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# Final Exam Study Guide Math 142

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This study guide covers everything for the final. Check out my other notes for the parts on series and parametric equations.
COURSE
Calculus II
PROF.
Vraciu
TYPE
Study Guide
PAGES
21
WORDS
CONCEPTS
Math, Calculus, Calc
KARMA
50 ?

## Popular in Mathematics (M)

This 21 page Study Guide was uploaded by Madeline Lacman on Wednesday April 27, 2016. The Study Guide belongs to Math 142 at University of South Carolina taught by Vraciu in Spring 2016. Since its upload, it has received 44 views. For similar materials see Calculus II in Mathematics (M) at University of South Carolina.

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Date Created: 04/27/16
Math 142 Final Exam Study Guide 2 Types of Integrals 1. Indefinite (antiderivative) f ( )x ∫ = all functions that have derivative equal to f(x) 2. Definite b ∫ f(x)dx = a number that represents area between graph and horizontal a axis, and x=a, x=b Fundamental Theorem of Calculus b b ∫ f(xdx=F(x) =F(b)-F(a) a a Basic Integrals to Memorize ∫ k dx k +c = ∫ e dx e x+c = x ax ∫ a dx = +c *works for any real number n ln(a) except -1 n+1 ∫ x dx = x +c n+1 1/2 ∫ 1 dx= ∫ −1/dx= x =2 √+c √x 1/2 Math 142 Final Exam Study Guide ∫ 1 dx |x| For n = -1, x = ln +c 1 Warning: expression is not necessarily a ln! (we may use u-substitution) sin xdx=−cosx+c ∫ ∫ cos x dx=sin x+c 1 = tan −1x+c ∫ x +a a a 1 =sin−1x+c ∫ √ a −x2 a ∫ a f( )±bg( )dx=a∫ f ( )x±b∫g x( ) 2 2 2 5/2 1/2 x +1dx= x + 1 dx= x dx+ 1 dx= x 3/dx+ x−1/dx= x +x +C ∫ √ x ∫ √x √ x ∫ √ x ∫ √x ∫ ∫ 5/2 1/2 U-Substitution Example : U=9x ∫ sin9 x dx Du=9dx Dx= sinudx=¿ 1∫ sinudu= (−cosu)= cos(9x)+c 1 9 9 9 9du ¿ ∫ PLUG INTO PREVIOUS EQUATION Math 142 Final Exam Study Guide Reduction/Recursive Formula: reduces Integral to one with a smaller exponent *don’t memorize the n−1 cos xdx= cos x∗sin x+n−1 (I) Reduction/Recursive formula, ∫ n n n−2 be able to show how reduction formula is found EXAMPLE: 6 N=6 ∫ cos (x)dx cos5x)∗sin?(x) 5 I6= + (I 4 6 6 3 I = cos x ∗sin (x)+ (I ) 4 4 4 2 cos(x∗sin (x) 1 I2= 2 + 2I 0 I =∫1dx=x+c 0 * If n was oddkeep doing reduction steps until I1 I1= ∫os x( )=sin x ( ) cos(x∗sin (x) x I2= 2 + 2c 8.3 Math 142 Final Exam Study Guide Trigonometry Algebra 2 2 2 2 2 sin x=1−cos x (a+b) =a +2ab+b 2 2 3 3 2 2 3 cos x=1−sin x (a+b) =a +3a b+3ab +b a 2 2 b tan x=sec x−1 ¿ ¿ ¿ sec x=1+tan x2 Differentials u=cos? (x) u=sin? (x) u=tan ?(x) u=sec ?(x) du=−sin (x)dx du=cos (x)dx du=sec (x dx du=sec (x ∗tan?(x) dx Powers of sin and cos: m n ∫ sin ( )cos x ( ) Case 1: “m” is ODD: Strategy use “u” substitution; make u=cos(x) and du=- sin(x)dx m=2 k+1 m=7=2∗3+1 sin 1−cos *method won’t work if m (¿¿2(x)) ∗sin?(x) isn’t odd k (¿ ¿2(x) ) ∗sin(x)=¿ sin2k+x=sin 2kx ∗sin (x=¿ Math 142 Final Exam Study Guide 1−cos k n (¿¿2(x)) ∗sin( )∗cos ( )x ¿ ∫ − (1−u ) ∗u dun = ∫ Do the Algebra Expand k-power and multiply by un , then integrate each Case 2: exponent of power of u cos is odd use u=sin(x) If both exponents are odd, use the smaller exponent m n This method also works when only one of sin (x) or cos (x) is present EXAMPLE: 7 5 5 2 2 2 2 ∫ sin( )∗cos (x)dx  cos x =(cos x) ∗cos x = (−sin (x))∗cos?(x) 7 2 2 = ∫ sin(x)∗(1−sin (x )∗cos (x)dx u=sin?(x) du=cos x dx 8 10 12 7 2 2 7 2 4 7 9 11 u u u ∫ u (1−u ) du= u∫ (1−2u +u du= ∫ −2u +u du= 8 −2 10 + 12 +c Replace u with sin(x) for final answer Case of Even Exponents: use Double Angle Identities 2 1−cos? (2x) Use when sin(x)= 2 exponents are the same 1+cos? (2x) cos2x = 2 Math 142 Final Exam Study Guide si(x∗cos (x= sin(2x) 2 EXAMPLE: ∫sin4(x∗cos4(x)dx= (in(x∗cos x )4dx 4 sin(2 ) sin (2x) 1 4 U=4x ∫( 2 ) dx=∫ 16 = 16∫ sin(2x)dx Du=4dx Dx= sin(2x)=(sin(2x )sin 2(x ) 1−cos?(4x) du 2 4 2 2 (−cos (4x ) = (1−cos(4 x) 2 4 1 ∫ (1−cos ( x)) dx= 1 ∫ 1−2 cos(4 ) +cos(4 ) dx=1 (x−2 sin(4 ))+∫ 1+cos?(8 x)dx 64 64 64 4 2 Eliminating the Square Root: 2 x 2 x x x ∫√ 1+cosxdx= ∫ 2cos ( )dx= ∫ √ 2∗ cos ( )dx= 2√ ∫ cos( )dx= 2√2sin () +c √ 2 √ 2 2 2 Double Angle U= 2 1+cos?(2x) cos (x= x 2 2 2 2cos (x)=1+cos?(2x) Du= dx 2cos2 x =1+cos? (x) 2 (2 Dx= du 4 Powers of Sec and Tan: When the exponent of sec is even, do “u” substitution Math 142 Final Exam Study Guide u=tanx du=sec xdx 2 2 2 Then use trig identity sec x=1+tan x=1+u EXAMPLE: 6 3 4 2 3 sec ∫ sec x∗tan xdx= ∫ec x∗sec x∗tan x dx (¿¿2 x) =(1+tan x)2 ¿ 2 2 3 ∫ (1+u ) ∗u du Expand the square, multiply out, and integrate each power of u Integration by Parts Examples: ln(2 x)dx ∫ u=ln (2x) v=x x 1 1 ¿xln2x )−∫ dx du= 2x = xx x dv=1dx ¿xln2x )−x+c UV− V∫du sec x∗tan x dx ∫ u=tanx du=sec xdx -If the exponent of sec is even, do -If the exponent of sec is odd, try integration by parts to reduce the exponent Products of sin and cos: 1 ∫ si(m )x∗sinn )x dx=2 [os(m−n )x−cos m+n x ] Math 142 Final Exam Study Guide 8.4: Trigonometric Substitutions Typical examples include: 2 2 2 2 2 2 √ x +a ,∨√x −a ,∨ a√−x EXPRESSION SUBSTITUTION TRIG IDENTITY DIFFERENTIAL x +a2 x=a∗tanθ a sec θ dx=a∗sec θ dθ 2 2 2 2 x −a x=a∗secθ a tan θ dx=a∗secθ∗a tθ dθ 2 2 2 2 a −x x=a∗sinθ a cos θ dx=a∗c θ θ For the final answer, you need to use the variable x! EXAMPLE: √x +1= √an θ+1= se√ θ=secθ 1 dx x=tanθ ∫ x2 x +1 √ dx=sec θ dθ 1 2 ∫ 2 ∗sec θ dθ tan θ∗secθ secθ secθ *Simplify: take out of the denominator and ‘lose’ one of the in 2 sec θ 1 ∫ secθ dθ secθ= cosθ tan θ tanθ= sinθ *Rewrite sec and tan in terms of sin cosθ and cos 1 1 2 cosθ cosθ∗cos θ ∫ 2 =∫ 2 dθ sin θ sin θ 2 cos θ Math 142 Final Exam Study Guide 1 cosθ sinθ∗cosθ −1 ¿∫ 2 dθ= ∫ =−cscθ= +c sin θ sinθ sinθ Put in terms of x=tanθ 1 θ=tan x ¿− −1 +c si(tan x) √x +1 2 2 2 1 +x =c x x sinθ= 2 2 √x +1 √ x +1=c θ 1 FINAL 2 ¿− 1 =− √ +1 +C x x 2 √x +1 8.5 Integration of Rational Functions: polynomial x +2x+7 RationalFunction= example: 3 polynomial x +x+2 1 ∫ dx=ln|x±a|+c x±a 1 −1 ∫ 2dx= +c (x±a ) x±a −n+1 1 dx=(x±a) ∫ (x±a) n −n+1 *does not apply to 1 ∫ 2 2dx x −a Math 142 Final Exam Study Guide ∫ 1 dx= tan −1x+c x +a2 a a Other Quadratics: x +x+1=u +a 2 (a+b )=a +2ab+b a=xb= 1 2 2 2 1 1 2 1 3 1 3 x ( )2 x +4=x +x+ +4 4x+ ) +2 4 1 1 ∫ 2 dxu=x+ 2u=dx x +x+1 1 3 ∫ dxusea= u +3 √4 4 ¿ 1 tan1 u +C 3 3 (4 √4 Quadratic Polynomials  reducible (can be factored) 2 2 x −a =(x−a )x+a ) x +a =Notable¿be factoreirreducib)e *Complete the Square for the quadratics that CANNOT be factored 2 -If expressionb −4ac<0 then the polynomial cannot be factored and you should complete the square METHOD: Math 142 Final Exam Study Guide 1. Long Division if necessary (if degree of numerator the degree of denominator)5 2 x x example:x +1 x +1 2. Factor out the denominator into linear and irreducible quadratics 1 1 ∫ 2 dx= ∫ dx x +7x+6 (x+6)(x+1) 3. Set up the simple fraction decomposition A,B= Unknown Coefficients 1 A B 2 = + x +7x+6 x+1 x+6 4. Solve for the Unknown Coefficients A x+6 +B x+1 =1 plug∈−1→5 A=1→ A= 1 ( ) ( ) 5 plug∈−6→−5B=1→B= −1 5 5. Integrate each simple fraction 1 −1 5 5 1 1 1 1 1 1 ∫ x+1+ ∫ x+6 =5 ∫ x+1 − 5∫ x+6 = 5n x+1 − 5n |x+6|+c 8.7: Numerical Integration b ∫a f(xdx=F b −F (a)F(x)=antiderivative Give up on finding a formula for the antiderivative. Instead, give a numerical approximation for the definite integral. Math 142 Final Exam Study Guide The former techniques have limited applicability (only work if function fits into a certain pattern) x2 2 There are functions like ∫ e dx∧ ∫in x dx that CANNOT be expressed as a formula *Numerical Approximations work for ANY function Trapezoidal Rule Step 1: Subdivide interval [a,b] into equal (n) parts Step 2: Draw vertical line through each part until they intersect the graph to make a trapezoid B 1 h∗B 1B 2 h 2 =areaof trapezoid B2 ∆ x T= (y0+2y 12 y 22y +3 )4 2 Math 142 Final Exam Study Guide For n parts: ∆ x T= (y0+2y 12 y 2…2 y n−1+yn) 2 Then, plug x values in to f(x) to find y coordinates **** NOTE THAT SIMPSON’S RULE WILL NOT BE ON THE EXAM SO IT IS NOT INCLUDED IN THIS STUDY GUIDE Error= |actual value−approximated value| M=upper bound for | f (x|ontheinterval[a,b ] 3 upperbounds forerror → | T M(b−a) 12n 2 8.8: Improper Integrals b Like a definite integral ∫ f(x)dx but it represents an area which is a ±∞ unbounded in the horizontal (bound of integration is ) or vertical (function has a vertical asymptote at 1 point inside the interval) direction They are either Convergent (finite value) OR Divergent (infinite value) We find this out by taking a limit ∞ 1 1 1 ∞ 1 ∫ 3dx= ∫ 3dx+ ∫ 3dx 1 x 0 x 1x Math 142 Final Exam Study Guide b lim x2 1 1 b→ ∞ 1 1 3=lim ∫ 3dx= −2 =¿− 2− −2 x b→∞1 x ∞ 2b ∫ ¿ CONVERG 1 1 1 1 1 1 −1 −1 1 ∫ x3=la→0∫ x3dx=la→ 02 x=la→ 0 2 + 2a2)=∞ 0 a consta nt DIVERGEN Goes to ∞ 1 ∫ 3dxisdivergent becauseat least1pieceof itisdivergent 0 x EXAMPLE: ∞ lnx ) b ln(x) ∫ 2 dx=lim ∫ 2 dx 1 x b→ ∞1 x −1 *Use integration by parts! u=ln(x)v=−x du= dxdv= 1 dx x x 2 b −ln?(x) b dx −ln (b) ln(b) 1 lim 1+∫ 2 =lim −b +1 =−lim −lim +1 b→∞[ x 1 x ] b→∞[ b ] b→ ∞ b b→∞b Goes to L’Hopital Rule It’s convergent and equals 0! 1 lim b −0+1 b→∞ 1 = 0 Math 142 Final Exam Study Guide Improper Integrals: How can we tell?! ±∞ 1. If is a bound for the integral 2. There’s an asymptote/division by zero DIRECT COMPARISON TEST: a,∞ Let f (x)≤g (x)on ¿ ∞ ∞ 1. If∫ g x dxconverges ,then ∫ f (x)dxconverges a a ∞ ∞ If f (x)dxdoesn tconverge,then g x dxdoesn tconvergeeither 2. ∫ ∫ a a 10.1: Sequence Definition: The number n factorial is n! N!=1*2*3*………*n 0!=1 Definition: A sequence of numbers is an ordered list { ,a ,a ,a ,… =} { } ∞ 1 2 3 4 nn=1 Indexed by the counting numbers/Natural numbers ∞ lim a =L(Lis anumber) 1. The sequence { n n=1 converges if n→ ∞ n ∞ lim a ≠L 2. The sequence { n n=1 diverges if n→ ∞ n for all numbers L Math 142 Final Exam Study Guide ∞ 3. The sequence { n n=1 diverges to ±∞ if lim an=±∞ n→ ∞ a ∞ b ∞ Let { n n=1 and { nn=1 be convergent sequences lim an=A∧lim b =Bn n→ ∞ n→ ∞ Then we have the following: lim ( nb =n)m a ±lin b =A±Bn 1. Sums/Difference: n→ ∞ n→ ∞ n→ ∞ lim c∗a =clim a =c∗A 2. Constant Multiple: n→ ∞ n n→∞ n lim(anb n)B 3. Product: n→ ∞ a lim n = A for B≠0 4. Quotient: n→ ∞(b) B n The Squeeze/Sandwich Theorem Let cn≤a ≤n n lim cn=lim b nLthenlim a =Ln If n→ ∞ n→∞ n→∞ Let f(x) be a continuous function ∞ { nn=1convergent¿L Math 142 Final Exam Study Guide n→ ∞f ( n)f l(n→∞ =n) (L) Math 142 Final Exam Study Guide 10.6: Alternating Series ∞ n ∞ n+1 ∑ (−1) ∗an ∑ (−1) ∗a n a n0 Alternating Series n=1 n=1 Test- if: lim an=0 1. n→ ∞ 2. an+1≤a n ∞ n Then ∑ −1 )∗an is convergent n=1 An alternating series could be absolutely convergent, conditionally convergent, or divergent ∞ Absolutely convergent: ∑ | n is convergent n=1 Conditionally convergent: not absolutely convergent, but alternating series test can be applied Divergent: usually you’ll see this from the divergence test EXAMPLES: ∞ n+1+n 3+n 23) ∑ −1 ) lim =1≠0 so it’s divergent n=1 5+n n→∞ 5+n ∞ ∞ ∞ ∞ n+1 n n n 1 n 1 19) ∑ −1 ) n +1 ∑ n +1∑ ∑n 3 n n +1 n 2 n=1 n=1 n=1 n=1 an bn p=2>1 So it’s absolutely convergent convergent Math 142 Final Exam Study Guide ∞ ∑ (−1 ) 1 lim 1 =0a n+1a n 1 ≤ 1 ln(n+1)>ln?(n) 10) n=2 ln(n) n→∞ln?(n) ln(n+1) ln( ) *need to Ln is an increasing check if it’s absolutely convergent function so ∞ 1 1 1 1 plugging a larger ∑ is divergent  > and ∑ is divergent so thein would n=2ln(n) ln( ) n n give a larger original sum is divergent too. This means that it is conditionally convergent. Error Estimate for Alternating Series ∞ ∑ (−1 )∗a n=1 n  Approximate by Partial Sums Sn=−a 1a …2 −1 )+a n an+10.0001 solve for n 1 1 2 2 2 < 1000(n+1 )3>1000→ n+1(>99)→n+1> 997→n>√0.4 (+1 )3 So n=31 10.7: Power Series ∞ n ∑ C n(x−a ) x=variable a=fixed number Cn=coefficient n=0 What values of x make the series convergent?  The Interval of Convergence Fact: 1. An interval that has a in the middle 2. Radius of convergence= R: interval of convergence is from a-R to a+R a. (a-R, a+R) 3. Endpoints may or may not be included (need to test on a case by case basis) a. [a-R, a+R) 4. The interval could be reduced to just one point, x=a a. R=0 Math 142 Final Exam Study Guide b. (a-R, a+R] −∞,∞ )R=∞ 5. The interval could be a. [a-R, a+R] Examples: ∞ ∑ x = 1 n=0 1−x  geometric series, r=x and a=1 Convergent precisely when -1<x<1 Interval of Convergence: (-1,1) ∞ 2n (x+1 ) 44) n=0 9n a=-1 power series centered at -1 The odd powers have coefficient 0 x+1 ) 9 2 ¿ x+1 ) ¿ which is geometric r= 9 ∞ ∑ ¿ n=0 2 (x+1) 2 Convergence: −1< 9 <1→−9< x(1 <)→ x+1|<3 | −3<x+1<3→−4<x<2 Interval of Convergence is (-4, 2) MAKE SURE TO TEST ENDPOINTS TO SEE IF T aylor and Maclaurin Series Goal: ∞ n When x is in For a given function f(x) find a power seriesan(x−a )= f (x) n=0 the interval of convergence, Taylor series of f at x=a Maclaurin Series = the Taylor series when a=0 Math 142 Final Exam Study Guide Taylor Polynomial of order n = the beginning of the Taylor Series 2 n a 0a 1(x−a )+a 2(x−a )+…a n(x−a ) We can use Taylor polynomials to be approximate functions To find the Coefficients in the Taylor Series f x )=a +a (x−a +a (x−a )+… 0 1 2 Plug in x = a  f(a) = a 0 ' 2 Take derivative then plug in x = a  f (x)=a 12a 2x−a )+3a 3x−a ) ' Plug in x = a  f a =a 1

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