Calculus Tutorial2-- Integrals

1.

C alculus Tutorial 2 ---- Integral C alculus S riniva san Ne nme li-K Introduction Inte gra l Ca lculus is the re ve rse proce ss of diffe re ntiation or finding the de rivative . S uppose y = f(x) We find the de rivative dy/dx by the me thods give n in the tutorial1. Le t dy/dx = g(x) , Now we re write this as follows:

anothe r function of x.

dy = g (x) dx

The proce ss of inte gration use s the inte gral symbol or sign: Le t us "inte grate " both side s

The re sult is : y = f(x) We ge t back the original function.! It is re ally that simple . The Inte gra l is ,the re fore ,calle d "Anti-de rivative ". A fe w te rms to le arn: The function f(x) is calle d 'the inte gral of g(x) dx" . The function g(x) (within the inte gral sign) is calle d the integrand. In ge ne ral we write :

He re C is calle d "the constant of inte gration" . Why we add this? You will le arn its me aning shortly. Note that : df(x)/dx= g(x) and dC/dx=0 The re fore the addition of C doe s not affe ct the proce ss at all, but se rve s a purpose .! Now le t us se e a fe w simple e xample s to be come familiar with the proce ss of inte gra tion,be fore we e xplore the 'physical and graphical inte rpre tation ' of the inte gral. Simple Integration Examples Example 1 Le t g(x) = k, a constant

say g(x) = 3

Inte gration is some what e asy,be cause we can always che ck our re sult by diffe re ntiating the inte gral;we should ge t back the inte grand.DO THIS CHECKING ALWAYS. f(x) = 3x+C df(x)/dx = 3 =g(x) This we ca n unde rstand at once : y= 3x+C is a straight line with slope of 3. The function g(x) = 3 which is the de rivative of y or dy/dx which is the slope . The re is nothing surprising or strange about this. Wha t doe s C signify he re ? we ll---C is the inte rce pt in this line e quation.. g(x) is the de riva tive a nd so, we are looking at the slope only.The inte rce pt could be any value .This inte gra l 3x+C re pre se nts all the straight line s with the same slope of 3 or the family of line s or paralle l line s with the same slope but with diffe re nt inte rce pts. Le t us find the inte gral : Give n a point on the line , x=2 and y=9 The n y= 9 = 3x+C= 6 +C --> C=3 --> y= 3x+3 Now you are spe cifying a particular line with inte rce pt C=3

This is the purpose or me aning of attaching C, the inte gration constant. Go ove r this e xample once again to gain an unde rstanding of this e sse ntial conce pt of inte gra tion. Example 2 Le t g(x) = x

The n

Che ck: Le t us unde rstand this re sult. y = g(x) = x is a straight line passing through the or igin with slope =1. The inte gral re pre se nts a family of parabolas. If it is give n that the inte gral re pre se nts a parabola pasing through (2,5) the n y = 5 = 4/2 + C C= 3 Now we a re re fe rring to a particular parabola with ve rte x at (0,3)

Try ske tching this parabola.! 3 Le t

The n:

Che ck: 4. Now we can ge ne ralize and find a formula: If

--------------------[1]

Che ck:Re ca ll that The re fore : You have to re me mbe r this formula for quick work. This formula works when n is negative ,[except when n = - 1] and also when n is a fraction. We ne e d two important re sults ,as we did for diffe re ntiation:

1.

If k is a constant, the n

2. [The inte gration is also a line ar ope rator like diffe re ntiation.] We illustra te this re sult with a fe w e xample s: Exa mple 3 Find

Example 4 Find

= This is calle d "inte rgration te rm by te rm". Applica tion Proble m 1: The ve locity of a car was give n by the e quation: v= 25 t + 10

whe re t is time in hours. Find its position,S , as a function of time . Ve locity = v = ds/dt = 25t +10 ds= (25t+10)dt

If the original position of the car was 5 mile s from home , find the e quation: Whe n t= 0, S =5=C [Note : In the e quation v=25t+10, the acce le ration is a=25, since v(t)=at+v(0)]

Example 5

Find Re call that

The re fore :

Example 6 Find He re n = -2 Using the formula [1] give n e arlie r:

+C

Example 7 He re n= 1/2

Find n+1 = 3/2 +C

Example 8

Find

Example 9 Find Expand the function (2x+1)^2 and inte grate te rm by te rm.

Pra ctice Proble ms

1 Find

2 Find

3 Find

4 Find

5 Find Physical and graphical interpretation of an Integral We inte rpre te d a de rivative as the slope for a curve at a give n point on the curve . The antide rivative or inte gral of a function is the are a unde r the curve of y= f(x) a nd the x axis. Take a simple function: y= 4x This is a straight line passing through the origin.Le t us take the triangular pie ce from x= 0 to x = 2. [The ve rtice s are (0,0)(0,2),(2,8).] The are a unde r the curve will be give n by are a= (1/2) base x he ight = (1/2) 2 x 8 = 8 units.In ge ne ral te rms, a re a = (1/2) x (4x) = 2x.x Le t us inte grate :

You ge t the same re sult. The re fore inte gration is adding up small strips of are a e ach with a width of dx along the x axis a nd the corre sponding y value s on the curve . Note that the inte gral is a parab ola and incre a se s rapidly as x is incre ase d---- the are a of the triangle forme d incre ase s in its are a, a s x square d. S ince the y value s are changing along the curve [e xce pt for y=k, a constant) we take instanta ne ous value of y and multiply with a small value of x,that is dx, and add all the se sma ll a re a s. Trigonometric Functions Re call the de rivative formulas.It is e asy to work out the se inte grals.

Example 10 Che ck:

d(-cosx)/dx= sin x

Example 9 Che ck:

d(sinx)/dx= cos x

Example 10

Example 11 [We shall take up othe r trig functions late r.] U- Substitution This is similar to chain rule for diffe re ntiation.We substitute a simple r e xpre ssion for a more comple x e xpre ssion, and fit into the formula give n e arlie r. This is a powe rful te chnique ....maste r it we ll!

Example 12 Le t Diffe re ntiating:

Find

du = 2x dx x.dx= du/2

The give n inte gral be come s:

= Che ck: We can che ck e asily by e xpanding the inte grand: g(x) = x^3+x

Note that the constant has change d.That doe s not make a difffe re nce for the inte gra l.

Example 13

Find

Le t Diffe re ntiating , we ge t:

The inte grand be come s:

The give n inte gral =

Example 14 Find Le t du = 3 dx

Example 15 Find Le t u= 2x

du = 2 dx

Example 16 Find Le t u= sin x

du = cosx dx

Che ck: 2sinxcosx= sin(2x) An alte rnate me thod:

Exa mple 17

Find

Re ca ll that d(tanx)/dx = se c^2 (x) Le t u=3x du = 3dx

Example 18

Find

Re call the trig ide ntity:

and

= =

Pra ctice Proble m: Find

Ans: x/2 - (1/2) sinx cosx + C

Example 19 Find I =

Le t

Example 20 Find ta n x = sinx/ cos x

Le t u=cos x

du =-sinx

Note that for ln(x) , x cannot be ne gative ,and so we write

Pra ctice Proble ms Find

Example 21

Ans: ln(sinx)+C

Find

Le t dx= u du

I= [S ubstitute for u and simplify.] Exponential and Logarithmic Functions It is e asy to inte grate an e xpone ntial function.

Example 22 Find +C Applie d Proble m The ra te of growth of a fish population in a lake is give n by: whe re t is in days and 100 is the initial population.Find the total fish population a t any time t and find the fish population afte r 100 days.

+C

If t= 100 da ys, Loga rithmic function You re call that the logarithmic function is an inve rse function of the e xpone ntial fu nction: If y = e xp(x) the n lny = x But inte gra tion of lnx is not straightforward.

To find Take y= xlnx - x The n dy/dx= lnx + x/x -1 = ln x

Example 23 Le t u= sinx

Find du = cosx dx

Example 24 Find Re call the ide ntity:

Pra ctice Proble ms 1 Find the inte grals:

Ans:-(1/6)e xp(-3x^2)+C

Ans: ln|e xp(x)-1|+C

Inte grate te rm by te rm

Hyperbolic Functions The following hype rbolic functions are de rive d from the e xpone ntial function:

Example 1

Find the inte gral:

+C

Pra ctice Proble ms: 1

2 Coth(x)=cosh(x)/sinh(x)

Find

3 Find I:

4 Find I:

Integration by parts This is one of the difficult portion of inte gral calculus to che w..This me htod is ve ry use ful for complicate d functions.You have to le arn this! Le t us re ca ll the 'product rule ':

Le t us inte grate both side s :

This me thod starts with one inte gration and re sults in anothe r inte gral to tackle .! -- A bit confusing,right!Hope fully the se cond inte gral on the right side will be e asy to find. Le t us se e a fe w e xample s.

Example 25

Find

Le t u=x

du=dx

dv= sinxdx

Using this me thod: I=

Example 26

Find

Earlie r I gave the inte gral by showing the answe r first and diffe re ntiating it.Now we shall find by this me thod--Inte gration by parts. Le t u= ln (x) dv = dx

Example 27

du = 1/x v= x

Find

Look a t this e quation. We se e m to be in a circular track since we got anothe r inte gra l to find. Apply "inte gration by parts" again:

The re fore

[Note :** The ne xt e xample is a bit complicate d. You can study it in the se cond round of your study.S kip for now.] ** Example 27 Find

The re fore

we find 2I now.

Example 28

Comme nt: It is important to practise "Inte gration by parts' we ll, particularly for AP Calculus. Integration by Partial Fractions Example 29 Find

S plit the inte grand by partial fractions:

Ne xt we find :

A=5/7

B=2/7

Example 30 Find: S plit the inte grand into partial fractions:

S olving

A=-1/2 B= -1/2 c= 1/2

[Note :

You ca n use the 'Table of Inte grals' found in the te xt books for quick re fe re nce .]

Pra ctice Proble ms Find the inte grals by partial fractions me thod: 1

2

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Definite Integrals We now move into de finite inte grals. A de finte inte gral has two limits of inte gration: the uppe r limit 'b' and the lowe r limit 'a' and is re pre se nte d by this sign:

The value of the inte gral is writte n as F(x), with uppe r case le tte r. [This is just conve ntion.You can all it g(x) too.] Le t the corre sponding inde finite inte gral which we have be e n doing in the pre vious paragra phs be F(x).

The n the de finite inte gral is simply the value of F(x) e valuate d at x=b ,the n F(x) at x=a and finding the diffe re nce :

--------------Equation [3] This re sult is calle d a "Fundame ntal The ore m of Calculus." S te p 1: Find the inte gral e xpre ssion for the inde finite inte gral first. S te p2: Eva luate the inte gral at the uppe r limit F(b) and at the lowe r limit F(a) S te p 3: Find the diffe re nce : F(b)-F(a) The physical me aning is that this inte gral is the are a be low the curve y=f(x) be twe e n two ve rtica l line s x=b and x=a and the x axis. Note that for de finite inte grals we ge t a numbe r and the constant of inte gration 'C' is now gone .We ge t a boundary to the inte gral and so we dont add that ubiquitous C of e arlie r se ctions!

Le t us se e a fe w simple e xample s: Example 31

Example 32

Now it is time for some applie d proble ms! ---------------------------------------------------------------------------------------------------------

Applied Problems

Proble m1 John sta rts his car from re st and acce le rate s for 6 mins.During this phase , the spe e d is give n by the e quation: v=90t whe re t is in hours.V is give n in mile s pe r hour[mph] .Afte r 6mins,John continue s to drive at a constant spe e d of 90 mile s pe r hour for the ne xt 2 4 minute s.Find the total distance trave lle d in this 30 minute s drive . Acce le ration phase : v= ds/dt ds=v dt We ha ve to inte grate 'ds' from time t=0 to t= 0.1 hour [6minute s] distance trave lle d in this phase =S 1 :

Constant Phase : v=constant=90 ds=90dt Inte grate from t=0.1 (6minute s) to t= 0.5 (30 minute s) Distance trave lle d in this phase = S 2:

Proble m 2 The stre ss ve rsus strain curve of a ste e l mate rial is give n by stre ss s= Ee whe re e is the strain and E is the e lastic modulus,[ in the e lastic re gion. ]Le t E =30 units. The e lastic e ne rgy store d [pe r unit volume ]W, is give n by the are a unde r the curve o f stre ss[y va lue s] and strain[x value s].Find the e alstic e ne rgy store d whe n the stain is incre a se d from 0 to 0.5.

Ela stic e ne rgy is give n by the following inte gral:

Proble m 3 The spe cific he at of a subtance is the he at e ne rgy re quire d in calorie s to he at the substa nce by one de gre e ce ntigrade . For most mate rials ,the spe cific he at C [at constant pre ssure ] is not constant but varie s with te mpe rature . For a ne w mate rial, C varie s with te mpe rature as follows: C= 3+0.02T whe re T is te mpe ra ture in de gre e s ce ntigrade . Find the he at e ne rgy re quire d to he at this substance from 25 de g Ce nt to 80 de g ce nt. The he a t e ne rgy [or e nthalpy]H re quire d is found by inte grating spe cific he at C with te mpe ra ture :

Proble m 4

The population of a town incre ase s according to the rate e quation: dp/dt= 75000e xp(0.03t)

Find the growth in population for the ne xt te n ye ars.Note that the curre nt populatio n at t=0 is 75000. dp = 75000e xp(0.03t)dt Inte grating from t=0 to t= 10, we ge t:

Proble m 5 Archimedes showe d that the are a of a parabolic arch is e qual to 2/3 rds of the base xhe ight. We can ge t this re sult using the inte gration of a parabola which is like a dome : Take y= 9 -x^2 The pe ak [ve rte x] occurs at y=9 The base e xte nds from x=-3 to x =3 . The base =

6 The inte gral which give s the are a unde r the parabolic arch=36 units. Using Archime de s' formula: are a=(2/3)(9)(6)=36 units!.

Proble m 6 Hooke 's law state s that force F(x) ne e de d to compre ss a coil spring or e alstic body is proportiona l to the le ngth compre sse d: F(x)= kx If k= 250,find the incre me ntal work to be done while e xte nding the spring from x=3 inche s to x= 6 inche s

Work done = Proble m 7 A force of 15 pounds is re quire d to stre ch a spring in an e xe rcise machine by 6inche s.Wha t is the work re quire d to stre tch the spring to 12 inche s. 180 ft lbs

Ans

Proble m 8 An a ircraft runs for 3600 fe e t on the runway be fore lift off, in 30 se conds.It starts with ze ro ve locity and runs with constant acce le ration.What is the spe e d at the lift-off mome nt? The initia l ve locity V(0)=0 The ve locity V(t) = a t whe re a is the acce le ration and is a consta nt. The dista nce trave lle d is S . V= ds/dt Now we have the e quations to solve this proble m.

Proble m 9 A nurse ry se lls its pine tre e s afte r 6 ye ars of growth.The spe e d of growth is give n by the e qua tion: v=dh/dt=1.5t+5 whe re h is the he ight and t in ye ars.At t=0,initial time of planting,the sapling has a he ight of 12 cms. Find the he ight afte r t ye ars. What would be the he ight a t 6 ye ars?

Proble m 10 The cost of owning a manufacturing plant[purchase +mainte nance costs] for t ye ars is give n by the re lation:

Find the cost upto 10 ye ars.

Proble m 11 The fue l consumption of an automobile is e xpre sse d in y[mile s pe r gallon].This parame te r is not constant ,but varie s with spe e d of the auto. The car manufacture r ga ve this re la tion: dy/ds= -0.012y whe re s is the spe e d in mile s pe r hour,for spe e ds > 50 mph.The value of y is 28mile s pe r gallon at s=50 mile s pe r hour. a) Find a n e quation for y (s) for s>50 mph. b)Use that e xpre ssion to find y for the spe e d at 70 mph.

To find C,use the initial condition: y= 28,whe n s=50

The re fore If S =70, y= 50.8e xp(-0.84)=50.8/2.32=21.9 mile s pe r gallon. [Note : We have actually solve d a simple diffe re ntial e quation.! dy/ds=-0.012s is a diffe re ntial e quation with initial value :y=28 for s= 50] Proble m 12 The Hanging Cable Hype rbolic functions are he lpful in re pre se nting a ha nging cable or chain be twe e n two pole s and also ce rtain arche s like the Gate way Arch in S t Louis, Missouri. An e le ctrical cable is hung from two towe rs with a distance of 200 fe e t be twe e n the m.The shape the cable assume s is calle d the Cate nary-- give n by the e quation:

Find its a rc le ngth.

Proble m 13 Ne wton's Law of Cooling This law is attribute d to Isaac Ne wton.This state s tha t the rate of cooling of an obje ct is proportional to the te mpe rature diffe re nce be twe e n the obje ct and its surrounding. For instance ,if you have a cup of coffe e at 150 de g F,and the room te mpe rature is 60 de g F,the ra te of cooling will be : dT/dt whe re T is the te mp and t the time (say, in minute s):

Note that at time t=o, the initial te mpe rature of your cup of cFind the e quation for va ria tion of te mpe raturoffe e is 150 de gF.Call it T(0)=150 Find the e quation for te mpe rature T as a function of time . Le t us inte grate :

We know that at t=0, T=150 150 - 60=90 =e xp(c)e xp(0)=e xp(C) Now ln90 =C The re fore ln(T-60)= kt +ln 90 ln(T-60) - ln90 = kt

Now to find k: To find k, we ne e d anothe r information; If we know that afte r 10 minute s, Twas 90 de gF 90= 60+90e xp(10K)

e xp(10K) = 30/90=1/3 10k= ln(1/3)=-1.1 k=-0.11 The final e quation be come s: Note tha t we have use d two pie ce s of information to solve this:1 the inital te mpe rature at t=0 a nd 2. te mp at 10 minute s to find k.This is calle d 'an initial value proble m'.If the initial te mp of your coffe e was diffe re nt, the n you will have a diffe re nt numbe r inste ad of 90 be fore the e xpone ntial function. Using this e quation, we can find its te mpe rature at any give n time .S ince k is ne gative , the te mp of your cup de cre ase s e xpone ntially and e xp(kt) te rm will go to ze ro and T be come s 60,the surrounding te mp.That is ,your coffe e cools slowly and re ache s the te mp of the surrounding air. Proble m 14 Wildlife conse rvation The rate of change of population of we ase ls follows the e qua tion: dP/dt=k(650-P) whe re t is time in ye ars. If the initial population whe n ye ar t is take n as 0 was 300 , and afte r two ye ars,the population incre ase d to 500, find the e quation for the population growth. Proce e d a s in the pre vious proble m . Inte gra ting At t=0, P=300 Whe n t=2, S olve for k:

ln(350)=C ln(650-P)-ln350=kt P=650-350e xp(kt) 500 = 650 - 350e xp(2k) k=-0.42 P= 650 - 350e xp(-0.42t)

Find the population figure for the third ye ar.What happe ns whe n t is ve ry large ? Try plotting P ve rsus t. [Ans: For t=3, P=552 ;Whe n t is large , P goe s to 650, the saturation value .]

-----------------------------------------------------------------------------------------------------------{Note : The following two se ctions can be skippe d on first study of this tutorial] Finding the average value of a function This me thod is one of the dire ct applications of inte gration and is e xtre me ly use fu l for scie ntists,e ngine e rs and statisticians. The de finite inte gral give s the are a unde r the curve y=f(x) be twe e n the curve and the x axis and be twe e n the two limits x=a and x=b,the two ve rtical line s. This are a can be take n a s the are a of a re ctangle with width as the inte rval (b-a) in the x axis and he ight as the ave rage value of the function.We re place the are a with a re ctangle of width (b-a).

-------------[Equation 4] Le t us illustrate with simple e xample s. Example 33 The spe e d of sound varie s with altitude .This data is important for aircr aft flights.The spe e d of sound S (h) (me te rs pe r se cond) is mode le d with the e quation for the altitude inte rval 0 to 11.5 km,as follows:

Find the a ve rage value of spe e d of sound-for this altitude range in which most of th e comme rcial flights are done .

Find the inde finite inte gral:

The ave rage value of spe e d of sound for this altitude range [0,11.5]=

Pra ctice Proble ms 1. The e le ctromotive force (voltage ) E in an e le ctric circuit is : E= 3 sin (2t) whe r e E is in volts a nd t in minute s.Find the value of E for the range of t from 0 to 0.5 se conds. Ans: 1.38 volts 2 The profit from a ne w product e arne d by a company varie s as follows: whe re P is in thousands of $ and t in months.Find the ave arge profit in the inte rva l of 1 month to 6 months by inte gration. Ans: 157000$ 3 A life insurance company use s the following e quation for the de ath rate (for 1000 pe rsons)of its custome rs in the age group 20 to 60 ye ars: Find the a ve rage de ath rate for pe ople in the age inte rval of 50 to 60 ye ars. Ans: 7.35 Finding Arc length Ta ke a cuve d line in the X-Y plane .We can find its le ngth,if we know the e quation: y= f(x) for the curve . Le t f'(x) be the de rivative : dy/dx=f'(x) The n the arc le ngth is give n by the de finite inte gral :Le t S be the arc le ngth from x=a to x=b.

Example 34

Find the arc le ngth of the parabola y = x^2

in the inte rval [2,5]

This inte gra l can be found by "Inte gration by parts":

Afte r some manipulation,we ge t: Taking the limits: S (5) -S (2)=21.91units Arc le ngth in parame tric form O fte n it is e asie r to work with parame tric form: x=f(t) and y=g(t) whe re t is the inte rme dia ry variable or the parame te r. The n arc le ngth e quation is:

Example 35 Find the arc le ngth of a circle ,give n that x=a cos t and y=asint for the inte rva l of t [0,2 ], whe re a is the radius.

S = circumfe re nce of a circle --- a cool re sult! Example Find the arc le ngth for x=t^2 , y= 2t in the inte rval: [0,2]

Use the table of inte grals for finding the above inte gral. Taking the limit t=2,find I Pra ctice Proble m 1 A cycloid is the trace of a curve whe n a point on a circle or whe e l rolls on a straight line The pa rame tric e quation for a cycloid is: x= a (t-sint) and y= a(1-cost) whe re a is the ra dius of the circle .Find the arc le ngth for the inte rval[0,2 ] Ans: 8a

Summary and comments 1 We have not include d all the inte gration me thods and formulae .You have to le arn ma ny more me thods to handle complicate d functions.We have not include d in this basic tutorial--- functions of trig-inve rse for instance .But the basic me thods are he re and must give you a good start. 2 The re are many inte gration proble ms which cannot be done by analytic me thods as give n he re .In such case s, we use nume rical me thods.For instance you cannot inte grate the e rror function by the se me thods.You may le arn simple me thods of nume rical inte gra tion in a Calculus course :me thods like re ctangular,midpoint,trape zoidal and simpson rule s/me thods. The re are more e ffe ctive me thods you would like to le arn in a course on nume rical analysis/me thods late r. S ome inte gration re quire s Comple x Analysis or use of Comple x Variable s which is a se pa rate subje ct by itse lf. 3 For complicate d functions, re fe r to a 'Table of Integrals' --you will find it in the appe ndix or inne r cove rs of large r te xt books.You can also che ck your re sults with such a table . 4 Le a rning the te chnique of inte gration is more important than pe rforming complicate d proble ms.The te chnique s are use ful in othe r fie lds too...e spe cially in physics. 5 The a pplications of inte gration are many.We have give n only the simple ,e le me ntary one s.Finding are as and volume s of comple x shape s by inte gration is a fascinating applica tion.You may le arn double and triple inte gration for two or thre e variable s. La stly, do se nd your fe e dback to me by e mail : nksrinivasan at hotmail dot com Suggested Books Whille the re are many e xce lle nt Calculus te xt books, I sugge st the following for additional study or e asy introduction: 1 Forgotte n Calculus by Barbara Le e Blou [Barron's se rie s] 2 Calculus Made Easy by S ilvanus Thompson {re vise d by Martin Gardne r} [This is a classic popular book] 3 How to Ace Calculus by Colin Adams and othe rs .{Wide ly praise d by math te ache rs and stude nts} 4 Te a ch Yourse lf Calculus by P Abbott--[a classic book from UK...may not be e asily availa ble in book store s --try your library.] Ple ase note that I am not e ndorsing any book but only sugge sting that you may look up

the se . -----------------------------------------------------------------------------------------------------------