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Calculus

Learn about various methods and rules for evaluating integrals, including integration by parts, trigonometric integrals, partial fractions, and numerical integration with Simpson's rule. Understand how to handle improper integrals. Examples provided for practice.

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Calculus

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  1. Calculus Chapter 8 Principles of Integral Evaluation By: Rhett Chien Edited: Anna Levina

  2. 8.1 An Overview of Integration methods • ∫dx = x + C • ∫xr dx = ((xr+1)/(r+1)) + C and x cannot = 1 • ∫ex dx = ex + C • ∫r dx = r ∫ dx = rx + C • ∫dx/x = ln|x| + C • ∫sin x dx = -cos x + C • ∫cos x dx = sin x + C • ∫sec2x dx = tan x + C • ∫csc2x dx = -cot x + C • ∫sec x tan x = sec x + C • ∫csc x cot x dx = -csc x + C

  3. 8.2 Integration by Parts • Method of integration based on the product rule for differentiation • ∫f(x)g(x) dx = f(x)G(x) - ∫f’(x)G(x) dx can also be written as: ∫u dv = uv - ∫v du u=f(x) du=f’(x) dx v=G(x) dv=g(x) dx

  4. Integration by Definite Parts b b b • ∫u dv = uv| - ∫v du a a a

  5. ∫x2ex dx u= x2 dv= ex dx du= 2x dx v= ex ∫x2ex dx = x2ex - ∫ex 2x dx u= 2x dv= exdx du= 2dx v= ex ∫x2ex dx = x2ex - 2xex - ∫2 ex dx = x2ex - 2xex - 2 ex + C 2 ∫x2lnx dx 1 u= lnx dv= x2 dx du= dx/x v= (1/3)(x)3 2 2 2 ∫x2lnx dx= (1/3)(x)3lnx| - ∫ ((1/3)(x)3)(1/x)dx 1 1 1 2 ∫x2lnx dx= (8/3)ln2 – (7/9) 1 ∫Calvin dx = Hobbes - ∫Calvin dx 2 ∫Calvin dx = Hobbes ∫Calvin dx = Hobbes/2 Examples~

  6. n is + and odd ∫sinnx dx = ∫sinx*sinn-1x dx Use sin2x=1-cos2x Pythagorean identities n is + and even ∫ sinnx dx Use half angle formulas Sin2x= (1-cos2x)/2 Cos2x= (1+cos2x)/2 m and n are even ∫ sinmx cosnx dx Use half angle formulas m is odd and both + ∫ sinmx cosnx dx = ∫ sinm-1x cosnx sinx dx sin2x=1-cos2x U-substitution 8.3 Trigonometric Integrals

  7. ∫sin4x dx = ∫ ((1-cos2x)/2)2 dx =(1/4) ∫1-2cos2x + cos22x dx =(1/4) ∫1-2cos2x + ((1+cos4x)/2) dx =(1/4) (x – sin2x + (1/2)x + (1/8)sin4x) + C ∫sin5x dx =∫ sin4x*sinx =∫(1- cos2x)2 sinx dx =∫sinx*(1 - 2cos2x + cos4x) dx Let u = cosx du = -sinx dx = -∫1 – 2u2 + u4 du = -cosx + (2/3)cos3x – (1/5)cos5x + C Examples~

  8. 8.4 Trigonometric Substitutions Evaluating integrals containing radicals by making substitutions involving trigonometric functions • x=a*sinθ x=a*tanθ x=a*secθ

  9. Examples~ ∫dx/(x2√x2-4) √x2-4 = 2 tan θ =∫(2sec θ2tan θ) / (2 sec θ)2(2tan θ) d θ =∫(1/4)cos θ d θ =(1/4)sin θ + C sin θ = (√x2-4)/x (1/4) [(√x2-4)/x] + C

  10. 8.5 Integrating Rational Functions by Partial Fractions Decompose a rational function into a sum of simple rational functions that can be integrated Proper Rational function = P(x)/Q(x) Rational Functions = F1(x) + F2(x) + … + Fn(x) P(x)/Q(x)= F1(x) + F2(x) + … + Fn(x) F1(x) + F2(x) + … + Fn(x) are in the forms of A/(ax+b)k or (Ax+B)/(ax2 + bx + c)k Denominators are factors of Q(x) Sum is called the partial fraction decomposition of P(x)/Q(x) and the terms are called partial fractions. Determine partial fraction decomposition: determining the exact form of the decomposition and finding the unknown constants.

  11. 8.5 continue • Linear Factor Rule For each factor of the form (ax+b)m the partial Fraction decomposition contains the following sum of m partial fractions A1/(ax+b) + A2/(ax+b)2 + … + Am/(ax+b)m A1 A2 Am are constants to be determined.

  12. 8.5 continue Quadratic Factor Rule For each factor of the form (ax2+bx+c)m the partial fraction decomposition contains the following sum of m partial fractions (A1x + B1)/(ax2+bx+c) + (A2x + B2)/(ax2+bx+c)2 + … + (Amx + Bm)/(ax2+bx+c)m

  13. Examples~ ∫(4x2 + 13x – 9)/(x3 + 2x2 - 3x) dx = (4x2 + 13x – 9)/(x3 + 2x2 - 3x) = (4x2 + 13x – 9)/((x)(x+3)(x-1) =A/x + B/(x+3) + C/(x-1) 4x2 + 13x – 9 = A(x+3)(x-1) + B(x)(x-1) + C(x)(x+3) Let x= 0 Let x = 1 Let x = -3 A=3 C=2 B= -1 ∫(4x2 + 13x – 9)/(x3 + 2x2 - 3x) dx = ∫3/x dx - ∫1/(x+3) dx +∫2/(x-1) dx =3ln|x| - ln|x+3| + 2ln|x-1| + C

  14. Section 8.7 Numerical Integration; Simpson’s Rule Riemann Sum b n ∫f(x) dx = lim ∑ f(xk*)∆x a n->∞ k=1 Trapezoidal approximation b ∫f(x) dx = ((b-a)/2n)(y0 + 2y1 + … + 2yn-1 + yn) a

  15. 8.7 continue Denoting errors of midpoint and trapezoidal approximations b |Em|=|∫f(x) dx - Mn| Midpoint approximation error a b |Et|=|∫f(x) dx - Tn| Trapezoidal approximation error a

  16. 8.7 continue Theorem: Let f be continuous on [a,b], and let |Em| and |Et| be the absolute errors that result from the midpoint and trapezoidal approximations of b using n subintervals ∫f(x) dx a 1.) If graph is concave up or down on (a,b) |Em| < |Et| 2.) If graph of f is concave down on (a,b), then b Tn < ∫f(x) dx < Mn a 3.) If graph of f is concave up on (a,b), then bMn < ∫f(x) dx < Tn a

  17. 8.7 continue Simpson’s rule – the combination of Trapezoidal and Midpoint approximations (best approximation for area) S2n=(1/3)(2Mn + Tn) =(1/3)((b-a)/2n)[yo + 4y1 + 2y2 + 4y3 + 2y4 + … + y2n]

  18. 8.7 continue

  19. 8.7 continue

  20. Examples~ Use the Simpson Rule to find the area 2 ∫e-x2 dx =??? 0

  21. Section 8.8 Improper Integrals • Let f be a function which is continuous on the closed interval [a,∞). We define If this limit exists and is finite then we say that the integral ∞ ∫ f(x) dx a is convergent; otherwise, we say that the integral is divergent.

  22. 8.8 continue • Let f be a function which is continuous on the closed interval (∞,b]. We define If this limit exists and is finite then we say that the integral b ∫f(x) dx - ∞ is convergent; otherwise, we say that the integral is divergent

  23. 8.8 continue Let f be a function which is continuous for all real numbers. If, for some real number c, both of Are convergent then we define and we say that the integral ∞ ∫ f(x) dx -∞ is convergent; otherwise, we say that the integral is divergent.

  24. Examples~ ∞ ∫dx/(x2+9) = 0 b b Lim ∫dx/(x2+9) = lim (1/3)tan-1(x/3)| b-> ∞ 0 b-> ∞ 0 =.524

  25. 0 ∫dx/(x-1)2 = -∞ 0 0 Lim ∫x-1)-2 dx = lim -1/(x-1)| b-> -∞ b b-> -∞ b =1

  26. QUIZ TIMEEEEEEEE

  27. 1. ∫(x-1)/(x2-2x) dx = a) (1/2)ln|x| + ln |x-2| + C c)ln|x-2| + ln|x| + C b) (1/2)ln|(x-2)/x| + C d)(1/2)ln|(x-2)(x)| + C e) None of these Answer : D 2. ∫(x3)(lnx) dx = a) (x3)(3lnx + 1) + C c) (x4/4)(lnx – 1) + C b) (x4/16)(4lnx – 1) + C d) 3x2(lnx – (1/2)) + C e) None of these Answer : B

  28. 3. .785 ∫cos2x dx= 0 a)1/2 c).643 b).393 d).893 e).143 Answer : C 4. Using the midpoint area with (n = 3) and trapezoidal area (n=6) find the area of the function y=6x-x2 a) Midpoint= 38 Trapezoid=35 c) Midpoint= 54 Trapezoid=60 b) Midpoint= 9 Trapezoid=30 d) Midpoint= 36 Trapezoid=36 e) Midpoint= 17.5 Trapezoid=17.5 Answer: A

  29. 5. Using Simpson’s rule, find the area for the function in the previous question a) 38 c) 28 b) 37 d) 30 e) 112 Answer : B 6. ∫ (ex) (cosx) dx a) (1/2)ex(cosx + cosx) + C c) ex sinx b) (1/2)ex(sinx + cosx) + C d) ex sinx + ex cosx e) (1/2)ex(cosx + sinx) + C Answer = B

  30. 7. ∞ ∫dx/(1+x2) = -∞ a) 3.142 c) 0 b) 1.571 d) .785 e) None of these Answer : A 8. ∫1/(x2√16-x2) dx a) (√16-x2 )/16x c) 16x/(√16-x2 ) b) 16x d) -(√16-x2 )/16x e) None of these Answer : D

  31. 9. ∞ ∫1/x2 dx = 1 a) 0 c) 1.5 b) 2 d) -1 e) None of these Answer : E, real answer is 1 10. ∫ (√4-x2)/x2 dx = a) x/( √4-x2) c) ( √4-x2)/x) b) sin-1(x/2) d) -( √4-x2)/x) – sin-1(x/2) + C e) ( √4-x2)/x) – sin-1(x/2) + C Answer : E

  32. Bibliography • http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/trigintdirectory/TrigInt.html • http://www.math.hmc.edu/calculus/tutorials/trig_substitution/ • http://ltcconline.net/greenl/courses/105/Antiderivatives/NUMINT.HTM • http://archives.math.utk.edu/visual.calculus/4/improper.2/index.html • AP book, barons

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