Antiderivatives Introduction

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1 Antierivatives 40. Introuction So far much of the term has been spent fining erivatives or rates of change. But in some circumstances we alreay know the rate of change an we wish to etermine the original function. For example, meters or ata loggers often measure rates of change, e.g., miles per hour or kilowatts per hour. If you know the velocity of an object, can you etermine the position of the object. This coul happen in a car, say, where the speeometer reaings were being recore. Can the position of the car be etermine from this information? Similarly, can the position of an airplane be etermine from the black box which recors the airspee? More generally, given f 0 (x) can we fin the function f (x). If you think about it, this is the sort of question I have aske you to o on labs, tests, an homework assignments where I gave you the graph of f 0 (x) an sai raw the graph of f (x). Or where I gave you the number line information for f 0 (x) an f 00 (x) an aske you to reconstruct the graph of f (x). Can we o this same thing if we start with a formula for f 0 (x)? Can we get an explicit formula for f (x)? We usually state the problem this way. DEFINITION 40.. Let f (x) be a function efine on an interval I. We say that F(x) is an antierivative of f (x) on I if F 0 (x) = f (x) for all x 2 I. EXAMPLE 40.. If f (x) =2x, then F(x) =x 2 is an antierivative of f because F 0 (x) =2x = f (x). But so is G(x) =x 2 + or, more generally, H(x) =x 2 + c. Are there other antierivatives of f (x) =2x besies those of the form H(x) = x 2 + c? We can use the MVT to show that the answer is No. The proof will require three small steps. THEOEM 40. (Theorem ). If F 0 (x) =0 for all x in an interval I, then F(x) =k is a constant function. This makes a lot of sense: If the velocity of an object is 0, then its position is constant (not changing). Here s the Proof. To show that F(x) is constant, we must show that any two output values of F are the same, i.e., F(a) =F(b) for all a an b in I. So pick any a an b in I (with a < b). Then since F is ifferentiable on I, then F is both continuous an ifferentiable on the smaller interval [a, b]. So the MVT

2 2 applies. There is a point c between a an b so that This means F(b) b F(a) a = F 0 (c) ) F(b) F(a) =F 0 (c)[b a] =0[b a] =0. F(b) =F(a), in other wors, F is constant. THEOEM 40.2 (Theorem 2). Suppose that F, G are ifferentiable on the interval I an F 0 (x) =G 0 (x) for all x in I. Then there exist k so that G(x) =F(x)+k. Proof. Consier the function G(x) F(x) on I. Then x (G(x) F(x)) = G0 (x) F 0 (x) =0. Therefore, by Theorem so G(x) F(x) =k G(x) =F(x)+k. THEOEM 40.3 (Theorem 3: Families of Antierivatives). If F(x) an G(x) are both antierivatives of f (x) on an interval I, then G(x) =F(x)+k. This is the theorem we want to show. Proof. If F(x) an G(x) are both antierivatives of f (x) on an interval I then G 0 (x) = f (x) an F 0 (x) = f (x), that is, F 0 (x) =G 0 (x) on I. Then by Theorem 2 G(x) =F(x)+k. DEFINITION If F(x) is any antierivative of f (x), we say that F(x)+c is the general antierivative of f (x) on I. Notation for Antierivatives Antiifferentiation is also calle inefinite integration. is the integration symbol f (x) is calle the integran x inicates the variable of integration f (x) x = F(x)+c. F(x) is a particular antierivative of f (x) an c is the constant of integration. We refer to f (x) x as an antierivative of f (x) or an inefinite integral of f.

3 math 30, ay 4 introuction to antierivatives 3 Here are several examples. 2xx= x 2 + c cos tt= sin t + c e z z = e z + c x = arctan x + c + x2 Antiifferentiation reverses ifferentiation so F 0 (x) x = F(x)+c an ifferentiation unoes antiifferentiation apple f (x) x = f (x). x Differentiation an antiifferentiation are reverse processes, so each erivative rule has a corresponing antiifferentiation rule. Differentiation x (c) =0 x (kx) =k Antiifferentiation 0 x = c kx= kx + c x (xn )=nx n x n x = xn+ n+ + c, n 6= x (ln x ) = x x x = x x = ln x + c x (sin x) =cos x cos xx= sin x + c x (cos x) = sin x sin xx= cos x + c x (tan x) =sec2 x sec 2 xx= tan x + c x (sec x) =sec x tan x sec x tan xx= sec x + c x (ex )=e x e x x = e x + c x (arcsin x) = p p x = arcsin x + c x 2 x 2 x (arctan x) = x = arctan x + c +x 2 +x 2 Variations an Generalizations Notice what happens when we use ax instea of x in some of these functions. We multiply by a when taking the erivative, so we have to ivie by a when taking the antierivative. Differentiation x (eax )=ae ax x (sin ax) =a cos ax x (tan ax) =a sec2 ax x (arcsin( x a )= x (arctan( x a )) = p a 2 x 2 a a 2 +x 2 Antiifferentiation e ax x = a eax + c cos ax x = a sin x + c sec 2 ax x = a tan ax + c p a 2 x = arcsin( x x 2 a )+c x = a 2 +x 2 a arctan( x a )+c Try filling in the rules for sin ax x an sec(ax) tan(ax) x.

4 4 EXAMPLE Here are a few examples Problems cos(4x) x = 4 sin(4x)+c e z/2 z = 2e z/22 + c 6 + x 2 x = 4 arctan x 4 + c. Determine antierivatives of the following functions. Take the erivative of your answer to confirm that you are correct. Why shoul you a +c to any answer? Basics: (a) 7x 6 (b) x 6 (c) 2x 6 () e x (e) 2e x (f )2e 2x (g) e 2x (h) x (j) cos x (k) 4 cos x (l) 4 cos(4x) (m) cos(4x) (n) (o) 2 sec x tan x (p) sin 4x (q) 4 sec 2 x (r) 2x (s) x (t) 8x (u) 6 x ln 6 (v) 6x 6 x ln 6 (w) 6 (x) 6+ sec 2 x (y) 2. Now try these. Think it through. (i) 6 + x 2 (z) 8 x 2 p x 2 (a) x (b) e 8x c (c) 2 sin 4x () cos 2q sec 2 2q (e) x 2 + 3x (f ) cos x (g) x 5/2 (h) x 3/5 (i) 2 (j) 4x 3 x (k) 4x(x 2 + 5) 6 (l) e x cos(e x ) (m) 2xe x2 +9 x + 4 cos x (n) 4p x 7 2(ln x)5 x Answers. Answers (a) x 7 + c (b) x7 7 + c (c) 2 7 x7 + c () e x + c (e) 2e x + c (f ) e 2x + c (g) 2 e2x + c (h) ln x + c (i) 8 ln x + c (j) sin x + c (k) 4 sin x + c (l) sin(4x)+c (m) sin(4x)+c (n) ln x + 4 sin x + c 4 (o) 2 sec x + c (p) 4 cos 4x + c (q) 4 tan x + c (r) x2 + c (s) 2 x2 + c (t) 4x 2 + c (u) 6 x + c (v) 3x 2 6 x + c (w) 6x + c (x) 6x + tan x + c (y) 6 arctan(x)+c (z) 2 arcsin(x)+c 2. Answers. (a) 9 x9 + 2x + c (b) 8 e8x cx + (c) 2 cos 4x + c () 2 sin 2q tan 2q + c 2 (e) 3 x x2 x + c (f ) sin x + c (g) x7/2 + c (h) 5 2 x2/5 + c 4 (i) 3 x 3/4 (j) ln(x 4 + 2)+c (k) (x 2 + 5) 7 + c (l) sin(e x )+c (m) 6e x c (n) 2(ln x) 6 + c

5 General Antierivative ules The key iea is that each erivative rule can be written as an antierivative rule. We ve seen how this works with specific functions like sin x an e x an now we examine how the general erivative rules can be reverse. FACT 4. (Sum ule). The sum rule for erivatives says (F(x) ± G(x)) = x x (F(x)) ± x (G(x)). The corresponing antierivative rule is ( f (x) ± g(x)) x = f (x) x ± g(x) x. FACT 4.2 (Constant Multiple ule). The constant multiple rule for erivatives says x (cf(x)) = c x (F(x)). The corresponing antierivative rule is cf(x) x = c f (x) x. Examples 8x 3 7 p xx= 8x 3 x 7x /2 x = 8 x 3 x 7 x /2 x = 8x4 4 7x 3/2 3/2 + c = 4x 3/2 2x4 + c 3 6 cos 2x 7 x + 2x /3 x = 6 cos 2xx 7 x x + 2 x /3 x = 6 2x2/3 sin 2x 7 ln x + 2 2/3 + c = 3 sin 2x 7 ln x + 3x2/3 + c. 3e x/2 8 p 9 x 2 x = 3 e x/2 x 8 p 9 x 2 x = 3 2ex/2 + 8 arcsin x 3 + c. = 6ex/2 + 8 arcsin x 3 + c. ewriting ewriting the integran can greatly simplify the antierivative process. 2 5p t 2 6 sec 2 tt= x x 2 x = 6x 2 (x 4 ) x = 6x 6 2 3p t = t 5 2t 2/5 6 sec 2 tt= 2t7/5 7/5 6 tan t + c = 0t7/5 7 x 2 + 2x 2 x = x x + c = x3 3 2x + c. 6x 2 x = 6x x3 3 + c = 6x x3 + c. 2t 5/3 2t 2/3 t = 2/3 + c = 3t 2/3 + c. 6 tan t + c. 8x p x = x 8x 3/2 + 7x /2 x = 8x5/3 5/3 + 7x/2 /2 + c = 24x5/3 + 4x /2 + c. 5

6 6 4.3 Evaluating c (Initial Value Problems) So far we have been calculating general antierivatives of functions. What this means is that if we know the velocity v(t) of the car we are riving in, we can etermine the position p(t) of the car, up to a constant c if we can fin an antierivative for v(t). If we have more information, the position of the car at a particular time say, then we are able to etermine the precise antierivative. EXAMPLE 4.3. Suppose that f 0 (x) = e x + 2x an f (0) = 3. Fin f (x). f (0) is sometimes calle the initial value an such questions are referre to as initial value problems. [How coul you interpret this information in terms of motion?] SOLUTION. f (x) must be an antierivative of f 0 (x) so f (x) = f 0 (x) x = e x + 2xx= e x + x 2 + c. Now use the initial value to solve for c: Therefore, f (x) =e x + x f (0) =e c = 3 ) + c = 3 ) c = 2. EXAMPLE 4.4. Suppose that f 0 (x) =6x 2 2x 3 an f () =4. Fin f (x). SOLUTION. Again f (x) must be an antierivative of f 0 (x) so f (x) = f 0 (x) x = 6x 2 2x 3 x = 2x 3 x c. Now use the initial value to solve for c: Therefore, f (x) =2x 3 f () =2 /2 + c = 3 ) c = 2.5. x EXAMPLE 4.5. Suppose that f 00 (t) =6t 2 (think acceleration) with f 0 () =8 (think velocity) an f () =3 (think position). Fin f (t). SOLUTION. First fin f 0 (t) which must be the antierivative of f 00 (t). So f 0 (t) = f 00 (t) t = 6x 2 t = 6t + c. Now use the initial value for f 0 (t) to solve for c: f 0 () = 6()+c = 8 ) c = 4. Therefore, f 0 (t) = 6t + 4. Now we are back to the earlier problem. f (t) = f 0 (t) t = 6t + 4 t = 6 ln t + 4t + c. Now use the initial value of f to solve for c: f () = 6 ln + 4()+c = 3 ) 6(0)+4 + c = 3 ) c =. So f (t) =6 ln t + 4t. In Class Practice EXAMPLE 4.6. Fin f given that f 0 (x) =6 p x + 5x 3 2 where f () =0. SOLUTION. f (x) must be an antierivative of f 0 (x) so f (x) = f 0 (x) x = 6 p x + 5x 3 2 x = 4x 3/2 + 2x 5/2 + c. Use the initial value to solve for c: f () = c = 0 ) c = 4. Therefore, f (x) =4x 3/2 + 2x 5/2 + 4.

7 math 30, ay 4 introuction to antierivatives 7 EXAMPLE 4.7. Fin f given that f 00 (q) =sin q + cos q where f 0 (0) = an f (0) =2. SOLUTION. First fin f 0 (q) which must be the antierivative of f 00 (q). So f 0 (q) = f 00 (q) q = sin q + cos q = cos q + sin q + c. Now use the initial value for f 0 (q) to solve for c: f 0 (0) = cos 0 + sin 0 + c = c = ) c = 2. Therefore, f 0 (q) = cos q + sin q + 2. f (q) = f 0 (q) q = cos q + sin q + 2 q = sin q cos q + 2q + c. Now use the initial value of f to solve for c: f (0) = sin 0 cos 0 + 2(0)+c = 0 + c = 2 ) c = 3. So f (q) = sin q cos q + 2q + 3.

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9 Motion Problems 4.4 Introuction Earlier in the term we interprete the first an secon erivatives as velocity an acceleration in the context of motion. So let s apply the initial value problem results to motion problems. ecall that s(t) =position at time t. s 0 (t) =v(t) velocity at time t. s 00 (t) =v 0 (t) =a(t) acceleration at time t. Therefore a(t) t = v(t)+c velocity. v(t) t = s(t)+c 2 position at time t. We will nee to use aitional information to evaluate the constants c an c 2. EXAMPLE 4.8. Suppose that the acceleration of an object is given by a(t) =2 for t 0 with cos t v(0) =, this is also enote v 0 s(0) =3, this is also enote s 0. Fin s(t). SOLUTION. First fin v(t) which is the antierivative of a(t). v(t) = a(t) t = 2 cos tt= 2t sin t + c. Now use the initial value for v(t) to solve for c : v(0) =0 0 + c = ) c =. Therefore, v(t) =2t s(t) = sin t +. Now solve for s(t) by taking the antierivative of v(t). v(t) t = 2t sin t + t = t 2 + cos t + t + c 2 Now use the initial value of s to solve for c 2 : So s(t) =t 2 + cos t + 2t + 2. s(0) =0 + cos 0 + c = 3 ) + c = 3 ) c = 2. EXAMPLE 4.9. If acceleration is given by a(t) = 0 + 3t function if s(0) = an s(2) =. 3t 2, fin the position SOLUTION. First v(t) = a(t) t = 0 + 3t 3t 2 t = 0t t2 t 3 + c. Now s(t) = 0t t2 t 3 + ct= 5t t3 4 t4 + ct +. But s(0) = = so =. Then s(2) = c + = so 2c = 0 ) c = 5. Thus, s(t) =5t t3 4 t4 5t +.

10 0 EXAMPLE 4.0. If acceleration is given by a(t) = sin t + cos t, fin the position function if s(0) = an s(2p) =. SOLUTION. First v(t) = a(t) t = sin t + cos tt= sin t + cos t + c. Now s(t) = sin t + cos t + ct= cos t sin t + ct +. But s(0) = = so = 2. Then s(p) = pc + 2 = so 2pc = 2 ) c = /p. Thus, s(t) = cos t sin t t p Constant Acceleration: Gravity In many motion problems the acceleration is constant. This happens when an object is thrown or roppe an the only acceleration is ue to gravity. In such a situation we have a(t) =a, constant acceleration with initial velocity v(0) =v 0 an initial position s(0) =s 0. Then v(t) = a(t) t = at= at + c. But So Next, At time t = 0, Therefore v(0) =a 0 + c = v 0 ) c = v 0. v(t) =at + v 0. s(t) = v(t) t = at + v 0 t = 2 at2 + v 0 t + c. s(0) = 2 a(0)2 + v 0 (0)+c = s 0 ) c = s 0. s(t) = 2 at2 + v 0 t + s 0. EXAMPLE 4.. Suppose a ball is thrown with initial velocity 96 ft/s from a roof top 432 feet high. The acceleration ue to gravity is constant a(t) = 32 ft/s 2. Fin v(t) an s(t). Then fin the maximum height of the ball an the time when the ball hits the groun. SOLUTION. ecognizing that v 0 = 96 an s 0 = 432 an that the acceleration is constant, we may use the general formulas we just evelope. v(t) =at + v 0 = 32t + 96 an s(t) = 2 at2 + v 0 t + s 0 = 6t t The max height occurs when the velocity is 0: v(t) = 32t + 96 = 0 ) t = 3 ) s(3) = = 576. The ball hits the groun when s(t) =0. s(t) = 6t t = 6(t 2 6t 27) = 6(t 9)(t + 3) =0. So t = 9 an (t 6= 3).

11 math 30, ay 4 introuction to antierivatives YOU TY IT 4.. A stone is thrown upwar with an initial velocity of 48ft/s from the ege of a cliff 64ft above a river. (emember acceleration ue to gravity is 32ft/s 2.) (o) Fin the velocity of the stone for t 0. (p) Fin the position of the stone for t 0. (q) Fin the time when it reaches its highest point (an the height). (r) Fin the time when the stone hits the groun. Answers: v(t) = 32t + 48, s(t) = 6t t + 64, max ht: 00 ft at EXAMPLE 4.2. A person rops a stone from a brige. What is the height (in feet) of t =.5s, hits groun at t = 4s. the brige if the person hears the splash 5 secons after ropping it? SOLUTION. Here s what we know. v 0 = 0 (roppe) an s(5) =0 (hits water). An we know acceleration is constant, a = 32 ft/s 2. We want to fin the height of the brige, which is just s 0. Use our constant acceleration motion formulas to solve for a. v(t) =at + v 0 = 32t an s(t) = 2 at2 + v 0 t + s 0 = 6t 2 + s 0. Now we use the position we know: s(5) =0. s(5) = 6(5) 2 + s 0 ) s 0 = 400. Notice that we i not nee to use the velocity function. YOU TY IT 4.2 (Extra Creit). In the previous problem i you take into account that soun oes not travel instantaneously in your calculation above? Assume that soun travels at 20 ft/s. What is the height (in feet) of the brige if the person hears the splash 5 secons after ropping it? EXAMPLE 4.3. Here s a variation an this time we will use metric units. Suppose a ball is thrown with unknown initial velocity v 0 ft/s from a roof top 49 meter high an the position of the ball at time t = 3 s is s(3) =0. The acceleration ue to gravity is constant a(t) = 9, 8 m/s 2. Fin v(t) an s(t). SOLUTION. This time v 0 is unknown but s 0 = 49 an s(3) =0. Again the acceleration is constant so we may use the general formulas for this situation. v(t) =at + v 0 = 9.8t + v 0 an But we know that which means s(t) = 2 at2 + v 0 t + s 0 = 4.9t 2 + v 0 t s(3) = 4.9(3) 2 + v 0 (3)+49 = 0 3v 0 = 4.9(9) 4.9(0) = 4.9 ) v 0 = 4.9/3. So an Interpret v 0 = 4.9/3. v(t) = 9.8t s(t) = 4.9t t EXAMPLE 4.4. Mo Green is attempting to run the 00m ash in the Geneva Invitational Track Meet in 9.8 secons. He wants to run in a way that his acceleration is constant, a, over the entire race. Determine his velocity function? (a will still appear as an unknown constant.) Determine his position function? There shoul be no unknown constants in your equation. What is his velocity at the en of the race? Do you think this is realistic?

12 2 SOLUTION. We have: constant acc = a m/s 2 ; v 0 = 0 m/s; s 0 = 0 m. So an v(t) =at + v 0 = at s(t) = 2 at2 + v 0 t + s 0 = 2 at2. But s(9.8) = 2 a(9.8)2 = 00, so a = 200 (9.8) 2 = m/s 2.Sos(t) =2.0825t 2. Mo s velocity at the en of the race is v(9.8) =a(9.8) =2.0825(9.8) =20.4 m/s... not realistic. EXAMPLE 4.5. A stone roppe off a cliff hits the groun with spee of 20 ft/s. What was the height of the cliff? SOLUTION. ecognizing that v 0 = 0 an s 0 is unknown an is the cliff height an that the acceleration is constant a = 32 ft/s, we may use the general formulas for motion with constant acceleration: v(t) =at + v 0 = 32t + 0 = 32t. Now we use the velocity an the one velocity we know v = 20 when it hits the groun so v(t) = 32t = 20 ) t = 20/32 = 5/4 when it hits the groun. Now remember when it hits the groun the height is 0. So s(5/4) =0. But we know s(t) = 2 at2 + v 0 t + s 0 = 6t 2 + 0t + s 0 = 6t 2 + s 0. Now substitute in t = 5/4 an solve for s 0. s(5/4) =0 ) 6(5/4) 2 + s 0 = 0 ) s 0 = 5 2 = 225. The cliff height is 225 feet. EXAMPLE 4.6. A car is traveling at 90 km/h when the river sees a eer 75 m ahea an slams on the brakes. What constant eceleration is require to avoi hitting Bambi? [Note: First convert 90 km/h to m/s.] SOLUTION. Let s list all that we know. v 0 = 90 km/h or = 25 m/s. Let s 0 = 0 an let time t represent the time it takes to stop. Then s(t )=75 m. Now the car is stoppe at time t, so we know v(t ) = 0. Finally we know acceleration is an unknown constant, a, which is what we want to fin. Now we use our constant acceleration motion formulas to solve for a. an v(t) =at + v 0 = at + 25 s(t) = 2 at2 + v 0 t + s 0 = 2 at2 + 25t. Now we use other velocity an position we know: v(t )=0 an s(t )=75 when the car stops. So v(t )=at + 25 = 0 ) t = 25/a an Simplify to get so s(t )= 2 a(t ) t = 2 a( 25/a)2 + 25( 25/a) = a 625 2a 2 = 625 a 2a The time it takes to stop is 350 2a = 625 2a a = = 25 6 m/s. t = 25 a = = 6. = 75 ) 50a = 625

13 math 30, ay 4 introuction to antierivatives 3 EXAMPLE 4.7. One car intens to pass another on a back roa. What constant acceleration is require to increase the spee of a car from 30 mph (44 ft/s) to 50 mph ( ft/s) in 5 secons? SOLUTION. Given: a(t) =a constant. v 0 = 44 ft/s. s 0 = 0. An v(5) = ft/s. Fin a. But v(t) =at + v 0 = at So v(5) =5a + 44 = ) 5a = = Thus a = 88 5.

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