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Integration

Integration. By : Wonhee Lee Edited by : Jiwoo Lee P.S : I definitely deserve an A for this. Index. 1. An Overview of the Area Problem 2. The Indefinite Integral; Integral Curves and Direction Fields 3. Integration by Substitution 4. Sigma Notation; Area as a Limit

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Integration

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  1. Integration By : Wonhee Lee Edited by : Jiwoo Lee P.S : I definitely deserve an A for this

  2. Index 1. An Overview of the Area Problem 2. The Indefinite Integral; Integral Curves and Direction Fields 3. Integration by Substitution 4. Sigma Notation; Area as a Limit 5. The Definite Integral 6. The Fundamental Theorem of Calculus 7. Rectilinear Motion Revisited; Average Value 8. Evaluating Definite Integrals by Substitution 9. Logarithmic Functions from the Integral Point of View

  3. An Overview of the Area Problem • The Antiderivative Method • Finding the area function by “undoing” a differentiation

  4. The Rectangle Method

  5. The Rectangle Method (Cont.) • Divide the Interval [a,b] into n subintervals • The Area of all the rectangles (Right) = ((b-a)/n)[(f(a+(b-a)/n) + (f(a+2(b-a)/n) + ... + (f(b-(b-a)/n) + f(b)] • As more and more subintervals are created, the calculation becomes increasingly accurate

  6. Example Problem • Use both the rectangle method and the antiderivate method to solve for the area under the curve y=x2 [0,2] (n=4,REA) • (2/4)[f(2/4) + f(4/4) + f(6/4) + f(8/4)] = 15/4 • Y(x) = X3/3 + C • ∴ Y(2)-Y(0) = 8/3

  7. The Indefinite Integral • Function F is the antiderivative of function f on interval I if F’(x) = f(x) • (d/dx)[∫f(x)dx] = f(x) • The integral of f(x) with respect to x is equal to F(x) plus a constant, C. • ∫f(x)dx = F(x) + C

  8. Integration Formulas • ∫cos x dx = sin x + C • ∫sin x dx = -cos x + C • ∫sec2x dx = tan x + C • ∫csc2x dx = -cot x + C • ∫sec x tan x dx = sec x + C • ∫csc x cot x dx = -csc x + C

  9. Integration Formulas (Cont.) • ∫ex dx = ex + C • ∫(1/x) dx = ln |x| + C • ∫(1/(1+x2)) dx = tan-1 x + C • ∫(1/(1-x2)^(1/2)) dx = sin-1 x + C • ∫(1/(x(x2-1)^(1/2))) dx = sec-1 |x| + C

  10. General Integration Theorems • ∫cf(x) dx = c ∫ f(x) dx • ∫[f(x) + g(x)]dx = ∫ f(x) dx + ∫ g(x) dx • ∫[f(x) – g(x)]dx = ∫ f(x) dx - ∫ g(x) dx • Try it out! • ∫5550x4 dx • = 5550∫x4 dx = 1110x5 + C • ∫5x4 + 4x3 + 3x2 + 2x + 1 dx • = 5∫x4 dx + 4∫x3 dx + 3∫x2 dx + 2∫x dx + ∫1 dx • = x5 + x4 + x3 + x2 + x + C

  11. Initial-Value Problem • dy/dx = f(x), y(x0) = y0 => Initial Condition • Ex) dy/dx = sinx , y(0) = 0 • Y = ∫ sinx dx = -cos x + C • 0 = -cos(0) + C , C = 1 • ∴ Y = -cosx + 1

  12. Direction Fields (Slope Fields) • y’ = e-y • Choose a rectangular gird of points • Calculate the slopes of the tangent lines • Draw small portions of the tangent lines

  13. Example Problems • Evaluate the following integrals • (a) ∫x2(1+x4) dx (b) ∫(sinx/cos2x) dx • (a) = ∫x2 + x6 dx = x3/3 + x7/7 + C • (b) = ∫tanxsecx dx = secx + C

  14. Integration by Substitution • (d/dx)[F(g(x))] = F’(g(x))g’(x) • ∫ F’(g(x))g’(x)dx = F(g(x)) + C • ∫ f(g(x))g’(x)dx = F(g(x)) + C • u = g(x), du = g’(x) dx • ∫ f(u)du = F(u) + C

  15. Integration by Substitution(Cont) • Find a function u = g(x) so that the equation ∫ f(g(x))g’(x) dx can be transformed to ∫ f(u) du • Evaluate the intgeral ∫ f(u) du in terms of u • Replace u = g(x) so that the final answer is in terms of x

  16. Some more integration formulas • ∫1/(1+x2) dx = tan-1x + C • ∫1/(1-x2)1/2 dx = sin-1x + C • ∫1/(x(x2-1)1/2) dx = sec-1|x| + C

  17. Example Problems • ∫(x2+10)20 • 2x dx u = x2+ 10 ∴ ∫(u)20 • du = u21/21 + C = ((x2+10)21)/21 + C • ∫sin7x dx u = 7x, du = 7 dx, dx = du/7 ∴ ∫sinu (1/7)du = -(1/7)cosu + C = -(1/7)cos7x + C

  18. Example Problems (Cont) • ∫ dx/(1+4x2) u = 2x, du = 2dx ∴(1/2)∫du/(1+u2) = (1/2)tan-1u + C = (1/2)tan-1(2x) + C • ∫ dy/(4-y2)1/2 = sin-1(y/2) + C

  19. Sigma Notation; Area as a Limit • The summation of the function f(k) from k = a to k = b • All numbers from a to b must be 0 or natural numbers • Also known as the summation notation

  20. Sigma Notation Theorems • A constant factor can be moved through a sigma sign. • Sigma distributes across sums. • Sigma distributes across differences.

  21. Common evaulations n ∑ k = 1 + 2 + … + n = n(n+1)/2 k=1 n ∑ k2 = 12 +22 + … + n2 = n(n+1)(2n+1)/6 k=1 Open form : Closed form

  22. Area under a curve • f must be continuous on [a,b] • f(x) ≥ 0 for all x in [a,b] If f(x) is not always greater than or equal to 0, the equation above becomes the Net Signed Area.

  23. Methods ofapproximation

  24. Example Problems • Write the following expression in sigma notation. 4•1 + 4•2 + 4•3 + … + 4•30 30 = ∑ 4k k=1 • Evaluate the sigma notation. 200 ∑ k = 200(200+1)/2 = 20100 k=1

  25. The Definite Integral b ∫ f(x) dx = a The Definite Integral of f from a to b If f is continuous on [a,b], f is integrable on [a,b]

  26. Duh! a ∫ f(x) dx = 0 a b a -∫ f(x) dx = ∫ f(x) dx a b

  27. Sooooo Obvious b c b ∫ f(x) dx = ∫ f(x) dx + ∫ f(x) dx a a c

  28. Example Problems • Find the area beneath the curve y = sinx from [0,∏] ∏ ∏ ∫ sinx dx = [-cosx] = -cos ∏ + cos 0 = 1 + 1 = 2 0 0 • Evaluate 4 ∫ 1 – (1/3)x2 dx 2 4 = [x-(1/9)x3] = 4-(64/9)-2+(8/9) = -(38/9) 2

  29. The Fundamental Theorem of Calculus (1) • If f is continuous on [a,b] and F is any antiderivative of f on [a,b], then b ∫ f(x) dx = F(b) – F(a) a

  30. The Mean-Value Theorem for Integrals b ∫ f(x) dx = f(x*)(b-a) a There exists at least one number x* in [a,b] such that it satisfies the equation above The two shaded regions are equal in area

  31. The Fundamental Theorem of Calculus (2) x x ∫ f(t) dt = F(x) , (d/dx)[∫ f(t) dt] = f(x) a a If f is continuous on interval I, then f has an antiderivative on I. If a is any number in I, then the function F is an antiderivative of f on I, that is, F’(x) = f(x) for each x in I, or in an alternative notation

  32. 1 ∫ (3x-2)3 dx 0 = F(1) – F(0) = [(3x-2)4/12]01 = -5/4 1 ∫ xex dx -1 = F(1) – F(-1) = [(xex – ex)]-11 = 2/e Example Problems

  33. Rectilinear Motion Revisited; Average Value • s(t) = ∫ v(t)dt • v(t) = ∫ a(t)dt • s(t) = s0 + v0t + (1/2)at2 • v(t) = v0 + at

  34. Distance vs Displacement • Distance : The length of the path that an object takes to get from one point to another • Displacement: The shortest distance possible to get from one point to another

  35. Distance vs Displacement (Cont.) t1 Displacement = ∫ v(t) dt t0 t1 |Distance| = ∫ |v(t)| dt t0

  36. Average Value b fave = (1/(b-a))∫ f(x) dx a t1 vave = (1/(t1-t0))∫ v(t) dt t0

  37. Example Problems • v(t) = -2t2 – t + 10, find both the displacement and the distance traveled between t=0 and t= 4 • ∫v(t)dt = s(t) = -(2/3)t3 – t2/2 + 10t + C • Displacement = s(4)-s(0) = -32/3 • |Distance| = (s(2) - s(0)) – (s(4) – s(2)) = -36 , ∴ Distance = 36

  38. Evaluating Definite Integrals by Substitution • Recall the process from Indefinite Integrals • d/dx[F(g(x))] = F’(g(x))g’(x) • ∫ F’(g(x))g’(x)dx = F(g(x)) + C • ∫ f(g(x))g’(x)dx = F(g(x)) + C • u = g(x), du = g’(x) dx • ∫ f(u)du = F(u) + C b g(b) ∫ f(g(x))g’(x) dx = ∫ f(u) du a g(a)

  39. Example Problems ∏ • Evaluate ∫ sinx cos2x dx 0 • g(x) = u = cosx, du = -sinx dx g(∏) -1 ∫ -u2 du = -[u3/3] = 2/3 g(0) 1

  40. Logarithmic Functions from the Integral Point of View x ln x = ∫ (1/t) dt , x > 0 1 (d/dx)[ln x] = (1/x) (x > 0) Fundamental Theorem of Calculus, Part 2, Remember?

  41. The Value of ln(x) in graphs

  42. Euler’s number • Recall that ∫ex dx = ex + C • ∴ (d/dx)[ex] = ex • lim (1 + x)1/x = e X-> 0

  43. Example Problem • Represent ln 10 on a graph and approximate it. 10 ln 10 = ∫ (1/t) dt ≈ 2.303 1

  44. Bibliography • http://www.sosmath.com/calculus/integ/integ01/integ01.html • http://en.wikipedia.org/ • http://www.sosmath.com/diffeq/slope/slope1.html • http://www.karlscalculus.org/notation.html#summation • http://home.alltel.net/okrebs/page136.html • http://tpub.com/math2/59.htm • http://www.geocities.com/pkving4math2tor7/7_app_of_the_intgrl/7_01_the_mean_val_thrm_for_intgrl.htm • http://hypertextbook.com/physics/mechanics/displacement/ • http://whyslopes.com/freeAccess/natural_logarithms_and_exponenti.html

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