7.1 Using Antiderivatives to find Area

Transcription

1 7.1 Using Antiderivatives to find Area Introduction finding te area under te grap of a nonnegative, continuous function f In tis section a formula is obtained for finding te area of te region bounded between te grap of a continuous, nonnegative function f and te x-axis. As mentioned in te previous capter, it is seen tat te antiderivatives of f play a crucial role in tis process. Let f be a function tat is continuous on [a, b]. Also suppose tat f is nonnegative, so tat its grap lies on or above te x-axis. In tis case, it makes sense to talk about te area under te grap of f; we seek te area between x = a and x = b. te area function; A(x) First, define: A(x) := te area under te grap of f, from a to x Observe tat A(a) = 0, and A(b) is te area being sougt. in te pictures, is positive A; a little piece of area Now, let x be a number between a and b, and let be a small positive number. In te exercises accompanying tis section, you will consider te case were is a small negative number. Focus attention on te little piece of area between x and x +, as sown below. Tis area can be obtained as follows: take te area under te grap from a to x +, and subtract off te area from a to x. Wat s left is te area under te grap between x and x +, as sown. Tus, tis little piece of area can be written in terms of te area function A as: A := Te symbol A is read as delta A and denotes a cange in A. EXERCISE 1 1. If is a small negative number, were is x + in relation to x? 2. Make a sketc sowing x and x +. Wat is te correct formula for A in tis case? 401

2 402 copyrigt Dr. Carol JV Fiser Burns ttp:// using te Max-Min Teorem By ypotesis, f is continuous on te entire interval [a, b], so it is also continuous on te subinterval [x, x + ]. Terefore, te Max-Min Teorem guarantees tat f attains a minimum value f(m) and a maximum value f(m) on [x, x + ], as illustrated below. Observe tat bot m and M come from te interval [x, x+]. under- and over-approximating te area wit rectangles Te actual area A of te little piece under inspection is under-approximated by te rectangle of eigt f(m) and widt. Also, A is over-approximated by te rectangle of eigt f(m) and widt. Tat is: f(m) A f(m) Division by te positive number yields f(m) A f(m), and substituting in te definition of A yields: f(m) f(m) Be aware! Te numbers m and M depend on: te function f te number x te number Wat is about to be said applies to bot m and M. For simplicity, it is stated only for m. It s important tat you understand tat te number m depends on: te function ( f ) tat you re working wit te small interval [x, x + ] tat is currently under investigation; tis interval depends on bot x and Cange any of tese ( f, x, or ) and te number m could cange! For tis reason, a name like m f,x, (wit tree subscripts) migt be better tan just m. But ten te notation would be so cumbersome tat it could make tings appear arder tan tey really are! So, we ll stick wit just m.

3 copyrigt Dr. Carol JV Fiser Burns ttp:// 403 EXERCISE 2 You sould ave discovered in te previous exercise tat if < 0, ten A = A(x) A(x + ), and te picture becomes te one sown below: 1. Wy is te area of te under-approximating rectangle given by te formula f(m) ( ) in tis case? 2. Wat is te formula for te area of te over-approximating rectangle? 3. Provide a justification for eac step in te matematical sentence below. Remember tat < 0, and A = A(x) A(x + ). f(m)( ) A f(m)( ) f(m) A f(m) A(x) A(x + ) f(m) f(m) f(m) f(m) Tus, precisely te same inequality is obtained as wen is positive. let approac 0; ten m must approac x Now let approac 0 (from te rigt-and side, since is positive). Remember tat m is trapped in te interval [x, x + ], so as approaces zero, m is forced to get close to x. Tat is, as 0 +, it must be tat m x +. Note: Here, we re olding x fixed and letting cange. Since is canging, m can cange! Te same label,, is used in all four sketces above, even toug is getting smaller. Te same label, m, is used, even toug it can cange. Tis can be confusing same labels, different numbers so be aware! EXERCISE 3 Rewrite te previous paragrap so tat it applies wen < 0. using te continuity of f By ypotesis, f is continuous at x. Terefore, wen te inputs are close to x, te corresponding outputs must be close to f(x). In particular, wen m is close to x, f(m) must be close to f(x). More precisely, as m x +, we must ave f(m) f(x).

4 404 copyrigt Dr. Carol JV Fiser Burns ttp:// as approaces 0, bot m and M must get close to x te quotient A(x+) A(x) is pinced between numbers tat are bot going to f(x) Similarly, since M is trapped between x and x +, as approaces 0, M must approac x. And as M gets close to x, te continuity of f at x tells us tat f(m) approaces f(x). Reconsider te previous inequality in ligt of our new information: f(m) f(m) As approaces 0 (from te rigt-and side), bot f(m) and f(m) are approacing f(x). So te quotient is pinced between numbers wic are bot going to te same number, f(x)! Terefore, A(x+) A(x) must also be getting close to f(x)! (Tis observation is sometimes formalized in a result called te Pincing Teorem for Limits.) Tat is, it must be tat: 0 + EXERCISE 4 Rewrite te necessary paragraps, and conclude tat: 0 te it 0 exists Now it is known tat ; 0 + and, if you ve been doing te exercises, it is also known tat: 0 Putting tese two pieces of information togeter, we conclude tat te two-sided it exists and equals f(x) : But wen te it 0 0 exists, it is given a special name: A (x)! So it is now known tat: A (x) Tat is, te area function A is a function wic, wen differentiated, yields f. Tat is, A is an antiderivative of f.

5 copyrigt Dr. Carol JV Fiser Burns ttp:// 405 A(x) is an antiderivative of f(x) now we know wat all te antiderivatives look like Te fact just discovered is so important tat it is wort repeating. Te area function A is an antiderivative of f. In particular, it as been sown tat wenever f is continuous and nonnegative on [a, b], an antiderivative of f always exists! Tis is an extremely beautiful and important result. Getting our ands on one antiderivative is always te ard part; now we know wat all te antiderivatives of f must look like tey must differ from A by at most a constant. Tat is, if F denotes any antiderivative of A, ten: A(x) = F (x) + C (*) solving for te constant C Remember tat we want to find A(b), since tis represents te area under te grap of f between a and b. Using te fact tat A(a) = 0, equation (*) yields 0 = A(a) = F (a) + C so tat C = F (a). Ten (*) can be rewritten as: Now, letting x equal b, we obtain: A(x) = F (x) F (a) desired area = A(b) = F (b) F (a) Tis is te formula for te desired area, given in terms of any antiderivative of f. Te result is summarized below. formula for te area beneat te grap of a nonnegative, continuous function f on [a, b] Let f be nonnegative and continuous on te interval [a, b]. Let F be any antiderivative of f on [a, b]. Ten: te area under te grap of f on [a, b] = F (b) F (a)

6 406 copyrigt Dr. Carol JV Fiser Burns ttp:// EXAMPLE testing te formula in a case were te answer is already known It s always a good idea to test a new result in a situation were you can find te answer by alternate means. So let s find te area under te grap of f(x) = 2x between x = 0 and x = 3. Calculus is certainly not needed, since te area is just a triangle: 1 2 (base)(altitude) = 1 2 (3)(6) = 9 Now, use te formula. An antiderivative of f(x) = 2x is needed; te easiest one is F (x) = x 2. Ten, wic agrees wit te first result. F (b) F (a) = F (3) F (0) = = 9, EXERCISE 5 1. Sow tat F (x) = x is an antiderivative of f(x) = 2x. 2. Find te area discussed in te previous example, using te antiderivative F (x) = x Wat appens to te 7? EXERCISE 6 Find te area under te grap of f(x) = 2x between x = 1 and x = 4 in two ways: 1. Sow tat te desired area is a trapezoid; find te area of tis trapezoid. 2. Use an antiderivative of f to find te area. EXAMPLE Problem: Find te area beneat te grap of f(x) = x 2 on [1, 3]. Solution: Here, te area of te region is not easily obtainable from geometry. However, we can get some roug bounds on te desired area, as follows. Te minimum value of f on [1, 3] is 1 2 = 1. Tus, te desired area is underapproximated by a rectangle of widt 3 1 = 2 and eigt 1. Te maximum value of f on [1, 3] is 3 2 = 9. Tus, te desired area is overapproximated by a rectangle of widt 2 and eigt 9. Togeter: (1)(2) actual area (9)(2) Te actual area must lie between 2 and 18. Also, from te sketc, we expect te actual area to be near te middle of tis range of numbers. applying te formula Now apply te formula. We need any antiderivative of f(x) = x 2 ; take F (x) =, since it s te simplest one. Ten: x 3 3 F (b) F (a) = F (3) F (1) = = = 82 3 Te answer is certainly believable, based on te earlier estimates.

7 copyrigt Dr. Carol JV Fiser Burns ttp:// 407 EXERCISE 7 1. Consider te function f(x) = x 2 on te interval [2, 5]. As in te previous example, get an under-approximation and an over-approximation of te area under f on [2, 5]. 2. Find te area, using an antiderivative of f. 3. Find te area, using a different antiderivative of f. EXERCISE 8 Use calculus to find te area under te grap of f(x) = x 2 on [ 2, 1]. Here, [a, b] = [ 2, 1], so a = 2 and b = 1. Make a sketc of te grap of f, and te area tat you are finding. EXERCISE 9 1. Grap f(x) = x 2. Sow te area trapped between te grap of f and te x-axis on [1, 3]. 2. Using any antiderivative F of f, compute F (3) F (1). How does your answer compare to te area under te grap of f(x) = x 2 on [1, 3]? 3. Make a conjecture, based on tis example. QUICK QUIZ sample questions 1. Suppose > 0, and f is continuous on te interval [x, x + ]. Wat does te Max-Min Teorem guarantee? 2. Under wat condition(s) does a function f ave te property tat as x a, f(x) f(a)? 3. Make a sketc tat illustrates a function f, and a D(f), for wic f(x) f(a) as x a. 4. Find te area under te grap of y = 3x 2 on te interval [0, 2]. 5. Suppose f is continuous and nonnegative on [c, d], and F is an antiderivative of f. Give a formula for te area under te grap of f on [c, d]. KEYWORDS for tis section Finding te area under te grap of a continuous, nonnegative function f on te interval [a, b]; a formula for tis area in terms of any antiderivative F of f. END-OF-SECTION EXERCISES In eac problem below, an area is described. Sketc te area tat is described. Approximate te area in any reasonable way. Use calculus to find te area. 1. area bounded between te grap of y = ln x and te x-axis on te interval [1, e] 2. area under te grap of y = 1 t on [1, 2] 3. area bounded by te grap of y = x, te x-axis, te line x = 1, and te line x = 4 4. area bounded by te grap of y = x 2 + 1, te line y = 1, te y-axis, and te line x = 1