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Calculus

Calculus. Chapter 8 Principles of Integral Evaluation By: Rhett Chien Edited: Anna Levina. 8.1 An Overview of Integration methods. ∫dx = x + C ∫x r dx = ((x r+1 )/(r+1)) + C and x cannot = 1 ∫e x dx = e x + C ∫r dx = r ∫ dx = rx + C ∫dx/x = ln|x| + C ∫sin x dx = -cos x + C

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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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