of integration, two integrals may be required with the order of subtraction of the functions changing from one to the other. EXAMPLE 7.16 Find the area between and on [0, 4]. SOLUTION Figure 7.5 shows a sketch of the two curves and the area between them. Figure 7.5 EXAMPLE 7.17 The graphs of g (x). and h (x) are shown in Figure 7.6...
- Stu Schwartz(Author)
- 2013(Publication Date)
- Research & Education Association(Publisher)
...The best way to remember the area between curve formulas is to follow these “recipes”: Vertical rectangles: Horizontal rectangles: EXAMPLE 1: Find the area between the curves y = 8 + x – x 2 and y = 2 x – 4. SOLUTION: This is not a calculator-active question, so some knowledge of functions and algebra is required. Without spending the time to actually graph the functions, you should know that y = 8 + x – x 2 is a parabola opening down and y = 2 x – 4 is a line with positive slope. So the graph (without axes or scale) must appear like this: With it being clear which function is on top and which is on the bottom, we need only find the limits of integration. We do that by setting the functions equal to each other. 8 + x – x 2 = 2 x – 4 0 = x 2 + x – 12 (x + 4)(x – 3) = 0 x = – 4, x = 3 Finally, we can set up the integral and use the Fundamental Theorem to calculate the area. TEST TIP Problems similar to the one above take a lot of time, usually involve fractions, and there are many opportunities to make careless errors. If calculators are not permitted, usually students will be asked to set up, but not actually calculate, the integral. But typically, area problems (combined with volume, covered later in this chapter) show up in a free-response question on the calculator-active section of the AP exam. Students are often asked to set up an integral representing the area between functions and then possibly calculate it. Be sure you know how to use your calculator to 1) graph functions, 2) find intersection of functions (2 nd CALC 5:intersect), and 3) find definite integrals, even if you can integrate using the Fundamental Theorem. Input your functions in Y1 and Y2 and use the command MATH 9:fnINT(Y1-Y2, X, lower limit, upper limit). EXAMPLE 2: (Calculator Active) Find the area bounded by the curves y = 6sin x and y = 4ln(x – 2) – 3. SOLUTION: This is a problem that will be done completely using the calculator...
- J. Rosebush, Stu Schwartz(Authors)
- 2016(Publication Date)
- Research & Education Association(Publisher)
...The area between f (x) and the x -axis on [ a, d ] is given by i. For instance, the area between f (x) = 1 – x 2 and the x -axis on [0, 3], is given by either or, equivalently,. 4. To find the area between f (x) and the y -axis on y = c to y = d, rewrite the equation in terms of y first. That is, rewrite the equation in the form x = g (y). If g (y) > 0 the area is represented by. If g (y) < 0 the area is represented by. If g (y) > 0 on [ a, b ] and g (y) < 0 on [ c, d ] the area between g (y) and the y -axis on [ a, d ] is given by or, equivalently,. i. For instance, the area between f (x) = e x and the y -axis from y = 1 to y = 2, is given by. Note that f (x) = e x → y = e x → x = ln(y). ii. Similarly, the area between f (x) = e x and the y -axis from to y = 1, is given by. iii. Also, the area between f (x) = e x and the y -axis from to y = 2, is given by Note that although these two methods are equivalent mathematically, the calculator is using a variation of Riemann sum calculations and thus an approximation, but it will always be accurate to 3 decimal places. B. Area Between Two Curves 1. The area between f (x) and g (x), where g (x). ≤ f (x) from x = a to x = b, is represented by. Loosely speaking, this is the integral of the top function minus the bottom function. If the answer is negative, then this is an indication that the order of the functions in the integrand is wrong and you must switch the functions around. i. For instance, the area between f (x) = x and g (x) = x 2 on [2, 3] is given by. 2. The area between f (y) and g (y), where g (y) ≤ f (y) from y = c to y = d, is represented by. Loosely speaking, this is the integral of the right function minus the left function...
eBook - ePub- Mike Aitken, Bill Broadhurst, Stephen Hladky(Authors)
- 2009(Publication Date)
- Garland Science(Publisher)
...The area formed by the two triangular regions is given. by Area = ∫ 0 3 y d x = ∫ 0 1 x d x + ∫ 1 3 (3 2 − x 2) d x = [ x 2 2 ] 0 1 + [ 3 x 2 − x 2 4 ] 1 3 = [ (1) 2 2 − (0) 2 2 ] + [ 3 × (3) 2 − (3) 2 4 − { 3 × (1[--=PLGO-. SEPARATOR=--]) 2 − (1) 2 4 } ] = [ 1 2 ] + [ 9 2 − 9 4 ] − [ 3 2 − 1 4 ] = 3 2. The result of this calculation can be checked by using the familiar 1/2 × base × height formula for the area of a triangle, which confirms that the final answer should indeed be (1/2) × 3 × 1 = 3/2. A graphical understanding of integration is also useful for illustrating the effect of setting the upper and lower limits of a definite integral equal to each other. This is equivalent to determining the area enclosed by a curve, the x axis, and two vertical lines that are superimposed on top of each other. Figure 6.22 attempts to illustrate this situation for the function y = 10, with the upper and lower boundaries set to x = 5. The enclosed area is clearly zero: Area = ∫ 5 5 10 d x = [ 10 x ] 5 5 = 50 − 50 = 0. Figure 6.22 When the lower and upper limits of a definite integral are the same, the integral corresponds to a vertical line with zero area. So far in this chapter, we have used integration to determine areas that are bounded by positive functions and the x axis between defined limits. We now need to consider what happens when a function becomes negative. As an example of a function that can be either positive or negative, Figure 6.23 shows data from a kinematics study of a swimming frog. Digital image technology was used to study the flow of water in the wake produced by the frog’s feet in order to identify factors that provide forward thrust in different styles of swimming. During a stroke of ‘asynchronous kicking’, the body of the frog is propelled forward, defining the direction for a positive velocity...
