Ch 5 Cal 1B.pdf - Calculus 1BChapter 5:Techniques of Integration This chapter helps you to learn a few different techniques of evaluating or solving

# Ch 5 Cal 1B.pdf - Calculus 1BChapter 5:Techniques of...

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This preview shows page 1 out of 43 pages. Unformatted text preview: Calculus 1B–Chapter 5:Techniques of Integration This chapter helps you to learn a few different techniques of evaluating or solving integrals. As we have seen on previous chapters (last quarter) integral operator is inverse operator of differential operator, so one should know differentiation of all the basic functions. (Polynomials, exponentials, rational and trigonometric functions) We will review the hyperbolic functions in this chapter. The techniques that we will learn in this chapter are as follow: 1- Basic integral, 2- U-sub, 3- Trigonometric identities, 4- Integrating by parts, 5Trigonometric substitutions, 6- Partial fraction, 7- Completing squares, 8- Roots eliminations, 9- Sine and Cosine eliminations. We will look at the evaluation of integration as sum of smaller religion which presents the idea of integration but we spend minimum time on it since to calculate the area or to evaluate the integral in this method requires some software. Techniques of Integration and its applications Differential operator is a linear operator ((() + ()) = (()) + (())) For example: (2 3 − 4) = 2 ( 3 ) − 4 () = 6 2 − 4 The inverse of Differential operator is called integral operator and it is also a linear operator. The symbol for integral which operate on a function f(x) is ∫ () = (). Note: the integrant [f(x)] is between the integral ∫ .The anti-derivative [F(X)] is noted by capital letter. ∫(() + ()) = ∫ () + ∫ () Let’s look at some functions and their derivative and anti-derivative. ′ () −1 () − () −1 + +1 1 + − + 1 + You must know differentiations of the following functions , , , , , , ℎ , ℎ , ℎ , ℎ , ℎ , −1 , −1 , −1 , −1 , −1 , −1 , , ln(), , √ , …. Practice: a) b) c) d) Make a chart with 3 columns and at least 22 rows. On 2nd column write all the basic functions that you have learned. Write the derivative of the functions on 1st column Figure out anti-derivative of all the function (that are possible) by inspection and write them on the 3rd column So if one knows derivative of any function, then one should know the integral of the derivative of the function with some arbitrary constant. ∫ () = () + (. . ∫ 2 ( + 5) = 2 + ) Differentiating 2 () = 2 + 5 decreases its exponent by one 1 () = 2 and by integrating it increases its exponent by one 2 () = 2 + .The differentiation takes a three dimensional function [ 2 , , 1] to a two dimensional function [, 1] . So the integration takes back to a three dimensional function. Imagine a point in a 3-D space (where its location is known (0 , 0 , 0 )) is compressed to a 2-D plane (let say (0 , 0 , 0)) and then extracted from 2-D to 3-D. Its location can be any points on the line (0 , 0 , )where C could be any constant. Let () ≥ 0 in the interval [a, b], then ∫ () = under the function from a to b. ∫ () = ()/ = () − () it is area from zero to b minus area from zero to a. Practice: 1- Find area for following functions in the interval given a) () = (2) [0, ] b) () = 3 [0, ln 3] 2- Find the area between the two functions a) () = 2 () = 4 b) () = 2 + 12 () = 8 U-Substitution Method You have solved problems like ( + 1)2 + 3( + 1) + 2 = 0 22 − 3 + 1 = 0 and you have used concept of substitution in algebra and trigonometry courses. Imagine that you are in a space that all x variables are x +1 variable then in that space you see the 1st equation like 2 + 3 + 2 = 0 and that can be solved easily X= -1 and X =-2, but in a space that all the x variables are x variables your answer will shift to x = -2 and x = -3. 1 1 Let’s solve this integral ∫0 Step1: Let = + 1 +1 and = . When x = 0 , U = 1 and when x = 1 , U = 2 21 ∫1 Step2: Find the anti-derivative of 1/U. [F(x) = ln U + C] Step3: Evaluate ln U at the upper limit minus its value at lower limit. ln /12 = ln 2 − ln 1 = 0.693 − 0 = 0.693 Before I explain this technique, I want you to take five minutes to give me the differentiation of the following functions. 1) = (()) 6) (()) 11) √() 2) = ln(()) 7) = (()) 12) = ln (√()) 3) = 4) = 8) = (()) 9) = () 13) = −1 (()) 5) = 10) = −1 14) = −1 (()) 15) = ℎ Now take another five minutes to find anti-derivative of the following functions. 1) = 2) = −1 3) = 7) 2 8) = 9) = 10) = 2 1 13) = 1+ 2 14) = 1 5) = 6) = 2 4) = 11) = ℎ 12) = ℎ2 1 15) = √ 2 √(1− 2 ) −1 U-sub technique can do two things either change the integral to a basic integral or change it to another integral that can be solve by another techniques. U-sub is all about ability to see a function if the derivative of the function is given. To solve an integral by U-Sub, all you have to do is to take part of the integral as U and the rest as dU. Then transform the integral to a function of U and integrate in terms of dU. Example ∫ 2 = = 2 Find the following U and dU of these integrals and then solve them ∫ 3 ∫ ln ∫ ln2 ∫ 2 −4 ∫ √2 + 2 Solve the following ∫ 2 ∫ 2 ∫ 1− 2 1 ∫ ln ∫ 3 ∫ ∫ 2 ∫ ℎ ℎ ∫ 3 1 ∫ ln Practice: 1 3 ) ∫ 2 0 +1 1 ) ∫ 0 √ + 1 1 2 ) ∫ 2 0 −1 3 ) ∫ 0 6 ) ∫ sin4 0 We have learnt that integral of derivative of any differentiable function is equal to the function plus some constant. (∫ () = () + ) Also we have seen differentiation of composite functions which chain rule was used to differentiate. ( [(())] = () (()) ∫ ()) S0, [(())] = ∫ (()) () () = ∫ (())() = ∫ (()) = (()) + () Now let’s use U-sub on chain rule to integrate functions which are not basic but composite. 1 ∫ ( 2 ) → 2 = () ℎ () = 2 → 2 () = Now substitute 1 1 back into the original integral ∫ ( 2 ) = 2 ∫ (()) () = 2 ∫ () This basic integral, you can integrate it by the chart that you have made. Practice −1 ∫ ∫ √ 2 √3 0 +1 4 3 √ 2 + 1 ∫ (ln )2 ∫ ∫ √ 2 −1 2 ∫ 0 1 ∫ ( 0 12 + 1) 4 2 ∫ 0 4 ∫ 0 4 2 ∫ 0 4 ∫ 0 Integral in the form ∫ Case I: If m is odd, ∫ −1 → = = Case II: If n is odd, ∫ −1 → = = Case III: If m and n are both even. Use identities to simplify then integral Integral in the form ∫ CaseI:If m is odd ∫ −1 −1 → = = Case II: If n is even, ∫ −2 2 → = = 2 Case III: If m is even and n is odd. Use identities to simplify then integral Practice ∫ 3 ∫ 2 3 ∫ 4 5 ∫ 3 2 ∫ 2 ∫ 2 ∫ 2 2 ∫ 3 ∫ 4 ∫ tan2 2 ∫ tan3 4 ∫ ∫ 3 ∫ tan2 ∫ tan2 3 ∫ tan3 3 Integrations by parts How about integrant which is neither basic nor composite, like integrant which are product of two functions? Let’s go back to product rule as we learned earlier (. . [()()] = [()]() + ()[()]). In another form [] = [] + [] with moving [] to the left side and [] to the right side. ([] = [] − []). And if we integrate from product rule, we get the rule for integration by parts.(∫ [] = [] − ∫ []) These are steps for integrations ∫ ()() → = () = () ℎ = () = () + ∫ ()() = ()() + () − ∫ ()() + ∫ () = ()() − ∫ ()() Example ) ∫ → = ) ∫ → = ) ∫ −1 → = sin−1 ) ∫ → = ) ∫ ln → = ln ) ∫ −1 → = −1 The rank order for choose function for U a) Inverse functions (ln , arcsin , arctan , … ) b) Polynomials ( ) c) Exponential and Sinusoidal have the same rank ( , , ) d) dx Why there is such rank order? Practice ∫( + 1)4 ( + 2)3 ∫ 3 ∫ −1 ∫(ln )2 “All sciences are mathematical physics or stamp collecting” Albert Einstein (14 March 1879–18 April 1955) We have dealt with differential operator acting on position vector in past. Now we are going to explore some interesting facts by using integral operator. A quick review We know that Velocity vector is rate of change of Position vector respect to time and Acceleration vector is rate of change of Velocity vector respect to time. Since Velocity is a vector (magnitude and direction), if any of these two changes the change introduce the acceleration vector. (i.e. Circular motion with constant speed create (classically) an acceleration toward the center of the circle) ⃗ = = ⃗ Classically time has one direction (increase in entropy) Now let’s look at the inverse of differential operator = ∫ 2 1 = ∫ 2 . 1 One can generate basic kinematic equation for constant acceleration. (Do it yourself) Three laws of Newtonian physics are 1) Inertia, 2) Net force, 3) Action and reaction (Classically) Force is defined as rate of change of momentum per time. ∑ = ⃗ = ⃗ (Momentum is product of mass and velocity) = ⃗ = ( ) = + Fdt mdv (For constant mass) this is Impulse equal change in Momentum and then dv we have F m ma dt (Mathematically) Work is defined as the area under the force function of position. = ∫ 2 ° and definition of Energy is ability to do work (Simple definition). 1 There are two types of Energy 1) Potential Energy or stored energy 2) Kinetic Energy or Energy of movement. Both Energy and Work are scalar due to dot product. To find gravitational potential energy: ℎ2 = ∫ ° = ∆ℎ ℎ1 ⃗⃗⃗ = where = To find spring potential energy: ℎ2 1 = ∫ ° = (22 − 12 ) 2 ℎ1 where = Hook’s law To find kinetic Energy: = ∫ ° = ∫ 2 1 ° = ∫ = ∫ = (22 − 12 ) 2 1 1 Where moment of inertia = ∑ 2 and KE For rotation is = 2 (22 − 12 ) Problems 1) Mathematically and graphically show that: a) The PE is the same for moving up an object from origin to h above the origin and from h to 2hfrom origin. b) The PE is different for compressing a horizontal spring from equilibrium to x and from x to 2x from the equilibrium. 2- Find the work done by the following forces from zero to L. () = () = 2 () = () = 3 () = 2 () = 2 () = () = 3 () = − 2 2 3 () = 1+ 2 3- Find the momentum of a particle that creates by such a force from zero to T. 2 () = ( + 1) () = √1 − 2 () = () = ℎ ℎ () = 2 () = ℎ √1 − 2 () = 2 () = 2 () = 1 1 + 2 () = 3 1 + 2 4- A spring is hung vertically with spring constant K=200 N/m and attached weight W=400 N (ignore the mass of spring) With Energy E joules you can compress the spring a distance of 1 m from equilibrium. a) How far can the spring be compressed from the equilibrium if twice the energy is used? b) How far can the spring be compressed from 1m from the equilibrium if twice the energy is used? 5- A bucket (M=10 Kg) contains water (m=40kg) is pulled a height of L=50m. If there is a leak of r=0.1kg/sec from the bucket and the rate of going up is R= 2m/sec, How many Joule energy required to pull up the bucket? Take g= 10 m/s/s. Graphical proof of kinematics equations A particle is at rest means that as the time passes by, the particle’s position would not change. Figure 1 shows that () = 0 . Figure 2 shows that as the time passes by, the particle’s position increases directly proportional to the time. The constant of proportionality is the slope of the line. = per elapse time for short it is called velocity. = ∆ ∆ is rate of change of position Where “dx” is very small change in position and “dt” is very small change in time. The above equation can be written as ∆ = ∆ or − 0 = ( − 0 ) 0 = 0, 0 = 0 then we have = . Figure 1 Figure 2 Figure 3 In Figure 3, Velocity vs. time the velocity is constant the area under the curve is the distance traveled by the particle. Hint: = ()() If the velocity is not constant and it is varying then the rate of change of velocity per elapse time is new quantity which is called acceleration = or = − 0 . Acceleration is the slope of graph of velocity as a function of time. 1 1 Figure 4 shows that the distance traveled is = 2 we have = → = 2 () = 1 2 2 1 Figure 5 shows that the distance traveled is = 2 ( + 0 ) we have 1 1 = + 0 → = [( + 0 ) + 0 ] = 2 + 0 2 2 Figure 4 Figure 5 You have learnt these equations in High school with the condition that acceleration is constant. Let’s look at the kinematic equations when acceleration is changing in time. 2 2 ⃗ () ≤ = ∫ , = ∫ > ≤ (), () >, ℎ 1 1 1 1 1 1 ℎ () ≤ = ∫ 2 ∫ 2 . ′, = ∫ 2 ∫ 2 . ′ > Practice: 1-Given acceleration of space is vector a(t) . Find an expression for velocity and position vector. ⃗ (0) ≤ 10, 5 > ) () ≤ 0. − 10 > 2 ) () ≤ −10. − 10 > 2 )() ≤ (). () > )() ≤ ln( + 1). 2 > )() ≤ . (0) ≤ 0, 0 > ⃗ (0) ≤ 20, 20 > (0) ≤ 0, 100 > ⃗ (0) ≤ −5, 0 > (0) ≤ 0, 0 > 2 2 ⃗ (0) ≤ −10, 10 > 1 > 2 +1 ⃗ (0) ≤ 0, 0 > (0) ≤ 5, −5 > (0) ≤ 0, 0 > 2- Find velocity and position of a particle at given time starting from origin (S =0) with zero speed (v=0) and for given acceleration. ) () = . () ) () = . () = ) () = . ln( + 1) = =2 3-Find potential energy on an object with mass of 10 kg in a gravitational field 1 () = − 2 moving from ) = 10 = 20 ) = 20 = 30 4- Find potential energy of spring with () = 100 − ; ) = 0 = 10 Homework (group work) part I A- Use the limit as → ∞ to find the area under the function 1) = 1 − 2 , 0≤≤1 2)^3 + , 0≤≤1 B- Evaluate the following integrals: 2 ln 1) ∫ 1 2) ∫ 6 0 3 sin2 3) ∫ sin −1 √1 − 2 4 4) ∫ ( 2 + 2 ) 3 2 5) ∫ 6) ∫ 0 √ 2 + 1 0 6 7) ∫ 3 4 8) ∫ 3 3 9) ∫ √ 4 0 10) ∫ 3 √3 √ 2 0 13) ∫ sin−1 16) ∫ 0 11) ∫ +1 1 + 4 2 3 √3 12) ∫ 14) ∫ −1 1 + 2 15) ∫ 3 1 + 2 17) ∫ 1 + 2 18) ∫ −1 +1 C- Find the area enclosed by tangent line to the curve = −1 at x = 2 and the coordinate axis. +1 D- Find the area under the curve = −1 where 2 ≤ ≤ 3 E-Find the limit, graph the function and evaluate the area. 1) lim ln →0 2) Graph () = ln from zero to infinity. 3) Find the area of () = ln from zero to two. 1 F- Show that = 2 2 + + 2∆ = 2 − 2 for a motion with a constant acceleration. G- Find the area between 1) () = 2 () = 4 2) Find the area between () = 2 () = 3 2. = ∫ () ∙ work is a scalar quantity ⃗ = ∫ () Momentum is a vector quantity H- Find the work done by the following forces from zero to L. 1) () = 2) () = 2 4) () = 2 5) () = 7) () = 2 8) () = 3 10) () = 3 1 + 2 3) () = 3 2 6) () = 9) () = − 11) () = − 2 12) () = ln( + 1) I- Find the momentum of a particle that creates by such a force from zero to T. 1) () = ( 2 + 1) 2) () = √1 − 2 3) () = 1 + 2 4) () = ℎ 5) () = 7) () = ℎ ℎ 8) () = 2 √1 − 2 6) () = 2 9) () = 3 1 − 2 Solutions by students Part I B-1) 2 ∫1 B-12) ∫ () 1 = , = u=x du = 1dx ∫ = 2 2 = lnx2 2 /1 2 = ~.693 dv = sin(x)dx v = -cos(x) −() − ∫ −() = −() + () + +1 C) Find the area enclosed by tangent line to the curve = −1 at x = 2 and the coordinate axis. ′ (2) = −2 (2) = 3 Equation of tangent line is = −2 + 7 The area enclosed is a right triangle with base = 3.5 and height =7, the area is 24.5 +1 D) Find the area enclosed by the curve = −1 at from x= 2 to x=3 and the coordinate axis. 3 +1 ∫2 −1 = −1 = ( + 1) = 1 3 2 + 2 − 2 2 − − ∫ 2 2 = − ∫ = 3 2 2 2 + 2 − + 2 2 − − 3 6 − 2 3 / 2 2 G-1): ∫04 2 () − ∫04 () = ( 4 ) − ( − ln(( 4 ))) = 1 − (−.35) = .65 1 1 2 2 G-2): ∫0 2 − ∫0 3 + ∫1 3 − ∫1 2 = 3 3 − 4 4 + 4 4 − 3 A= ~2.34 3 H-7): Find the work done by the following forces from zero to L. ∫0 2 ()() ∙ = ∫ ∙ = 2 2 + = = (), = 2 () ∙ ()2 2 /0 = tan()2 2 − (0)2 2 = ()2 2 H-9) Find the work done by the following forces from zero to L. ∫ () − = − = () = −() = - − () − ∫ − () = () = () = − − = − = − − − − () + − − ∫ − − ()=− − () + − + ∫ () − 1 1 (− − () + − ) + 2 2 I-5) sec()+tan() ∫ sec()dt =∫ () (sec()+tan()) =∫ 2 ()+()() ()+() = () + tan(t) , = 2 () + ()() ∫ = (() + ())/ = -ln1 + (() + ()) = 5-Trigonometric substitutions (() + ()) This technique is due to nature of a right triangle and Pythagorean’s theorem. You have seen relations between trigonometric functions using a right triangle or a unit circle. Base on this relation which converts two terms in to on term (1 − 2 = 2 ) if there are two terms in integrand as 2 − ()2 we use () = as substitution. Example: ∫ 2 √4− 2 the substitution = 2 and = 2 will change the 1 1 integral to ∫ 42 = ∫ 4 2 = − 4 + Now this answer is in domain and we change it back to x domain by using either the identities or a triangle. = √4 − 2 √4 − 2 ℎ = , ℎ ∫ =− + 2 4 2 √4 − 2 Similarly for 1 + 2 = 2 2 + ()2 we use () = as substitution. Example: ∫ √1+ 2the substitution = = 2 will change the integral 2 ∫ = ∫ = ln| − | + to Now this answer is in domain and we change it back to x domain by using either the identities or a triangle. If = then = − = − √1 + 2 So, the solution is ∫ √1 + 2 = ln |√1 + 2 − 1| − + Similarly for 2 = 2 − 1 ()2 − 2 we use () = as substitution. Example: ∫ 2 −9 the substitution = 3 = 3 will change the integral to ∫ 92 1 1 = 9 ∫ = 9 ln| − | + Now this answer is in domain and we change it back to x domain by using either the identities or a triangle. If = 3 ℎ = √ 2 −9 3 1 So the solution is ∫ 2 −9 = 9 ln| + √ 2 − 9| + 1 6- Partial Fractions Method of Partial Fraction helps us to decompose a fraction into simpler fractions. Use techniques of partial fractions which you have learned in previous courses to break down an integrant of an integral into simpler integrant so that the integral can be written as a few simpler integrals and then integrate it term by term. Note that: i) If the degree of () is not lower than the degree of () in () () , use long division to obtain the proper form. (Degree of the polynomial at nominator must be less than the degree of polynomial at denominator) ii) Express () as a product of linear factors + or irreducible quadratic factors 2 + + and collect repeated factors so that () is a product of different factors of the form ( + ) ( 2 + + ) for a nonnegative integer n. iii) Apply the following rules a) For each factor ( + ) with ≥ 1, and with ( + ) irreducible, the partial fraction decomposition contains a sum of partial fractions in the form + + ⋯+ 2 ( + ) ( + ) ( + ) b) For each factor ( 2 + + ) with ≥ 1, and with ( 2 + + ) irreducible, the partial fraction decomposition contains a sum of partial fractions in the form + + + + + ⋯ + ( 2 + + ) ( 2 + + ) ( 2 + + )2 The power of the numerator is always smaller than the power of the denominator. 2 −−21 3+1 −6 3 1 5 ∫ 2 3 − 2 +8−4 = ∫ 2 +4 + ∫ 2+1 = 2 ln( 2 + 4) + 2 −1 (2) − 2 ln|2 + 1| + Practice: ∫ 2 +3−5 2+3 7- Competing the squares We have look at some integrant with quadratic expressions 2 + + , if b =0 then 1 > 0, = 0, > 0 2 = ( ) 1 > 0, = 0, < 0 2 = ( ) 1 < 0, = 0, > 0 2 = ( ) And also in some special cases, one could employ the technique of partial fractions. If b is not zero, in some special cases, partial fraction can be used but in general one can take an advantage of concept of competing squares. Example: ∫ − 2 −10+75 = ∫ 100−( 2 +10+25) = ∫ 100−(+5)2 From here a U-sub and a trigonometric substitution change it to 1 ln | 10 15+ √100−(+5)2 1 10 ∫ = |+ Practice: ∫ √ 2 + 2 8-Eliminating roots As the number of operations on integral increases, there will be more substitution to simplify it. Consider the integral ∫ +1. This integral could be solved easily if √ there were no radical in the integrant. (Like this one∫ ). +1 How does one eliminate the radical? Follow the rules below; a) Find the common multiply for the roots in the integrant b) Let = where n is the common multiplier c) Substitute and simplify, then integrate Example: ∫ 1 1 so = 6 = 65 this transformation change the integral to 65 2 + 3 63 ∫ 3 +2 = ∫ +1 = 6 ∫ (3 +1−1) +1 = [6 ∫ 3 +1 +1 1 − 6 ∫ +1 ] = 6 ∫(2 − + 1) − 6 ∫ +1 This integral can be easily solved and converted back in terms of x. 9-Elimating Sine and Cosine We have worked with Trigonometric identities technique and also when we used Trigonometric substitutions we end up using Trigonometric identities to solve it at the end. Partial fraction deals with Fractions (With polynomials). What about if the integrant is rational expression which involve with sine and cosine? This transformation involves with half and double angle identities. 2 (2) = ℎ = 1+2 1 2 2 (2) , = Or = 2 2 ( ) 2 2 = 1+2 1 2 = 2 (2) (2) = 2 (√1+2 ) (√1+2 ) = 1+2 1 2 1 − 2 = 2 ( ) − 2 ( ) = − = 2 2 1 + 2 1 + 2 1 + 2 1−2 2 2 Example: ∫ + We have = 1+2 , = 1+2 and = 1+2 these substitutions transform the integral into ∫ ∫ 2 ) 1+2 2 1−2 ( ) + 1+2 1+2 ( 2 2 = ∫ 2+1−2 = ∫ 2−(−1)2 2 2√2 = ∫ = ∫ √2 = √2| + | + 2 − 2 2 − 22 − 1 + √2 = √2 | |+ √2 − ( − 1)2 Example: There are some integrals involve with Sine and Cosine which can be solved with easier techniques (Trigonometric Identities). Given the following integrals, ∫ +1 and ∫ √(2)+1 1− 1− ∫ +1 = ∫ 1+ ∙ (1−) = ∫ = − + 2 ∫ √(2)+1 = ∫ √22 = ∫ √2 = = ∫ 2 − ∫ √2 ∫ 2 = √2 | 2 + | + A quick review of Hyperbolic functions To explain these relations one should understand complex variables and geometry of hyperbolic surfaces. Without any mathematical and logical explanations, I give you these relations. sinh = − − cosh = 2 sinh − −...
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