Calculus Tutorial2-- Integrals

8/14/2019 Calculus Tutorial2-- Integrals

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Calculus Tutorial 2 ---- Integral Calculus

Srinivasan Nenmeli-K

IntroductionIntegral Calculus is the reverse process of differentiation or finding the derivative.

Suppose y = f(x)We find the de rivative dy/dx by the methods given in the tutoria l1.

Let dy/dx = g(x) , ano ther function of x.Now we rewrite this as follows:

dy = g (x) dx

The process of integration uses the integral symbol or sign:Let us "integrate" both sides

The result is : y = f(x)We ge t back the o rig inal function.! It is rea lly that simple .The Integral is ,therefo re ,called "Anti-der ivative".

A few terms to learn: The function f(x) is called 'the integral of g(x) dx" .The function g(x) (within the integral sign) is called the integrand.

In general we write:

Here C is called "the constant of integration" . Why we add this? You will learn itsmeaning shortly.

Note that : df(x)/dx= g(x) and dC/dx=0

Therefore the addition of C does not affect the process at all, but serves a purpose.!

Now let us see a few simple examples to become familiar with the process of integration,before we explore the 'physical and graphical interpretation ' of the integral.

Simple Integration Examples

Example 1 Let g(x) = k, a constant say g(x) = 3

Integration is somewhat easy,because we can always check our result by

diffe ren tiating the integra l;we should get back the integrand.DO THIS CHECKING

ALWAYS.

f(x) = 3x+C df(x)/dx = 3 =g(x)This we can understand at once: y= 3x+C is a straight line with slope of 3. The functiong(x) = 3 which is the der ivative of y or dy/dx which is the slope.There is nothing surprising or strange about this.

What does C signify here? well---C is the intercept in this line equation.. g(x) is thederivative and so, we are looking at the slope only.The intercept could be any value.Thisintegra l 3x+C represents all the stra ight lines with the same slope o f 3 or the family of lines or parallel lines with the same slope but with different intercepts.

Let us find the integral : Given a point on the line, x=2 and y=9Then y= 9 = 3x+C= 6 +C --> C=3 --> y= 3x+3

Now you are specifying a particular line with intercept C=3



This is the purpose or meaning of attaching C, the integra tion constant.Go over this example once again to gain an understanding of this essential concept of integration.

Example 2 Let g(x) = x Then

Check:

Let us understand this result. y = g(x) = x is a straight line passing through the origin

with slope =1.The integral represents a family of parabolas.If it is given that the integra l repre sents a parabola pa sing through (2,5) then

y = 5 = 4/2 + C C= 3Now we are referring to a particular parabola with vertex at (0,3)

Try sketching this parabola.!

3 Let Then:

Check:

4. Now we can generalize and find a formula:If

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

Check:Recall that

Therefore:

You have to remember this formula for quick work.This formula works when n is negative ,[except when n = -1] and also when n is afraction .

We need two important results ,as we did for diffe ren tiation:

1. If k is a constant, then

2.[The integration is also a linear operator like differentiation.]

We illustra te this re sult with a few examples:

Example 3 Find

Example 4 Find

=This is called "intergration term by term".

Application Problem 1: The velo city of a car was given by the equation: v= 25 t + 10



where t is time in hours.

Find its position,S , as a function of time.

Velocity = v = ds/dt = 25t +10ds= (25t+10)dt

If the or iginal position o f the car was 5 miles from home, find the equation:When t= 0, S=5=C

[Note: In the equation v=25t+10, the acceleration is a=25, since v(t)=at+v(0)]

Example 5 Find

Recall that

Therefore:

Example 6 FindHere n = -2 Using the formula [1] given ea rlier:

+C

Example 7 Find

Here n= 1/2 n+1 = 3/2

+C

Example 8 Find

Example 9 FindExpand the function (2x+1)^2 and integrate term by term.

Practice Problems

1 Find

2 Find

3 Find



4 Find

5 Find

Physical and graphical interpretation of an Integral

We interpreted a derivative as the slope for a curve at a given point on the curve.The antiderivative or integral of a function is the area under the curve of y= f(x) andthe x axis.

Take a simple function: y= 4x This is a stra ight line pa ssing through the origin.Let ustake the triangular pie ce from x= 0 to x = 2. [The vertices are (0,0)(0,2),(2,8).] Thearea under the curve will be given by area= (1/2) base x height = (1/2) 2 x 8 = 8units.In general terms,

area = (1/2) x (4x) = 2x.xLet us integrate:

You get the same result.Therefore integration is adding up small strips of area each with a width of dx along thex axis and the corresponding y values on the curve. Note that the integral is a parabolaand increases rapidly as x is increased---- the area of the triangle formed increases in

its are a, as x squared.

Since the y values are changing along the curve [except for y=k, a constant) we takeinstantaneous value of y and multiply with a small value of x,that is dx, and add all thesesmall are as.

Trigonometric FunctionsRecall the derivative formulas.It is easy to work out these integrals.

Example 10

Check: d(-cosx)/dx= sin x

Example 9

Check: d(sinx)/dx= cos x

Example 10

Example 11[We shall take up other trig functions later.]

U- Substitution

This is similar to chain rule for differentiation.We substitute a simpler expression for amore complex expression, and fit into the formula g iven ea rlier . This is a power fultechnique....master it well!

Example 12 Find

LetDifferentiating: du = 2x dx

x.dx= du/2

The given integral becomes:



=Check: We can check easily by expanding the integrand: g(x) = x 3̂+x

Note that the constant has changed.That does not make a difffe rence for the integra l.

Example 13 Find

LetDifferentiating , we get:

The integrand becomes:

The given integral =

Example 14 Find

Let

du = 3 dx

Example 15 Find

Let u= 2x du = 2 dx

Example 16 Find

Let u= sin x du = cosx dx

Check: 2sinxcosx= sin(2x)An alternate method:

Example 17 Find

Recall that d(tanx)/dx = sec^2 (x)

Let u=3x du = 3dx



Example 18 Find

Recall the trig identity: and

=

=

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

Example 19 Find I =

Let

Example 20 Find

tan x = sinx/ cos x Let u=cos x du =-sinx

Note that for ln(x) , x canno t be negative ,and so we write

Practice Problems

Find Ans: ln(sinx)+C

Example 21 Find

Let

dx= u du

I=

[Substitute for u and simplify.]

Exponential and Logarithmic Functions

It is easy to integrate an exponential function.



Example 22 Find

+C

Applied Problem

The rate of growth of a fish population in a lake is given by:where t is in days and 100 is the initial population.Find the total fish population at anytime t and find the fish population after 100 days.

+C

If t= 100 days,

Logar ithmic functionYou recall that the loga rithmic function is an inverse function of the exponential fu nction:

If y = exp(x) then lny = xBut integration of lnx is not straightforward.

To find

Take y= xlnx - xThen dy/dx= lnx + x/x -1 = ln x

Example 23 Find

Let u= sinx du = cosx dx

Example 24 FindRecall the identity:

Practice Problems

1 Find the integrals:



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

Ans: ln|exp(x)-1|+C

Integrate term by term

Hyperbolic Functions

The following hyperbolic functions are derived from the exponential function:

Example 1 Find the integra l:

+C

Practice Pro blems:

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 integral calculus to chew..This mehtod is very usefulfor complicated functions.You have to learn this!Let us recall the 'product rule ':

Let us integrate both sides :



This method starts with one integration and results in another integral to tackle.! -- A bitconfusing,right!Hopefully the second integra l on the right side will be easy to find.

Let us see a few examples.

Example 25 Find

Let u=x du=dx dv= sinxdx

Using this method: I=

Example 26 Find

Earlier I gave the integral by showing the answer first and differentiating it.Now we shall

find by this method--Integration by parts.

Let u= ln (x) du = 1/xdv = dx v= x

Example27 Find

Look at this equation. We seem to be in a circular track since we got another integral tofind.Apply "integration by parts" again:

Therefore

[Note:** The next example is a bit complicated. You can study it in the second round of your study.Skip for now.]** Example27 Find



Therefore we find 2I now.

Example 28

Comment: It is important to practise "Integration by parts' well, particularly for APCalculus.

Integration by Partial FractionsExample 29Find

Sp lit the integrand by partial fractions:

Next we find : A=5/7 B=2/7

Example 30

Find:

Sp lit the integrand into partial fractions:



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

[Note:

You can use the 'Table of Integrals' found in the text books for quick reference .]

Practice ProblemsFind the integrals by partial fractions method:1

2

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Definite Integrals

We now move into de finite integrals.

A definte integral has two limits of integration: the upper limit 'b' and the lower limit 'a'and isrepre sented by this sign:

The value of the integral is written as F(x), with upper case letter. [This is justconvention.You can all it g(x) too.]

Let the corresponding indefinite integral which we have been doing in the previousparagraphs be F(x).

Then the definite integral is simply the value of F(x) evaluated at x=b ,then F(x) at x=aand finding the d ifference:

--------------Equation [3]

This result is called a "Fundamental Theorem of Calculus."

Step 1: Find the integral expre ssion for the inde finite integra l first.Step2: Evaluate the integral at the upper limit F(b) and at the lower limit F(a)Step 3: Find the difference: F(b)-F(a)The physical meaning is that this integral is the area below the curve y=f(x) between twovertical line s x=b and x=a and the x axis.

Note that for de finite integra ls we get a number and the constant of integra tion 'C' is nowgone.We get a boundary to the integral and so we dont add that ubiquitous C of earlier

sections!



Let us see a few simple examples:

Example 31

Example 32

Now it is time for some applied problems!

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Applied ProblemsProblem1John starts his car from rest and accelerates for 6 mins.During this phase, the speed is

given by the equation: v=90t where t is in hour s.V is given in miles per hour[mph] .After6mins,John continues to drive at a constant speed of 90 miles per hour fo r the next 24minutes.Find the total distance travelle d in this 30 minutes drive.

Accele ra tion phase: v= ds/dt ds=v dtWe have to integra te 'ds' from time t=0 to t= 0.1 hour [6minutes]

distance travelled in this phase=S1 :

Constant Phase: v=constant=90 ds=90dt

Integrate from t=0.1 (6minutes) to t= 0.5 (30 minutes)

Distance travelled in this phase= S2:

Problem 2

The stre ss versus stra in curve of a stee l materia l is given by stress s= Ee where e isthe strain and E is the elastic modulus,[ in the elastic region. ]Let E =30 units.The elastic ene rgy stored [per un it volume]W, is given by the area unde r the curve o f stress[y values] and strain[x values].Find the ealstic energy stored when the stain isincreased from 0 to 0.5.



Elastic energy is given by the following integra l:

Problem 3The specific hea t of a subtance is the hea t ene rgy required in calories to hea t thesubstance by one degree centigrade. For most materials ,the specific heat C [at constantpressure] is not constant but varies with temperature.For a new materia l, C varies with temperature as follows: C= 3+0.02T where T istemperature in degrees centigrade.Find the hea t ene rgy required to hea t this substance from 25 deg Cent to 80 de g cent.The heat energy [or enthalpy]H required is found by integrating specific heat C withtemperature:

Problem 4The population of a town increases according to the rate equation:

dp/dt= 75000exp(0.03t)

Find the growth in population fo r the next ten years.Note that the current popu lation att=0 is 75000.

dp = 75000exp(0.03t)dtIntegra ting from t=0 to t= 10, we get:

Problem 5 Archimedes showed that the area of a parabolic arch is equal to 2/3 rds of the basexheight.We can get this result using the integration of a para bola which is like a dome:

Take y= 9 -x^2

The peak [vertex] occurs at y=9 The base extends from x=-3 to x =3 . The base =

6

The integral which gives the area under the parabolic arch=36 units.Using Archimedes' formula: are a=(2/3)(9)(6)=36 un its!.



Problem 6Hooke's law states that force F(x) needed to compre ss a coil spring or ea lstic body is

proportiona l to the length compre ssed:F(x)= kx

If k= 250,find the incremental work to be done while extend ing the spring f rom x=3inches to x= 6 inches

Work done =Problem 7

A force of 15 pounds is required to strech a spring in an exercise machine by6inches.What is the work required to stre tch the spring to 12 inches. Ans180 ft lbs

Problem 8An aircra ft runs for 3600 fee t on the runway befo re lift off, in 30 seconds.It star ts

with zero velocity and runs with constant acceleration.What is the speed at the lift-off moment?

The initial velocity V(0)=0 The velo city V(t) = a t where a is the a ccele ration and is aconstant.

The distance travelled is S. V= ds/dt Now we have the equations to solve this problem.

Problem 9

A nursery sells its pine trees after 6 years of growth.The speed of g rowth is given bythe equation: v=dh/dt=1.5t+5 where h is the he ight and t in years.At t=0,initial time o f planting,the sapling has a height of 12 cms. Find the height after t years. What would be

the height at 6 yea rs?

Problem 10

The cost of owning a manufacturing plant[purchase+maintenance costs] for t years isgiven by the relation:

Find the cost upto 10 years.

Problem 11The fue l consumption o f an automobile is expressed in y[miles per ga llon].This

parameter is not constant ,but varies with speed of the auto. The car manufacturer gavethis rela tion: dy/ds= -0.012y where s is the speed in miles per hour,for speeds > 50mph.The value of y is 28miles per ga llon a t s=50 miles per hour.a) Find an e qua tion fo r y (s) for s>50 mph. b)Use that expression to find y for thespeed at 70 mph.



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

ThereforeIf S=70, y= 50.8exp(-0.84)=50.8/2.32=21.9 miles pe r ga llon.[Note: We have actually solved a simple differential equation.! dy/ds=-0.012s is adiffe ren tial equa tion with initial value :y=28 for s= 50]

Problem 12 The Hang ing Cable Hyperbo lic functions are he lpful in repre senting ahanging cable o r chain between two poles and a lso certain arches like the GatewayArch in St Lou is, Missouri.

An electrical cable is hung from two towers with a distance of 200 feet betweenthem.The shape the cable assumes is called the Catenary-- given by the equa tion:

Find its arc length.

Problem 13 Newton 's Law of Coo ling This law is attributed to Isaac Newton.This statesthat the rate o f cooling of a n object is propo rtiona l to the temperature differencebetween the object and its surrounding.

For instance,if you have a cup of coffee at 150 deg F,and the room temperature is 60deg F,the rate of cooling will be:dT/dt where T is the temp and t the time(say, in minutes):

Note that at time t=o, the initial temperature of your cup of cFind the equation forvariation of temperaturoffee is 150 degF.Call it T(0)=150

Find the equation for tempera ture T as a function of time.Let us integrate:

We know that at t=0, T=150150 - 60=90 =exp(c)exp(0)=exp(C)

Now ln90 =CThere fore ln(T-60)= kt +ln 90

ln(T-60) - ln90 = kt

Now to find k:To find k, we need another information;If we know that after 10 minutes, Twas 90 degF90= 60+90exp(10K)



exp(10K) = 30/90=1/310k= ln(1/3)=-1.1

k=-0.11

The final equation becomes:

Note that we have used two pieces of information to solve this:1 the inital temperature att=0 and 2. temp at 10 minutes to find k.This is called 'an initial value prob lem'.If theinitial temp of your coffee was different, then you will have a different number insteadof 90 before the exponential function.Using this equation, we can find its temperature at any given time.Since k is negative,the temp of your cup de creases expone ntially and exp(kt) term will go to zero and T

becomes 60,the surrounding temp.That is ,your coffee cools slowly and reaches the tempof the surrounding air.

Problem 14 Wildlife conservation The rate of change of population of weasels followsthe equation: dP/dt=k(650-P)where t is time in years. If the initial population when year t is taken as 0 was 300 , andafter two years,the population increased to 500, find the equation for the populationgrowth.Proceed as in the previous problem .

IntegratingAt t=0, P=300 ln(350)=C

ln(650-P)-ln350=ktP=650-350exp(kt)

When t=2, 500 = 650 - 350exp(2k)So lve for k: k=-0.42

P= 650 - 350exp(-0.42t)

Find the population figure for the third year.What happens when t is very large? Tryplotting P versus t.

[Ans: For t=3, P=552 ;When t is large, P goes to 650, thesaturation value.]

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{Note: The following two sections can be skipped on first study of this tutorial]Finding the average value of a function

This method is one of the dire ct applications of integration and is extremely useful forscientists,engineers and statisticians.The definite integral gives the area under the curve y=f(x) between the curve and the xaxis and between the two limits x=a and x=b,the two vertical lines. This area can betaken as the a rea of a rectang le with width as the interval (b-a ) in the x axis and heightas the average va lue of the function.We replace the a rea with a re ctang le of width(b-a).

-------------[Equation 4]

Let us illustrate with simple examples.Example 33 The speed of sound varies with altitude .This data is important for aircra ftflights.The speed of sound S(h) (meters per second) is mode led with the equa tion fo rthe a ltitude interval 0 to 11.5 km,as follows:



Find the average va lue of speed of sound-for this altitude range in which most of thecommercial f lights are done.

Find the indefinite integral:

The average value of speed of sound for this altitude range [0,11.5]=

Practice Problems1. The e lectromotive force (voltage ) E in an e lectric circuit is : E= 3 sin (2t) wher e E is involts and t in minutes.Find the value of E for the range of t from 0 to 0.5 seconds.

Ans: 1.38 volts

2 The profit from a new product earned by a company varies as follows:

where P is in thousands of $ and t in months.Find the avea rge pro fit inthe interval of 1 month to 6 months by integration. Ans: 157000$3 A life insurance company uses the following equation for the death rate (for 1000persons)of its customers in the age group 20 to 60 yea rs:

Find the a verage dea th ra te fo r people in the age interval o f 50 to 60 years.Ans: 7.35

Finding Arc lengthTake a cuved line in the X-Y plane.We can find its length,if we know the equation: y=

f(x) for the curve.Let f'(x) be the de rivative: dy/dx=f'(x)Then the arc length is given by the de finite integra l :Let S be the arc length from x=a tox=b.

Example 34 Find the arc length of the parabola y = x^2 in the interval [2,5]

This integral can be found by "Integration by parts":

After some manipulation,we get:

Taking the limits:S(5) -S(2)=21.91units

Arc length in parametric formOften it is easier to work with parametric form: x=f(t) and y=g(t) where t is theintermediary variable or the parameter.Then arc length equation is:

Example 35 Find the a rc length of a circle,given that x=a cos t and y=asint for theinterval of t [0,2 ], where a is the radius.



S= circumference o f a circle--- a cool re sult!

Example Find the arc length for x=t̂ 2 , y= 2t in the in terval: [0,2]

Use the table of integrals for finding the above integral.Taking the limit t=2,find I

Practice Problem1 A cycloid is the trace of a curve when a po int on a circle or whee l rolls on a straight

lineThe parametric equation for a cycloid is: x= a (t-sint) and y= a(1-cost) where a is theradius of the circle.Find the a rc length for the interval[0,2 ]Ans: 8a

Summary and comments

1 We have not included all the integration methods and formulae.You have to learnmany more methods to handle complicated functions.We have not included in this basictutorial--- functions of trig-inverse for instance.But the basic methods are here and mustgive you a good start.2 There a re many integration p rob lems which cannot be done by analytic methods asgiven here.In such cases, we use numerical methods.For instance you cannot integratethe error function

by these methods.You may learn simple methods of numericalintegration in a Calculus course:methods like rectangular,midpoint,trapezoida l andsimpson rules/methods. There are more effe ctive methods you would like to learn in acourse o n numerical ana lysis/methods later.Some integra tion requires Complex Ana lysis or use o f Complex Variab les which is aseparate subject by itself.3 For complicated functions, refer to a 'Table of Integrals' --you will find it in theappendix or inner covers of larger text books.You can also check your results with sucha table.4 Learn ing the technique o f integration is more important than performing complicatedproblems.The techniques are useful in other fields too...especially in physics.5 The app lications of integration are many.We have given only the simple,e lementaryone s.Finding are as and volumes of complex shapes by integration is a fascinatingapplication.You may learn double and triple integration for two or three variables.

Lastly, do send your feedback to me by email : nksrinivasan at hotmail dot com

Suggested BooksWhille there are many excellent Calculus text books, I suggest the following foradditiona l study o r easy introduction:1 Forgo tten Calculus by Barbara LeeBlou [Barron 's ser ies]

2 Calculus Made Easy by S ilvanus Thompson {revised by Martin Gardne r} [This is aclassic popular book]3 How to Ace Calculus by Colin Adams and others .{Widely praised by math teachers andstudents}4 Teach Yourself Calculus by P Abbott--[a classic book from UK...may not be easilyavailable in book stores --try your library.]

Please no te that I am not endorsing a ny book but only suggesting that you may look up



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Calculus Tutorial2-- Integrals