1 / 56

Chapter 6

Chapter 6. Integration. Section 6.1. Antiderivatives and Indefinite Integration. Definition of an Antiderivative. A function F is an antiderivative of f on an interval I if F’(x) = f (x) for all x in I. Theorem – Representation of Antiderivtives.

angeni
Download Presentation

Chapter 6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6 Integration

  2. Section 6.1 Antiderivatives and Indefinite Integration

  3. Definition of an Antiderivative A function F is an antiderivative of f on an interval I if F’(x) = f(x) for all x in I.

  4. Theorem – Representation of Antiderivtives If F is an antiderivative of f on an interval I, then G is an antiderivative of f on the interval I if and only if G is on the form: G(x) = F(x) + C, for all x in I where C is a constant

  5. Terminology C - is called the constant of integration G(x) = x2 + c is the general solution of the differential equation G’(x) = 2x Integration is the “inverse” of Differentiation

  6. Differential Equation Is an equation that involves x, y and derivatives of y. EXAMPLES y’ = 3x and y’ = x2 + 1

  7. Notation for Antiderivatives When solving a differential equation of the form dy/dx = f(x) you can write dy = f(x)dx and is called antidifferentiation and is denoted by an integral sign ∫

  8. Integral Notation y = ∫ f(x)dx = F(x) + C f(x) – integrand dx – variable of integration C – constant of integration

  9. Basic Integration Rules y = ∫ F’(x)dx = F(x) + C f(x) – integrand dx – variable of integration C – constant of integration

  10. Basic Integration Rules Differentiation formula Integration formula • d/dx [c] = 0 • d/dx [kx] = k • d/dx [kf(x)] = kf’(x) • d/dx [f(x) ± g(x)] = f’(x) ± g’(x) • d/dx [xn] = nxn-1 • ∫ 0 dx = c • ∫ kdx = kx + c • ∫ kf(x)dx = k ∫ f(x) dx • ∫ f(x) ± g(x)]dx = ∫ f(x) dx ± ∫ g(x) dx • ∫ xndx = (xn+1)/(n+1) + c, n≠ - 1 • (Power Rule)

  11. Examples • ∫ 3x dx • ∫ 1/x3dx • ∫ (x + 2) dx • ∫ (x + 1)/√x dx

  12. Finding a Particular Solution EXAMPLE F’(x) = 1/x2, x > 0 and find the particular solution that satisfies the initial condition F(1) = 0

  13. Finding a Particular Solution • Find the general solution by integrating, - 1/x+ c. x > 0 • Use initial condition F(1) = 0 and solve for c, F(1) = -1/1 + c, so c = 1 • Write the particular solution F(x) = - 1/x + 1, x > 0

  14. Solving a Vertical Motion Problem A ball is thrown upward with an initial velocity of 64 ft/sec from an initial height of 80 ft. • Find the position function giving the height s as a function of the time t • When does the ball hit the ground?

  15. Solution • Let t = 0 represent the initial time; s(0) = 80 and s’(0) = 64 • Use -32 ft/sec as the acceleration due to gravity, then s”(t) = - 32 • ∫ s”(t) dt = ∫ -32 dt = -32 t + c • s’(0) =64 = -32(0) = c, so c = 64 • s(t) = ∫ s’(t) = ∫ (-32t + 64) dt = -16t2 + 64t + C • s(0) = 80 = -16(02) + 64(0) + C, hence C = 80 • s(t) = -16t2 + 64t + 80 = 0, solve and t = 5

  16. Section 6.2 Area

  17. Sigma Notation The sum of n terms a1, a2, a3…,anis written as n ∑ ai= a1 + a2 + a3+ …+ an i = 1 Where i the index of summation, a1 is the ith term of the sum, and the upper and lower bounds of summation are n and 1, respectively.

  18. EXAMPLES 6 ∑ i= 1 + 2 + 3 + 4 + 5 + 6 i = 1 5 ∑ (i + 1)= 1 + 2 + 3 + 4 + 5 + 6 i = 0 7 ∑ j2 = 9 + 16 + 25 + 36 + 49 j = 3

  19. SUMMATION FORMULAS n • ∑ c = cn i = 1 n • ∑ i = n(n + 1)/2 i = 1 n • ∑ i2 = n(n + 1)(n + 2)(2n + 1)/6 i= 1 n • ∑ i3 = n2(n + 1)2/4 i = 1

  20. PROPERTIES OF SUMMATION nn • ∑ kai= k ∑ ai i = 1 i = 1 nnn • ∑ (ai± bi) = ∑ ai± ∑ bi i = 1 i = 1 i = 1

  21. EXAMPLE Find the sum for n = 10 and n = 100 n • ∑ ( i + 1)/n2 i = 1

  22. AREA OF A PLANE REGION Find the area of the region lying between the graph of f(x) = - x2 + 5 and the x-axis between x = 0 and x = 2 using five rectangles to find an approximation of the area. You should use both inscribed rectangles and circumscribed rectangles. In doing so you will be able to find a lower and upper sum.

  23. Limits of the Lower and Upper Sums Let f be continuous and nonnegative on the interval [a,b]. The limits as n→∞ of both the lower and upper sums exist and are equal to each other. That is, n lim s(n) = lim ∑ f(mi)x and n→∞ n→∞ i = 1

  24. Limits of the Lower and Upper Sums n lim s(n) = lim ∑ f(Mi)x n→∞ n→∞ i = 1 n lim S(n) = lim ∑ f(Mi)x n→∞ n→∞ i = 1

  25. Definition of the Area of a Region in the Plane Let f be continuous and nonnegative on the interval [a,b]. The area of the region bounded by the graph of f, the x-axis, and the vertical lines x= a and x = b is n Area = lim ∑ f(ci)x, xi -1  ci xi n→∞ i = 1 Where x = (b-a)/n

  26. EXAMPLE Fine the area of the region bounded by the graph of f(x) =x3, the x-axis, and the vertical lines x=0 and x =1. • Partition the interval [0,1] into n subintervals each of width 1/n = x • Simplify using the formula below and A = 1/4 n Area = lim ∑ f(ci)x, xi -1  ci xi n→∞ i = 1 Where x = (b-a)/n

  27. Section 6.3 Riemann Sums and Definite Integrals

  28. Definition of Riemann Sum Let f be defined on the closed interval [a, b] and let  be a partition of [a,b] given by a = xo < x1 < x2 < …<xn-1<xn =b Where xi is the width of the ith subinterval. If ci is any point in the ith subinterval, then the sum n f(ci) xi, xi-1  ci  xi is called a Riemann i = 1 sum of f for the partition 

  29. Definition of the Norm The width of the largest subinterval of a partition  is the norm of the partition and is denoted by  . If every subinterval is of equal width, the partition is regular and the norm is denoted by   =  x = (b – a)/n

  30. Definite Integrals If f is defined on the closed interval [a, b] and the limit n lim∑ f(ci)x    → 0i = 1 exists, then f is integrable on [a,b] and the limit is b ∫ f(x) dx a The number a is the lower limit of integration, and the number b is the upper limit of integration

  31. Continuity Implies Integrability If a function f is continuous on the closed interval [a,b], then f is integrable on ]a,b]

  32. Evaluating a Definite Integral Evaluate the definite integral 1 ∫ 2xdx -2

  33. The Definite Integral as the Area of a Region If f is continuous and nonnegative closed interval [a, b] the the area of the region bounded by the graph of f, the x-axis, and the vertical lines x =a and x = b is given by b ∫ f(x) dx a

  34. Examples Evaluate the Definite Integral 0 ∫ (x + 2)dx 3 Sketch the region and use formula for trapezoid

  35. Additive Interval Property If f is integrable on the three closed intervals determined by a, b and c, then , b c b ∫ f(x) dx = ∫ f(x) dx= ∫ f(x) dx a a c

  36. Properties of Definite Integrals If f and g are integrable on [a,b] and k is a constant, then the functions of kf and f ± g are integrable on [a,b], and b b 1. ∫kfdx = k ∫f(x) dx a a

  37. Properties of Definite Integrals If f and g are integrable on [a,b] and k is a constant, then the functions of kf and f ± g are integrable on [a,b], and b bb 1. ∫[f(x) ± g(x)] dx = ∫f(x) dx± ∫f(x) dx a aa

  38. Section 6.4 THE FUNDAMENTAL THEOREM OF CALCULUS

  39. The Fundamental Theorem of Calculus If a function f is continuous on the closed interval [a,b] and F is an antiderivative of f on the interval [a,b], then b ∫f(x) dx = F(b) – F(a) a

  40. Using the Fundamental Theorem of Calculus • Find the antiderivative of f if possible • Evaluate the definite integral Example: ∫ x3dx on the interval [1,3]

  41. Using the Fundamental Theorem of Calculus to Find Area Find the area of the region bounded by the graph of y = 2x2 – 3x +2, the x-axis, and the vertical lines x=0 and x= 2. • Graph • Find the antiderivative • Evaluate on your interval

  42. Mean Value Theorem for Integrals If f is continuous on the closed interval [a,b], then there exists a number c in the closed interval [a,b] such that ∫ f(x)dx = f(c)(b-a)

  43. Average Value of a Function on an Interval If f is integrable on the closed interval [a,b], then the average value of f on the interval is b 1/(b-a) ∫f(x)dx a

  44. The Second Fundamental Theorem of Calculus If f is continuous on an open interval I containing a, then, for every x in the interval x d/dx[∫f(t) dt] = f(x) a

  45. Section 6.5 INTEGRATION BY SUBSTITUTION

  46. Antidifferentiation of a Composite Function Let g be a function whose range is an interval I, and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then ∫ f(g(x))g’(x)dx = F(g(x)) + C If u = g(x), then du = g’(x)dx and ∫ f(u)du = F(u) + C

  47. Change in Variable You completely rewrite the integral in terms of u and du. This is useful technique for complicated intergrands. ∫ f(g(x))g’(x)dx = ∫ f(u) du = F(u)+ C

  48. Example Find ∫ (2x – 1).5dx Let u = 2x - 1, then du/dx = 2dx/dx Solve for dx and substitute back to obtain the antiderivative. Check your answer.

  49. Power Rule for Integration If g is a differentiable function of x, then, ∫ ((g(x))ng’(x)dx = ∫ (g(x))n+1/(n+1) + C

  50. Change of Variables for Definite Integrals If the function u = g(x) has a continuous derivative on the closed interval [a,b] and f is continuous on the range of g, then, b g(b) ∫ (g(x)g’(x)dx = ∫ f(u)du a g(a)

More Related