Introduction to Integration Part 1: AntiDifferentiation


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1 Mathematics Learning Centre Introuction to Integration Part : AntiDifferentiation Mary Barnes c 999 University of Syney
2 Contents For Reference. Table of erivatives New notation Introuction 2 2. How to use this book Objectives Assume knowlege Test yourself Solutions to Test yourself Definition of the Integral as an AntiDerivative 5 4 Some Rules for Calculating Integrals 7 5 Integrating Powers of an Other Elementary Functions 9 6 Things You Can t Do With Integrals 3 7 Using the Chain Rule in Reverse 5 8 Applications 2
3 Mathematics Learning Centre, University of Syney For Reference. Table of erivatives Function (f()) Derivative ( f () i.e. (f())) n n n (n any real number) e e ln ( >0) sin cos tan cot sec csc sin tan cos sin sec 2 csc 2 sec tan csc cot ( 2 ) + 2 ( < ).2 New notation Symbol f() Meaning The inefinite integral of f() with respect to i.e. a function whose erivative is f(). Note that... acts like a pair of brackets aroun the function. Just as a lefthan bracket has no meaning unless it is followe by a closing righthan bracket, the integral sign cannot stan by itself, but nees tocomplete it. The integral sign tells us what operation to perform an the tells us that the variable with respect to which we are integrating is. New terms Antierivative Primitive function Inefinite integral Meaning These are all ifferent ways of saying a function whose erivative is...
4 Mathematics Learning Centre, University of Syney 2 2 Introuction This booklet is intene for stuents who have never one integration before, or who have one it before, but so long ago that they feel they have forgotten it all. Integration is use in ealing with two essentially ifferent types of problems: The first type are problems in which the erivative of a function, or its rate of change, or the slope of its graph, is known an we want to fin the function. We are therefore require to reverse the process of ifferentiation. This reverse process is known as antfferentiation, orfining a primitive function, orfining an inefinite integral. The secon type are problems which involve aing up a very large number of very small quantities, (an then taking a limit as the size of the quantities approaches zero while the number of terms tens to infinity). This process leas to the efinition of the efinite integral. Definite integrals are use for fining area, volume, centre of gravity, moment of inertia, work one by a force, an in many other applications. This unit will eal only with problems of the first type, i.e. with inefinite integrals. The secon type of problem is ealt with in Introuction to Integration Part 2  The Definite Integral. 2. How to use this book You will not gain much by just reaing this booklet. Have pencil an paper reay to work through the eamples before reaing their solutions. Do all the eercises. It is important that you try har to complete the eercises on your own, rather than refer to the solutions as soon as you are stuck. If you have one integration before, an want to revise it, you shoul skim through the tet an then o the eercises for practice. If you have any ifficulties with the eercises, go back an stuy the tet in more etail. 2.2 Objectives By the time you have worke through this unit, you shoul: Be familiar with the efinition of an inefinite integral as the result of reversing the process of ifferentiation. Unerstan how rules for integration are worke out using the rules for ifferentiation (in reverse). Be able to fin inefinite integrals of sums, ifferences an constant multiples of certain elementary functions. Be able to use the chain rule (in reverse) to fin inefinite integrals of certain epressions involving composite functions. Be able to apply these techniques to problems in which the rate of change of a function is known an the function has to be foun.
5 Mathematics Learning Centre, University of Syney Assume knowlege We assume that you are familiar with the following elementary functions: polynomials, powers of, an the trigonometric, eponential an natural logarithm functions, an are able to ifferentiate these. We also assume that you can recognise composite functions an are familiar with the chain rule for ifferentiating them. In aition you will nee to know some simple trigonometric ientities: those base on the efinitions of tan, cot, sec an csc an those base on Pythagoras Theorem. These are covere in sections of the Mathematics Learning Centre booklet Trigonometric Ientities. Other trigonometric ientities are not neee for this booklet, but will be neee in any course on integration, so if you are preparing for a course on integration you shoul work through the whole of Trigonometric Ientities as well as this booklet. Finally, knowlege of the inverse trigonometric functions, sin, cos, an tan an their erivatives woul be a help, but is not essential. 2.4 Test yourself To check how well you remember all the things we will be assuming, try the following questions, an check your answers against those on the net page.. Fin the erivatives of the following functions: i 0 i iv Fin f () for each of the functions f() given: i i iv f() =e f() =ln() f() =cos + sin f() =cot 3. Fin erivatives of: i (2 +) 2 sin 3 i e 2 iv 2 3 v cos( 3 ) vi ln(sin )
6 Mathematics Learning Centre, University of Syney 4 4. Simplify the following epressions: i tan csc sec 2 *5. Fin the erivatives of: i sin tan *Omit this question if you have not stuie inverse trigonometric functions. 2.5 Solutions to Test yourself. i (0 )=0 9 i iv ( )= ( 2 )= 2 2 = 2 ( )= ( )=( ) 2 = 2 (53 3 )= ( 2) 3 = i f () =e i iv f () = f () = sin + cos f () = csc 2 3. i +)2 = 2(2 +) 2=24(2 +) (sin 3) =cos 3 3=3cos3 i iv v vi (e2 )=e 2 2 =2e 2 ( )= 2 3 ((2 3) )=( )( 2 3) 2 2 = (cos(3 )) = sin( 3 ) 3 2 = 3 2 sin( 3 ) (ln(sin )) = cos = cot sin 4. i tan csc = sin = = sec cos sin cos Since sec 2 =+tan 2, sec 2 = tan 2 *5. i (sin )= 2 (tan )= ( 2 3) 2 If you ha ifficulty with many of these questions it may be better for you to revise ifferentiation an trig ientities before going on with this booklet.
7 Mathematics Learning Centre, University of Syney 5 3 Definition of the Integral as an AntiDerivative If (F ()) = f() then f() = F (). In wors, If the erivative of F () isf(), then we say that an inefinite integral of f() with respect to is F (). For eample, since the erivative (with respect to ) of 2 is 2, wecan say that an inefinite integral of 2 is 2. In symbols: (2 )=2, so 2 = 2. Note that we say an inefinite integral, not the inefinite integral. This is because the inefinite integral is not unique. In our eample, notice that the erivative of 2 +3is also 2, so 2 +3is another inefinite integral of 2. In fact, if c is any constant, the erivative of 2 + c is 2 an so 2 + c is an inefinite integral of 2. We epress this in symbols by writing 2 = 2 + c where c is what we call an arbitrary constant. This means that c has no specifie value, but can be given any value we like in a particular problem. In this way we encapsulate all possible solutions to the problem of fining an inefinite integral of 2 in a single epression. In most cases, if you are aske to fin an inefinite integral of a function, it is not necessary to a the +c. However, there are cases in which it is essential. For eample, if aitional information is given an a specific function has to be foun, or if the general solution of a ifferential equation is sought. (You will learn about these later.) So it is a goo iea to get into the habit of aing the arbitrary constant every time, so that when it is really neee you on t forget it. The inverse relationship between ifferentiation an integration means that, for every statement about ifferentiation, we can write own a corresponing statement about integration. For eample, (4 )=4 3, so 4 3 = 4 + c.
8 Mathematics Learning Centre, University of Syney 6 Eercises 3. Complete the following statements: (i) (sin ) = cos, so cos = sin + c. () (5 ) =, so = (i) (e ) =, so = (iv) ( ) 2 =, so = (v) (vi) =, so =, so = = The net step is, when we are given a function to integrate, to run quickly through all the stanar ifferentiation formulae in our mins, until we come to one which fits our problem. In other wors, we have to learn to recognise a given function as the erivative of another function (where possible). Trytoothe following eercises by recognising the function which has the given function as its erivative. Eercises 3.2 i ( sin ) 3 2 i 2 iv sec 2 v vi ( 2 )
9 Mathematics Learning Centre, University of Syney 7 4 Some Rules for Calculating Integrals Rules for operating with integrals are erive from the rules for operating with erivatives. So, because (cf()) = c (f()), for any constant c, we have Rule (cf()) = c f(), for any constant c. For eample 0 cos =0 cos =0sin + c. It sometimes helps people to unerstan an remember rules like this if they say them in wors. The rule given above says: The integral of a constant multiple of a function is aconstant multiple of the integral of the function. Another way of putting it is You can move a constant past the integral sign without changing the value of the epression. Similarly, from we can erive the rule (f()+g()) = (f()) + (g()), Rule 2 (f()+g()) = f() + g(). For eample, (e +2) = e + 2 = e c. In wors, the integral of the sum of two functions is the sum of their integrals. We can easily eten this rule to inclue ifferences as well as sums, an to the case where there are more than two terms in the sum. Eamples Fin the following inefinite integrals: i ( sin ) (3 cos 2 e )
10 Mathematics Learning Centre, University of Syney 8 Solutions i ( sin ) = ( sin ) = cos + c. Note: We have written + sin as ( sin ) because ( sin ) isthe erivative of cos. (3 cos 2 e ) = 3 cos 2 e = 3sin 2 e + c. You will fin you can usually omit the first step an write the answer immeiately. Eercises 4 Fin the following inefinite integrals: i (cos + sin ) (e ) i ( ) iv (3 sec )
11 Mathematics Learning Centre, University of Syney 9 5 Integrating Powers of an Other Elementary Functions We can now work out how to integrate any power of by looking at the corresponing rule for ifferentiation: (n )=n n, so n n = n + c. Similarly ( ) n+ =(n +) n, so (n +) n = n+ + c. Therefore n = = = n + (n +)n (n +) n n + n + n+ + c. notice that when we cancel n+ (n +)isjust take n+ outsie the sign We shoul now look carefully at the formula we have just worke out an ask: for which values of n oes it hol? Remember that the ifferentiation rule (n )=n n hols whether n is positive or negative, a whole number or a fraction or even irrational; in other wors, for all real numbers n. We might epect the integration rule to hol for all real numbers also. But there is one snag: in working it out, we ivie by n +. Since ivision by zero oes not make sense, the rule will not hol when n +=0,that is, when n =. So we conclue that Rule 3 n = n + n+ + c for all real numbers n, ecept n =. When n =, n becomes =. We on t nee to worry that the rule above oesn t apply in this case, because we alreay know the integral of. Since (ln ) =, we have =ln + c.
12 Mathematics Learning Centre, University of Syney 0 Eamples Fin i i 3 2 Solutions i 3 = i = 2 = (3 + ) 4 + c = c. replacing n by 3inthe formula c = + c. replacing n by 2 inthe formula 2 = c = c. replacing n by = Eercises 5.. Fin antierivatives of the following functions: i 5 9 i 4 iv 2 v vi 3 v 2 vi i π 2. Fin the following integrals: i 3 i iv v vi v ( ) ( ) ( ) 2 ( ) ( ) 2 2 Hint: multiply out the epression Hint: ivie through by the enominator Hint: ivie through by the enominator
13 Mathematics Learning Centre, University of Syney At this stage it is very tempting to give a list of stanar integrals, corresponing to the list of erivatives given at the beginning of this booklet. However, you are NOT encourage to memorise integration formulae, but rather to become VERY familiar with the list of erivatives an to practise recognising a function as the erivative of another function. If you try memorising both ifferentiation an integration formulae, you will one ay mi them up an use the wrong one. An there is absolutely no nee to memorise the integration formulae if you know the ifferentiation ones. It is much better to recall the way in which an integral is efine as an antierivative. Every time you perform an integration you shoul pause for a moment an check it by ifferentiating the answer to see if you get back the function you began with. This is a very important habit to evelop. There is no nee to write own the checking process every time, often you will o it in your hea, but if you get into this habit you will avoi a lot of mistakes. There is a table of erivatives at the front of this booklet. Try to avoi using it if you can, but refer to it if you are unsure. Eamples Fin the following inefinite integrals: i (e ) (5 csc 2 +3sec 2 ) Solutions i (e ) = e = e c 2 = e c = e c. (5 csc 2 +3sec 2 ) = 5 ( csc 2 ) +3 = 5 cot +3tan + c. sec 2
14 Mathematics Learning Centre, University of Syney 2 Eercises 5.2 Integrate the following functions with respect to : i 0e 5 sin i ( 2 + +) 5 ( 2 ) + Hint: Multiply through by, an write with fractional eponents. iv v tan sin cos Hint: Divie through by + 2, an consult table of erivatives. Hint: Write tan as sin cos an simplify. vi tan 2 Hint: Remember the formula + tan 2 = sec 2. Youmay use the table of erivatives if you like. (If you are not familiar with inverse trig functions, omit parts i an iv.) Hint: In orer to get some of the functions above into a form in which we can recognise what they are erivatives of, we may have to epress them ifferently. Try to think of ways in which they coul be change that woul be helpful.
15 Mathematics Learning Centre, University of Syney 3 6 Things You Can t Do With Integrals It is just as important to be aware of what you can t o when integrating, as to know what you can o. In this way you will avoi making some serious mistakes. We mention here two fairly common ones.. We know that cf() = c f(), where c is a constant. That is, you are allowe to move a constant past the integral sign. It is often very tempting to try the same thing with a variable, i.e. to equate f() with f(). If we check a few special cases, however, it will become clear that this is not correct. For eample, compare the values of an. = 2 = c while These epressesions are obviously ifferent. = 2 2 = c. So the rule we trie to invent oes not work! It is unlikely that anyboy woul try to fin 2 in the way shown above. However, if aske to fin sin, one might very easily be tempte to write sin = cos + c. Although we o not yet know a metho for fining sin, wecan very easily show that the answer obtaine above is wrong. How? By ifferentiating the answer, of course! If we on t get back to sin, wemust have gone wrong somewhere. Notice that cos is a prouct, so we must use the prouct rule to ifferentiate it. ( cos + c) = ( sin )+( ) cos = sin cos. The first term is correct, but the secon term shouln t be there! So the metho we use was wrong. f() is not equal to f(). In wors, you cannot move a variable past the integral sign.
16 Mathematics Learning Centre, University of Syney 4 2. Again, we know that (f()+g()) = f() + g(). That is, the integral of a sum is equal to the sum of the integrals. It may seem reasonable to woner whether there is a similar rule for proucts. That is, whether we can equate f() g() with f() g(). Once again, checking a few special cases will show that this is not correct. Take, for eample, sin. Now sin = 2 2 ( cos )+c = 2 2 cos + c. But ( ) 2 2 cos + c = 2 2 ( sin )+( ) cos = 2 2 sin cos. prouct rule again! an this is nothing like the right answer! (Remember, it ought to have been sin.) So we conclue that f()g() isnot equal to f() g(). In wors, the integral of the prouct of two functions is not the same as the prouct of their integrals. The other important point you shoul have learne from this section is the value of checking any integration by ifferentiating the answer. If you on t get back to what you starte with, you know you have gone wrong somewhere, an since ifferentiation is generally easier than integration, the mistake is likely to be in the integration. Eercises 6 Eplain the mistakes in the following integrations, an prove that the answer obtaine in each case is wrong, by ifferentiating the answers given. i 2 e = 3 3 e + c ( 2 ) = ( 2 ) = sin + c You will learn how to integrate these functions later.
17 Mathematics Learning Centre, University of Syney 5 7 Using the Chain Rule in Reverse Recall that the Chain Rule is use to ifferentiate composite functions such as cos( 3 +),e 2 2, (2 2 +3), ln(3+). (The Chain Rule is sometimes calle the Composite Functions Rule or Function of a Function Rule.) If we observe carefully the answers we obtain when we use the chain rule, we can learn to recognise when a function has this form, an so iscover how to integrate such functions. Remember that, if y = f(u) an u = g() so that y = f(g()), (a composite function) y then = y u u. Using function notation, this can be written as y = f (g()) g (). In this epression, f (g()) is another way of writing y u an g () isanother way of writing u where u = g(). This last form is the one you shoul learn to recognise. where y = f(u) an u = g() Eamples By ifferentiating the following functions, write own the corresponing statement for integration. i sin 3 (2 +) 7 i e 2 Solution i sin 3 = cos 3 3, so cos 3 3 = sin 3 + c. i (2 +)7 = 7(2 +) 6 2, so ( e 2) = e 2 2, so 7(2 +)6 2 = (2 +) 7 + c. e 2 2 = e 2 + c.
18 Mathematics Learning Centre, University of Syney 6 Eercises 7. Differentiate each of the following functions, an then rewrite each result in the form of a statement about integration. i (2 4) 3 sin π i e 3 5 iv ln(2 ) v 5 3 vi tan 5 v ( 5 ) 4 vi sin( 3 ) i e cos 5 i tan ( 2 +) ln(sin ) The net step is to learn to recognise when a function has the forms f (g()) g (), that is, when it is the erivative of a composite function. Look back at each of the integration statements above. In every case, the function being integrate is the prouct of two functions: one is a composite function, an the other is the erivative of the inner function in the composite. You can think of it as the erivative of what s insie the brackets. Note that in some cases, this erivative is a constant. For eample, consier e 3 3. We can write e 3 as a composite function. 3isthe erivative of 3 i.e. the erivative of what s insie the brackets in e (3). This is in the form f (g()) g () with u = g() =3, an f (u) =e u. Using the chain rule in reverse, since (f(g())) = f (g()) g () wehave f (g()) g () = f(g()) + c. In this case e 3 3 = e 3 + c. If you have any oubts about this, it is easy to check if you are right: ifferentiate your answer! Now let s try another: cos( 2 +5) 2. cos( 2 +5)isacomposite function. 2 is the erivative of 2 +5,i.e. the erivative of what s insie the brackets.
19 Mathematics Learning Centre, University of Syney 7 So this is in the form f (g()) g () with u = g() = 2 +5an f (u) =cos u. Recall that if f (u) =cos u, f(u) =sin u. So, cos( 2 +5) 2 = sin( 2 +5)+c. Again, check that this is correct, byifferentiating. People sometimes ask Where i the 2 go?. The answer is, Back where it came from. If we ifferentiate sin( 2 +5)weget cos( 2 +5) 2. So when we integrate cos( 2 +5) 2 we get sin( 2 + 5). Eamples Each of the following functions is in the form f (g()) g (). Ientify f (u) an u = g() an hence fin an inefinite integral of the function. i (3 2 ) 4 6 sin( ) 2 Solutions i (3 2 ) 4 6 is a prouct of (3 2 ) 4 an 6. Clearly (3 2 ) 4 is the composite function f (g()). So g() shoul be is the other part. This shoul be the erivative of what s insie the brackets i.e. 3 2, an clearly, this is the case: (32 ) = 6. So, u = g() =3 2 an f (u) =u 4 giving f (g()) g () =(3 2 ) 4 6. If f (u) =u 4, f(u) = 5 u5. So, using the rule f (g()) g () = f(g()) + c we conclue (3 3 ) 4 6 = 5 (32 ) 5 + c. You shoul ifferentiate this answer immeiately an check that you get back the function you began with.
20 Mathematics Learning Centre, University of Syney 8 sin( ) 2 This is a prouct of sin( ) an 2. Clearly sin( )isacomposite function. The part insie the brackets is,sowewoul like this to be g(). The other factor 2 ought to be g (). Let s check if this is the case: g() = = 2,sog () = 2 2 = = So we re right! Thus u = g() = an f (u) =sin u giving f (g()) g () =sin( ) 2. Now, if f (u) =sin u, f(u) = cos u. So using the rule f (g()) g () = f(g()) + c we conclue sin( ) 2 = cos( )+c. Again, check immeiately by ifferentiating the answer. Note: The eplanations given here are fairly lengthy, to help you to unerstan what we re oing. Once you have graspe the iea, you will be able to o these very quickly, without neeing to write own any eplanation. Eample Integrate sin 3 cos. Solution sin 3 cos = (sin ) 3 cos. So u = g() =sin with g () =cos. An f (u) =u 3 giving f(u) = 4 u4. Hence sin 3 cos = 4 (sin )4 + c = 4 sin4 + c.
21 Mathematics Learning Centre, University of Syney 9 Eercises 7.2 Each of the following functions is in the form f (g()) g (). Ientify f (u) an u = g() an hence fin an inefinite integral of the function. i i (ln )2 iv e v sin( 3 ) 3 2 vi cos ( ) π π 2 2 v (7 8) 2 7 vi sin(ln ) i ( ) cos sin e tan sec 2 i e sec 2 (5 3) 5 i (2 ) 3 2 iv sin cos The final step in learning to use this process is to be able to recognise when a function is not quite in the correct form but can beput into the correct form by minor changes. For eample, we try to calculate We notice that 4 +is a composite function, so we woul like to have u = g() = 4 +. But this woul mean g () =4 3, an the integran (i.e. the function we are trying to integrate) only has 3.However, we can easily make it 4 3,asfollows: = 4 Note: The 4 an the 4 cancel with each other, so the epression is not change. So u = g() = 4 +, g () =4 3 An f (u) =u 2 f(u) = 2u So, = = ( 4 + ) c. 3 Note: We may only insert constants in this way, not variables. We cannot for eample evaluate e 2 by writing e 2 2, because 2 the in front of the integral sign oes not cancel with the which has been inserte in the integran. This integral cannot, in fact, be evaluate in terms of elementary functions.
22 Mathematics Learning Centre, University of Syney 20 The eample above illustrates one of the ifficulties with integration: many seemingly simple functions cannot be integrate without inventing new functions to epress the integrals. There is no set of rules which we can apply which will tell us how to integrate any function. All we can o is give some techniques which will work for some functions. Eercises 7.3 Write the following functions in the form f (g()) g () an hence integrate them: i cos 7 e 2 i 2 2 iv 2 (4 3 +3) 9 v sin( + 3) vi sin v ( 2 ) vi e 3 i tan 6 Hint: Write tan 6 in terms of sin 6 an cos 6.
23 Mathematics Learning Centre, University of Syney 2 8 Applications Applications of antiifferentiation arise in problems in which we know the rate of change of a function an want to fin the function itself. Problems about motion provie many eamples, such as those in which the velocity of a moving object is given an we want to fin its position at any time. Since velocity is rate of change of isplacement, we must antiifferentiate to fin the isplacement. Eamples. A stone is thrown upwars from the top of a tower 50 metres high. Its velocity in an upwars irection t secon later is 20 5t metres per secon. Fin the height of the stone above the groun 3 secons later. Solution Let the height of the stone above groun level at time t be h metres. We are given two pieces of information in this problem: i the fact that the tower is 50m high tells us that when t =0(that is, at the moment the stone leaves the thrower s han), h = 50, the fact that the velocity at time t is 20 5t tells us that h t =20 5t. We begin with the secon statement, which tells us about rate of change of a function. By antiifferentiating, we obtain h = (20 5t)t = 20t 5 2 t2 + c. () Note: it is vitally important not to forget the +c (the constant of integration) in problems like this. Now we can make use of the first statement, which is calle an initial conition (it tells us what things were like at the start) to fin a value for the constant c. By substituting h =50an t =0into the equation (), we obtain 50 = c. So h =20t 5 2 t (2) Finally, let us go back to the problem, an rea it again, to check eactly what we are aske to fin: Fin the height of the stone above the groun 3 secons later. To fin this, all we have to o is substitute t =3in the epression we have just erive. When t =3,h = =87.5, so the height of the stone 3 secons later is metres.
24 Mathematics Learning Centre, University of Syney 22 Let us look back at the structure of this problem an its solution. We are given information about the rate of change of a quantity, an we antfferentiate (i.e. integrate) to get a general epression for the quantity, incluing an arbitary constant. We are given information about initial conitions an use this to fin the value of the constant of integration. We now have a precise epression for the quantity, an can use that to answer the questions aske. 2. When a tap at the base of a storage tank is turne on, water flows out of the tank at the rate of 200e 5 t litres per minute. If the volume of water in the tank at the start is 000 litres, fin how much is left after the tap has been running for 0 minutes. Solution Let the volume of water in the tank t minutes after the tap is turne on be V litres. Since water is running out of the tank, V will be ecreasing, an so V must be t negative. V Hence = 200e 5 t t ( 200e an so V = t) 5 t ( = 200 ( 5) e 5 t ) t 5 = 000e 5 t + c. Now when t =0,V = 000, so 000 = c, hence c =0. So at any time t, V = 000e 5 t Thus when t = 0, V = 000e litres. Eercises 8. When a stone is roppe into smooth water, circular ripples sprea out from the point where it enters the water. If the area covere by ripples increases at a rate of 2πt square metres per secon, fin the total area of isturbe water t secons after the stone hits the water. What is the area covere by ripples after 3 secons? [Note: when t =0,the stone is just entering the water, so the area of isturbe water is 0.]
25 Mathematics Learning Centre, University of Syney A population of animals, uner certain conitions, has a growth rate given by 500π cos 2πt where t is the time in years. If the initial size of the her is 3000, fin the size of the population at time t. What are the maimum an minimum numbers in the her uring the course of a year? 3. During an eperiment, the height of a growing plant increase at a rate of t+4 cm per ay, where t represents the number of ays since the start of the eperiment. If the plant was 20 cm high at the beginning, what woul its height be after 2 ays? 4. An oral ose of a rug was aministere to a patient. t hours later, the concentration of the rug in the patient s bloo was changing at a rate give by 5e t e 0.2t. If none of the rug was present in the bloo at the time the ose was given, fin the concentration of the rug in the patient s bloo t hours later. How long after aministration will the concentration be greatest? 5. An object is propelle along the groun in such a way that its velocity after t secons is metres per secon. If it starts 2 metres from a fie point, an moves in a t+ straight line irectly towars the point, how long will it take to reach the point?
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