Calculus 2 Cheat Sheet: Quick Review, Cheat Sheet of Calculus

Download Calculus 2 Cheat Sheet: Quick Review and more Cheat Sheet Calculus in PDF only on Docsity! B Veitch Calculus 2 Study Guide This study guide is in no way exhaustive. As stated in class, any type of question from class, quizzes, exams, and homeworks are fair game. 1. Some Algebra Review (a) Logarithm Properties i. y = logb x is equivalent to x = b y. ii. logb b = 1 iii. logb 1 = 0, ln 1 = 0 iv. logb(b r) = r, ln(er) = r v. logb(x r) = r logb(x), ln(x r) = r ln(x) vi. logb(MN) = logb(M) + logb(N) vii. logb ( M N ) = logb(M)− logb(N) viii. The domain of logb(u) is u > 0 ix. All the rules above hold for lnx. Remember that lnx = loge x x. Change of Base loga(M) = lnM ln a = logbM logb a This is useful if you’re asked to find the derivative of y = log5(x+ 1) (b) Exponential Properties i. anam = an+m ii. (an)m = anm iii. (ab)n = anbn iv. an am = an−m = 1 am−n v. a−n = 1 an vi. 1 a−n = an vii. am/n = (am)1/n = (a1/n)m (c) Properties of Radicals i. n √ a = a1/n ii. n √ am = am/n 2. Limits (a) Notation i. General Limit Notation: lim x→a f(x) = L ii. Left Hand Limit: lim x→a− f(x) = L iii. Right Hand Limit: lim x→a+ f(x) = L iv. If lim x→a− f(x) = lim x→a+ f(x) = L then lim x→a f(x) = L 1 B Veitch Calculus 2 Study Guide (b) Limits at ±∞. Assume all polynomials are in descending order. i. lim x→∞ a xr = 0, r > 0 ii. If n > m, then lim x→∞ axm + ... bxn + ... = 0 iii. lim x→∞ axn + ... bxn + ... = a b iv. If m > n, then lim x→∞ axm + ... bxn + ... = ±∞ (c) Evaluation Techniques i. If f(x) is continuous, then lim x→a f(x) = f(a) lim x→3 √ x2 + 4 = √ 32 + 4 = √ 13 ii. Factor and Cancel If you evaluate a rational function and get 0 0 , then try to factor. lim x→3 x2 − 9 x− 3 = lim x→3 (x− 3)(x+ 3) x− 3 = lim x→3 (x+ 3) 1 = 6 iii. Rationalizing Numerators / Denominators Try this technique if you have radicals. Multiply top and bottom by the conjugate. lim x→4 √ x− 4 x− 16 = lim x→4 √ x− 4 x− 16 · √ x+ 4√ x+ 4 = lim x→4 x− 16 (x− 16)( √ x+ 4) = lim x→4 1√ x+ 4 iv. Combine by Using Common Denominators Try this when you need to combine fractions within fractions lim x→a 2 x+4 − 2 a+4 x− a = lim x→a 2 x+4 − 2 a+4 x− a · (x+ 4)(a+ 4) (x+ 4)(a+ 4) = lim x→a 2(a+ 4)− 2(x+ 4) (x− a)(x+ 4)(a+ 4) = lim x→a 2a− 2x (x− a)(x+ 4)(a+ 4) = lim x→a −2(x− a) (x− a)(x+ 4)(a+ 4) = lim x→a −2 (x+ 4)(a+ 4) = −2 (a+ 4)(a+ 4) v. L’Hospital’s Rule and limits can be found later in this guide. (d) Piecewise Functions - Know how to graph and evaluate Piecewise Functions. These are good ones to test your understanding of left-hand, right-hand, and general limits. They are also used to test your understanding of continuity. (e) Definition of Continuity A function f is continuous at x = a if the following three conditions are satisfied: i. f(a) must exist ii. lim x→a f(x) must exist iii. lim x→a f(x) = f(a) 2 B Veitch Calculus 2 Study Guide iv. The absolute max is the largest function value and the absolute min is the smallest function value. (l) L’Hospitals Rule i. If lim x→a f(x) = 0 and lim x→a g(x) = 0, then lim x→a f(x) g(x) = lim x→a f ′(x) g′(x) , g′(x) 6= 0 ii. If lim x→a f(x) = ±∞ and lim x→a g(x) = ±∞, then lim x→a f(x) g(x) = lim x→a f ′(x) g′(x) iii. Indeterminate Form of Type 0 · ∞ Given lim x→a f(x)g(x), you do the following to put it in the form 0 0 or ∞ ∞ A. Rewrite f(x)g(x) as f(x) 1/g(x) B. Rewrite f(x)g(x) as g(x) 1/f(x) iv. Indeterminate Form of 00, ∞0, or 1∞ Given lim x→a f(x)g(x), rewrite the limit as lim x→a eg(x) ln(f(x)) then take the limit of the exponent lim x→a g(x) ln(f(x)) This should put the limit in the Indeterminate Form of Type 0 · ∞ v. Indeterminate Form of ∞−∞ The first thing to try to combine the functions into One Big Fraction. Then try L’Hospitals Rule. Another option would be to factor and turn it into a product of two functions. (m) Inverse Derivative Theorem Let f(x) be a one-to-one function. ( f−1(b) )′ = 1 f ′(f−1(b)) 5 B Veitch Calculus 2 Study Guide 4. Summary of Curve Sketching − Always start by noting the domain of f(x) (a) x and y intercepts i. x-intercepts occur when f(x) = 0 ii. y-intercept occurs when x = 0 (b) Find any vertical, horizontal asymptotes, or slant asymptotes. i. Vertical Asymptote: Find all x-values where lim x→a f(x) = ±∞. Usually when the denomi- nator is 0 and the numerator is not 0. Rational function MUST be reduced. ii. Horizontal Aymptotes: Find lim x→∞ f(x) and lim x→−∞ f(x). There are shortcuts based on the degree of the numerator and denominator. iii. Slant Asymptotes: Occurs when the degree of the numerator is one larger than the denom- inator. You must do long division to determine the asymptotes. (c) Find f ′(x) i. Find the critical values, all x-values where f ′(x) = 0 or when f ′(x) does not exist. ii. Plot the critical values on a number line. iii. Find increasing / decreasing intervals using number line iv. Use The First Derivative Test to find local maximums / minimums (if any exist). Remember to write them as points. A. Local Max at x = c if f ′(x) changes from (+) to (−) at x = c. B. Local Min at x = c if f ′(x) changes from (−) to (+) at x = c. C. Note: If f ′(x) does not change signs, it’s still an important point. It may be a place where the slope is 0, a corner, an asymptote, a vertical tangent line, etc. (d) Find f ′′(x) i. Find all x-values where f ′′(x) = 0 or when f ′′(x) does not exist. ii. Plot these x-values on a number line. iii. Find intervals of concavity using the number line iv. Find points of inflection A. Must be a place where concavity changes B. The point must exist (i.e, can’t be an asymptote, discontinuity) (e) Sketch i. Draw every asymptote ii. Plot all intercepts iii. Plot all critical points (even if they are not relative extrema). They were critical points for a reason. iv. Plot all inflection points. v. Connect points on the graph by using information about increasing/decreasing and its concavity. 6 B Veitch Calculus 2 Study Guide 5. Integrals (a) Definitions i. Antideriative: An antiderivative of f(x) is a function F (x), where F ′(x) = f(x). ii. General Antiderivative: The general antiderivative of f(x) is F (x) +C, where F ′(x) = f(x). Also known as the Indefinite Integral∫ f(x) dx = F (x) + C iii. Definite Integral: ∫ b a f(x) dx = F (b)− F (a) (b) Approximation Integration Techniques Given the integral ∫ b a f(x) dx and n (for Simpson’s Rule n must be even), with ∆x = b− a n and xi = a+ i∆x, then Left-hand ∫ b a f(x) dx = ∆x [f(x0) + f(x1) + f(x2) + f(x3) + ...+ f(xn−1)] Right-hand ∫ b a f(x) dx = ∆x [f(x1) + f(x2) + f(x3) + f(x4) + ...+ f(xn)] Midpoint ∫ b a f(x) dx = ∆x [f(x∗1) + f(x ∗ 2) + f(x ∗ 3) + f(x ∗ 4) + ...+ f(x ∗ n)] , where x ∗ i is midpoint in [xi−1, xi] Trapezoid ∫ b a f(x) dx = ∆x 2 [f(x0) + 2f(x1) + 2f(x2) + 2f(x3) + ...+ 2f(xn−2) + 2f(xn−1) + f(xn)] Simpson’s ∫ b a f(x) dx = ∆x 3 [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + ...+ 2f(xn−2) + 4f(xn−1) + f(xn)] (c) Common Integrals i. ∫ k dx = kx+ C ii. ∫ xn dx = xn+1 n+ 1 + C, n 6= −1 iii. ∫ 1 x dx = ln |x|+ C iv. ∫ 1 kx+ b dx = 1 k ln |kx+ b|+ C v. ∫ ex dx = ex + C vi. ∫ ekx dx = 1 k ekx + C vii. ∫ ekx+b dx = 1 k ekx+b + C viii. ∫ ax dx = ax ln a + C ix. ∫ akx dx = 1 k ln a akx + C x. ∫ akx+b dx = 1 k ln a akx+b + C xi. ∫ lnx dx = x lnx− x+ C xii. ∫ cosx dx = sinx+ C xiii. ∫ sinx dx = − cosx+ C xiv. ∫ sec2 x dx = tanx+ C xv. ∫ csc2 x dx = − cotx+ C xvi. ∫ secx tanx dx = secx+ C xvii. ∫ cscx cotx dx = − cscx+ C xviii. ∫ tanx dx = ln | secx|+ C xix. ∫ cotx dx = ln | sinx|+ C xx. ∫ cscx dx = ln | cscx− cotx|+ C 7 B Veitch Calculus 2 Study Guide SA = ∫ b a 2πy √ 1 + ( dy dx )2 dx SA = ∫ d c 2πy √ 1 + ( dx dy )2 dy The first integral is in terms of x. That means the y in front of the square root should be in terms of x. For example, y = x2 + 2. You use x2 + 2 instead of y. (b) Rotated around the y-axis. The formula doesn’t change much. SA = ∫ b a 2πx √ 1 + ( dy dx )2 dx SA = ∫ d c 2πx √ 1 + ( dx dy )2 dy The second integral is in terms of y. That means the x in front of the square root should be in terms of y. For example, if y = x2, solve for x (x = √ y) and use √ y instead of x. 9. Integration Techniques (a) u Substitution Given ∫ b a f(g(x))g′(x) dx, i. Let u = g(x) ii. Then du = g′(x) dx iii. If there are bounds, you must change them using u = g(b) and u = g(a)∫ b a f(g(x))g′(x) dx = ∫ g(b) g(a) f(u) du (b) Integration By Parts ∫ u dv = uv − ∫ v du Example: ∫ x2e−x dx u = x2 dv = e−x dx du = 2x dx v = −e−x∫ x2e−x dx = −x2e−x − ∫ −2xe−x dx You may have to do integration by parts more than once. When trying to figure out what to choose for u, you can follow this guide: LIATE 10 B Veitch Calculus 2 Study Guide L Logs I Inverse Trig Functions A Algebraic (radicals, rational functions, polynomials) T Trig Functions (sinx, cosx) E Exponential Functions (c) Products of Trig Functions i. ∫ sinn x cosm x dx A. m is odd (power of cosx is odd). Fac- tor out one cosx and place it in front of dx. Rewrite all remaining cosx as sinx by using cos2 x = 1−sin2 x. Then let u = sinx and du = cosx dx B. n is odd (power of sinx is odd). Fac- tor out one sinx and place it in front of dx. Rewrite all remaining sinx as cosx by using sin2 x = 1−cos2 x. Then let u = cosx and du = − sinx dx C. If n and m are both odd, you can choose either of the previous methods. D. If n and m are even, use the following trig identities sin2 x = 1 2 (1− cos(2x)) cos2 x = 1 2 (1 + cos(2x)) sin(2x) = 2 cosx sinx ii. ∫ tann x secm x dx A. m is even (power of secx is even). Fac- tor out one sec2 x and place it in front of dx. Rewrite all remaining secx as tanx by using sec2 x = 1 + tan2 x. Then let u = tanx and du = sec2 x dx B. n is odd (power of tanx is odd). Fac- tor out one secx tanx and place it in front of dx. Rewrite all remain- ing tanx as secx by using tan2 x = sec2 x − 1. Then let u = secx and du = secx tanx dx C. If n odd and m is even, you can use either of the previous methods. D. If n is even and m is odd, the previous methods will not work. You can try to simplify or rewrite the integrals. You may try other methods. (d) Partial Fractions Use this method when you are integrating ∫ p(x) q(x) dx, the degree of p(x) must be smaller than the degree of q(x). Factor the denominator q(x) into a product of linear and quadratic factors. There are four scenarios. Unique Linear Factors: If your denominator has unique linear factors x (2x− 1)(x− 3) = A 2x− 1 + B x− 3 To solve for A and B, multiply through by the common denominator to get x = A(x− 3) +B(2x− 1) You can find A by plugging in x = 1 2 . Find B by plugging in x = 3. 11 B Veitch Calculus 2 Study Guide Repeated Linear Factors: Every power of the linear factor gets its own fraction (up to the highest power). x (x− 3)(3x+ 4)3 = A x− 3 + B 3x+ 4 + C (3x+ 4)2 + D (3x+ 4)3 Unique Quadratic Factor: 3x− 1 (x+ 4)(x2 + 9) = A x+ 4 + Bx+ C x2 + 9 To solve or A, B, and C, multiply through by the common denominator to get 3x− 1 = A(x2 + 9) + (Bx+ C)(x+ 4) Repeated Quadratic Factor: Every power of the quadratic factor gets its own fraction (up to the highest power). 2x (3x+ 4)(x2 + 9)3 = A 3x+ 4 + Bx+ C x2 + 9 + Dx+ E (x2 + 9)2 + Fx+G (x2 + 9)3 (e) Trig Substitution If you see Substitute Uses the following Identity √ a2 − x2 x = a sin(θ) 1− sin2(θ) = cos2(θ) √ a2 + x2 x = a tan(θ) 1 + tan2(θ) = sec2(θ) √ x2 − a2 x = a sec(θ) sec2(θ)− 1 = tan2(θ) Example: ∫ x3√ 16− x2 dx Let x = 4 sin θ, dx = 4 cos θ dθ. So x3 = 64 sin3 θ ∫ x3√ 16− x2 dx = ∫ 64 sin3 θ√ 16− 16 sin2 θ · 4 cos θ dθ = ∫ 64 sin3 θ · 4 cos θdθ 4 cos θ = ∫ sin3 θ dθ Now you have an integral containing powers of trig functions. You can refer to that method to solve the rest of this integral. ∫ sin3 dθ = cos3 θ 3 − cos θ + C To get back to x, we need to use a right triangle with the original substitution x = 4 sinx. If cos θ = √ 16− x2 4 , then ∫ x3√ 16− x2 dx = (√ 16−x2 4 )3 3 − √ 16− x2 4 + C 12 B Veitch Calculus 2 Study Guide 13. Estimating a Series (a) Remainder Estimate for Integral Test: If ∑ n=1 an is convergent, f(n) is continuous, positive, and decreasing, then RN ≤ ∫ ∞ N f(n) dn If you’re asked to estimate ∑ n=1 an to 4 decimals (or have an error less than 0.0001), then solve integrate and solve for N . ∫ ∞ N f(n) dn < 0.0001 (b) Alternating Series Estimation Theorem: Given a convergent alternating series ∑ n=1 an = ∑ n=1 (−1)nbn, then |RN | ≤ bN+1 14. Power Series (a) A power series centered at 0∑ n=0 cnx n = c0 + c1x+ c2x 2 + c3x 3 + ... (b) A power series centered at a∑ n=0 cn(x−a)n = c0+c1(x−a)+c2(x−a)2+c3(x−a)3+... (c) Radius and Interval of Convergence: Perform the Ratio Test or Root Test (usu- ally Ratio Test) For a given a power series ∞∑ n=1 cn(x− a)n, there are only three possibil- ities: i. The series converges only when x = a. The interval of convergence is {a} and the radius of convergence is R = 0. This happens when the Ratio Test gives L > 1. ii. The series converges for all x. The in- terval of convergence is (−∞,∞) and the radius of convergence is R = ∞. This happens when the Ratio Test gives L = 0. iii. The series converges on an interval (a− R, a+R). This happens when the Ratio Test gives K|x − a|, where you need to solve K|x− a| < 1. 15. Writing Functions as a Power Series Given f(x) = ∞∑ n=0 cn(x− a)n = c0 + c1(x− a) + c2(x− a)2 + c3(x− a)3 + ... with a radius of conver- gence R > 0, then f ′(x) = ∞∑ n=1 ncn(x− a)n−1 = 1c1 + 2c2(x− a) + 3c3(x− a)2 + 4c4(x− a)3 ∫ f(x) dx = ∞∑ n=0 cn(x− a)n+1 n+ 1 = c0(x− a) + 1 2 c1(x− a)2 + 1 3 c2(x− a)3 + 1 4 c3(x− a)4 + ... You use this technique if you’re asked to find a power series of a function like 1 (1 + x)2 , ln(1 + x), tan−1 x, etc. Your goal is to differentiate or integrate your function f(x) until it’s in the proper form A 1− u 15 B Veitch Calculus 2 Study Guide A and u can be anything. Once it’s in the proper form, you can write it as a power series. A 1− u = ∑ n=0 A · (u)n (a) If you had to integrate to get the proper form, then differentiate the power series to get back to the original f(x). (b) If you differentiated to get the proper form, then integrate the power series to get back to the original f(x). 16. Taylor Series / Macluarin Series (a) Taylor Series is centered at a f(x) = ∑ n=0 cn(x−a)n = c0+c1(x−a)+c2(x−a)2+... with cn = fn(a) n! (b) Maclaurin Series is centered at 0 f(x) = ∑ n=0 cn(x) n = c0 + c1x+ c2x 2 + ... with cn = fn(a) n! It may helfpul to determine the coefficients c0, c1, c2, ... by using the table n fn(x) fn(a) cn = fn(a) n! 0 f(x) f(a) c0 = f(a) 0! 1 f ′(x) f ′(a) c1 = f ′(a) 1! 2 f ′′(x) f ′′(a) c2 = f ′′(a) 2! 3 f ′′′(x) f ′′′(a) c3 = f ′′′(a) 3! (a) If you need to find the Taylor / Maclaurin Series, find a pattern for cn. Then you write the following series with the appropriate cn f(x) = ∑ n=0 cn(x− a)n (b) If you need to find the n-th degree Taylor / Maclaurin Polynomial, then just find the coefficients you need. For example, a 3rd degree Taylor Polynomial requires the coefficients up to c3. f(x) = c0 + c1(x− a) + c2(x− a)2 + c3(x− a)3 16