Fall 204: the Area Problem Example Calculate the area under the graph f (x) =x 2 over the interval 0 apple x apple.
Velocity Suppose that a particle travels a distance d(t) at time t. Then Average velocity = distance traveled time elapsed = d t d Instantaneous velocity =lim h!0 h =lim h!0 d(t + h) h d(t)
Limits AnumberL is a limit of a sequence of numbers a, a 2,...,a n,... if the numbers a n approach the number L as n!.inthiscase,wewrite: lim a n = L. n! Example lim n! n =0 lim n! 2 n =0. Example Let f (x) =x 2.Then,sincef (x) is continuous, lim f (x) =f (3) = x!3 32 =9.
Sum of an infinite series of numbers - Zeno s paradox Consider the sequence: Then a n = 2 + 4 + 8 + 6 +...+ 2 n = 2n 2 n = lim a n =. n! 2 n.
Definition of Derivative Definition Let f : R! R be a real valued function f (x). Then the derivative f 0 (x) is given by the following formula if the limit exists: Example f (x) =x 2 f 0 (x) = lim h!0 f (x + h) f (x) h lim h!0 f (x + h) h f (x) apple (x + h) 2 x 2 =lim = x 2 +2hx + h 2 x 2 h!0 h h apple 2hx + h 2 =lim =2x + h =2x h!0 h
Exponential Rule For f (x) =e x, f 0 (x) =e x Example If f (x) =e x 4x 2,findf 0 (x) by using the sum and constant rule: (f (x)+c g(x)) 0 = f 0 (x)+c g 0 (x). f 0 (x) =(e x 4x 2 ) 0 =(e x ) 0 4(x 2 ) 0 = e x 2 4x = e x 8x
Product and quotient rules Example Compute f 0 (x) forf (x) = (f (x)g(x)) 0 = f 0 (x)g(x)+f(x)g 0 (x) apple 0 f (x) = f 0 (x)g(x) f (x)g 0 (x) g(x) (g(x)) 2 ex +x. 2 Solution. dy dx = ( + x 2 ) d dx (ex ) e x d dx ( + x 2 ) ( + x 2 ) 2 = ( + x 2 )e x e x (2x) ( + x 2 ) 2 = ex ( x) 2 ( + x 2 ) 2.
Chain Rule (f g) 0 (x) =f 0 (g(x)) g 0 (x) Example Find F 0 (x) iff (x) =e x 2 +. Solution: Let f (x) =e x and g(x) =x 2 +. Then F (x) =f g(x). So, F 0 (x) =f 0 (g(x)) g 0 (x) =e x 2 + (2x)
Derivatives of classical functions (x n ) 0 = nx n. sin 0 (x) = cos(x) cos 0 (x) = (e x ) 0 = e x sin(x) ln 0 (x) = x,whereln(x) is the natural log. log 0 a(x) = x ln(a), where log a(x) is the log in base a. d dx (sin (x)) = p x 2 d dx (tan (x)) = +x 2
Antiderivatives Definition A function F is called an antiderivative of f on an interval I if F 0 (x) =f (x) forallx in I. Example Let f (x) =x 2.ThenanantiderivativeF (x) forx 2 is F (x) = x 3 3. Theorem If F is an antiderivative of f on an interval I =(a, b), then the most general antiderivative of f on I is where C is an arbitrary constant. F (x)+c
Table of Anti-di erentiation Formulas Function Particular antiderivative c f (x) c F (x) f (x)+g(x) F (x)+g(x) x n x (n 6= ) n+ n+ ln x x e x cos x e x sin x
Table of Anti-di erentiation Formulas Function Particular antiderivative sin x cos x sec 2 x tan x sec x tan x sec x p x 2 sin x +x 2 tan x
Example A ball is thrown upward with a speed of 48 ft/s from the edge of a cli 432 ft above the ground. Find its height above the ground t seconds later. When does it reach its maximum height? When does it hit the ground? Solution: At time t the distance above the ground is s(t) andthevelocity v(t) is decreasing. Therefore, the acceleration must be negative: a(t) = dv dt = 32 Taking antiderivatives v(t) = 32t + C To determine C, usethatv(0) = 48. This gives 48 = 0 + C, so v(t) = 32t + 48 The maximum height is reached when v(t) = 0, that is after.5 s. Since s 0 (t) =v(t), we anti-di erentiate and obtain s(t) = 6t 2 + 48t + D.
Continuation of Solution: Using that s(0) = 432, we have 432 = 0 + D and so, s(t) = 6t 2 + 48t + 432 The expression for s(t) is valid until the ball hits the ground. This happens when s(t) = 0: or, equivalently, 6t 2 + 48t + 432 = 0 t 2 3t 27 = 0 t = 3 ± 3p 3 2 We reject solution with the minus sign since it gives a negative value for t. Therefore, the ball hits the ground after 3( + p 3)/2 6.9 s.
Area under a graph - approximation by rectangles
Area under y = x 2 from 0 to. Example Use 4 rectangles to estimate the area under the parabola y = x 2 from 0 to.
Area estimate using right end points R 4 = 4 2 + 4 4 Note area A < 5 32 =.46875 2 + 2 4 2 3 + 4 4 2 = 5 32
Area estimate using left end points L 4 = 4 02 + c 4 4 + 4 2 + 2 4 2 3 = 7 4 32 =.2875 Note area A satisfies.2875 apple A apple.46875
General calculation using right end points
Area definition using right end points Definition The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles: A = lim n! R n = lim n! [f (x ) x + f (x 2 ) x + + f (x n ) x] Note in the above definition that if I =[a, b], then x = b a n, where n is the number of rectangles or divisions.
Sigma ( ) notation for summing and A = Area nx f (x i ) x = f (x ) x + f (x 2 ) x + + f (x n ) x i= = lim n! when f is continuous. A = lim n! = lim n! nx i= nx f (x i ) i= x nx f (x i ) x i= f (x i ) x, x i 2 [x i, x i ],
Definition of a Definite Integral Definition If f is a continuous function defined for a apple x apple b, wedividethe interval [a, b] into n subintervals of equal width x =(b a)/n. We let x 0 (= a), x, x 2,...,x n (= b) be the endpoints of these subintervals and we let x, x 2,...,x n be any sample points in these subintervals, so x i lies in the i-th subinterval [x i,xi ]. Then the definite integral of f from a to b is Z b a f (x) dx = lim n! nx i= f (x i ) x
Midpoint Rule Z b f (x) dx a i= nx f (x i ) x = x[f (x )+ + f (x n )] where x = b a n and x i = 2 (x i + x i )=midpoint of [x i, x i ]
Properties of the Integral Z b a Z b cdx= c(b [f (x)+g(x)] dx = a), where c is any constant Z b f (x) dx + Z b a a a g(x) dx Z b a c f (x) dx = c Z b a f (x) dx, where c is any constant Z b [f (x) g(x)] dx = Z b a a f (x) dx Z b a g(x) dx
The Fundamental Theorem of Calculus, Part Theorem (Fundamental Theorem of Calculus, Part ) If f is continuous on [a, b], then the function g defined by g(x) = Z x a f (t) dt a apple x apple b is continuous on [a, b] anddi erentiable on (a, b), and g 0 (x) =f (x).
The Fundamental Theorem of Calculus, Part 2 Theorem (Fundamental Theorem of Calculus, Part 2) If f is continuous on [a, b], then Z b a f (x) dx = F (b) F (a), where F (x) is any antiderivative of f (x), that is, a function such that F 0 (x) =f (x).
Application of FTC Example Evaluate Solution: Z 6 3 x dx. An antiderivative for x So, by FTC, is F (x) =lnx. Z 6 3 dx = F (6) F (3) = ln 6 ln 3. x
Example Evaluate Solution: Z 3 e x dx. Note that an antiderivative for e x is F (x) =e x. So, by FTC, Z 3 e x dx = F (3) F () = e 3 e.
Example Find area A under the cosine curve from 0 to b, where0apple b apple 2. Solution: Since an antiderivative of f (x) = cos(x) isf (x) =sin(x), we have A = Z b cos(x) dx = 0 0 b sin(x) = sin(b) sin(0) = sin(b).
The Fundamental Theorem of Calculus Theorem (Fundamental Theorem of Calculus) Suppose f is continuous on [a, b]. If g(x) = Z b a Z x a f (x) dx = F (b) f (t) dt, then g 0 (x) =f (x). f (x), that is, F 0 (x) =f (x). F (a), where F (x) is any antiderivative of
Notation: Indefinite integral R f (x) dx = F (x) means F 0 (x) =f (x). We use the notation R f (x) dx to denote an antiderivative for f (x) and it is called an indefinite integral. A definite integral has the form: Z b a Z f (x) dx = f (x) dx b a = F (b) F (a)
Table of Indefinite Integrals R c f (x) dx = c R f (x) dx R [f (x)+g(x)] dx = R f (x) dx + R g(x) dx R kdx= kx + C R x n dx = xn+ n+ + C (n 6= ) R x dx =ln x + C R e x dx = e x + C
Table of Indefinite Integrals R sin xdx= cos x + C R cos xdx=sinx + C R sec 2 xdx=tanx + C R csc 2 xdx= cot x + C R sec x tan xdx=secx + C R csc x cot xdx= R R csc x + C x 2 + dx =tan x + C p x 2 dx =sin x + C
Example Find areas. Z 2 0 Solution: 2x 3 6x + 3 x 2 dx and interpret the result in terms of + FTC gives Z 2 0 =2 x 4 4 2x 3 6x + 3 x 2 dx + 6 x 2 2 2 +3tan x. 0 = 2 x 4 3x 2 +3tan x = 2 (24 ) 3(2 2 )+3tan 2 0 = 4 + 3 tan 2 2 0
The Net Change Theorem Theorem (Net Change) The integral of a rate of change is the net change: Z b a F 0 (x) dx = F (b) F (a).
Example A particle moves along a line so that its velocity at time t is v(t) =t 2 t 6 (measured in meters per second). Find the displacement of the particle during the time period apple t apple 4. Solution: By the Net Change Theorem, the displacement is s(4) s() = apple t 3 = 3 Z 4 v(t)dt = t 2 2 6t Z 4 4 (t 2 t 6) dt = 9 2. This means that the particle moved 4.5 m toward the left.
Example A particle moves along a line so that its velocity at time t is v(t) =t 2 t 6 (measured in meters per second). Find the distance traveled during the time period apple t apple 4. Solution: Note v(t) =t 2 t 6=(t 3)(t + 2) and so v(t) apple 0 on the interval [, 3] and v(t) 0 on [3, 4]. From the Net Change Theorem, distance traveled is Z 4 Z 3 v(t) dt = Z 3 [ v(t)] dt + Z 4 Z 4 3 v(t) dt = ( t 2 + t + 6) dt + (t 2 t 6) dt 3 apple t 3 = 3 + t2 3 apple t 3 2 +6t t 2 4 + 6t 3 2 = 6 6 0.7m 3
The Substitution Rule The Substitution Rule is one of the main tools used in this class for finding antiderivatives. It comes from the Chain Rule: [F (g(x))] 0 = F 0 (g(x))g 0 (x). So, Z F 0 (g(x))g 0 (x) dx = F (g(x)).
Substitution Rule: If u = g(x) isadi erentiablefunction whose range is an interval I and f is continuous on I,then Z Z f (g(x))g 0 (x) dx = f (u) du. Example Find R x 3 cos(x 4 + 2) dx. Solution: Make the substitution: u = x 4 + 2. Get du =4x 3 dx. Z Z x 3 cos(x 4 + 2) dx = cos u 4 du = Z 4 cos udu = 4 sin u + C = 4 sin(x 4 + 2) + C. Note at the final stage we return to the original variable x.
Example Evaluate R p 2x +dx. Proof. Solution: Let u =2x +. Then du =2dx, sodx = du 2. The Substitution Rule gives Z Z p2x pu du +dx = 2 = Z 2 u 2 du = 2 u 3 2 3 2 + C = 3 u 3 2 + C = 3 (2x + ) 3 2 + C.
Substitution Rule: If u = g(x) isadi erentiablefunction whose range is an interval I and f is continuous on I,then Z Z f (g(x))g 0 (x) dx = f (u) du. Example Calculate R e 5x dx. Solution: If we let u =5x, thendu =5dx, sodx = 5 du Therefore Z e 5x dx = Z e u du = 5 5 eu + C = 5 e5x + C.
Substitution Rule: If u = g(x) isadi erentiablefunction whose range is an interval I and f is continuous on I,then Z Z f (g(x))g 0 (x) dx = f (u) du. Example Calculate R tan xdx. Solution: Z Z tan xdx= sin x cos x dx. This suggests substitution u = cos x, since then du = sin xdx and so, sin xdx= du: Z Z Z sin x du tan xdx= cos x dx = u = ln u + C = ln cos x + C.
The Substitution Rule for Definite Integrals Theorem (Substitution Rule for Definite Integrals) If g 0 is continuous on [a, b] andf is continuous on the range of u = g(x), then Z b a f (g(x))g 0 (x) dx = Z g(b) g(a) f (u) du.
The Substitution Rule for Definite Integrals: If g 0 is continuous on [a, b] andf is continuous on the range of u = g(x), then Z b a f (g(x))g 0 (x) dx = Z g(b) g(a) f (u) du. Example Calculate R e Solution: ln x x dx. We let u =lnx because its di erential du = dx x occurs in the integral. When x =, u = ln = 0; when x = e, u =lne =. Thus Z e ln x x dx = Z 0 udu= u2 2 0 = 2.
Areas between curves
Area between curves The area A of the region S bounded by the curves y = f (x), y = g(x), and the lines x = a, x = b, wheref and g continuous and f (x) g(x) for all x in [a, b], is A = Z b a [f (x) g(x)] dx
Example Find the area A of the region enclosed by the parabolas y = x 2 and y =2x x 2.
Example Find the area A of the region enclosed by the parabolas y = x 2 and y =2x x 2. Solution: We first find the points of intersection of the parabolas by solving their equations simultaneously. This gives x 2 =2x x 2,or 2x 2 2x =0. Thus 2x(x ) = 0, so x = 0 or. The points of intersection are (0, 0) and (, ). So the total area is: A = R 0 (2x 2x 2 ) dx =2 R 0 (x x 2 ) dx =2 =2 2 3 = 3 h x 2 2 i x 3 3 0
Areas between curves The area between the curves y = f (x) andy = g(x) andbetweenx = a and x = b is A =. Z b a f (x) g(x) dx The area between the curves x = f (y) andx = g(y) andbetweeny = a and y = b is A =. Z b a f (y) g(y) dy
Example Find the area enclosed by the line y = x y 2 =2x + 6. and the parabola
Example Find the area enclosed by the line y = x y 2 =2x + 6. and the parabola Solution: Solving the two equations we find points of intersection (, 2) and (5, 4). Next solve the equation of the parabola for x: x L = 2 y 2 3 x R = y + We must integrate between the appropriate y-values, y = y = 4. Thus A = Z 4 Z 4 2 (x R x L ) dy = ( 2 2 y 2 + y + 4) dy = y 3 + y 2 4 2 3 2 +4y 2 Z 4 2 [(y + ) ( 2 y 2 3)] dy = 2and = 6 (64)+8+6 4 3 +2 8 = 8
Volumes Volume = A distance Cylinder: V = Ah Circular Cylinder: V = r 2 h Rectangular Box: V = lwh
Volumes V = volume P n i= A(x i ) x
Definition of volume Definition (Definition of Volume) Let S be a solid that lies between x = a and x = b. Suppose the cross-sectional area of S in the plane P x, through x and perpendicular to the x axis, is A(x), where A is a continuous function. Then the volume of S is V = lim n! nx i= A(x i ) x = Z b a A(x) dx
Computing volume of a sphere
Computing volume of a sphere Example Show the volume of a sphere of radius r is V = 4 3 r 3. Solution: x 2 + y 2 = r 2 y = p r 2 x 2. The cross sectional area is A(x) = y 2 = (r 2 V = R r r A(x) dx = R r r (r 2 h i r =2 r 2 x x 3 3 =2 0 r 3 r 3 3 x 2 ). So: x 2 ) dx =2 R r 0 (r 2 x 2 ) dx = 4 3 r 3
Computing a volume of revolution Example Find the volume V of the solid obtained by rotating about the x-axis the region under the curve y = p x from 0 to.
Computing a volume of revolution Example Find the volume V of the solid obtained by rotating about the x-axis the region under the curve y = p x from 0 to. Solution: The area of the cross section is: A(x) = ( p x) 2 = x. Z Z So, V = A(x) dx = x dx= x 2 = 0 0 2 0 2
Volume of a solid paraboloid Example Find the volume V of the solid obtained by rotating the region bounded by y = x 3, y =8andx = 0 about the y-axis.
Volume of a solid paraboloid Example Find the volume V of the solid obtained by rotating the region bounded by y = x 3, y =8andx = 0 about the y-axis. Solution: Note that x = 3 p y. The area of cross section is: A(y) = x 2 = ( 3 p y) 2 = y 2 3. So, V = Z 8 0 A(y) dy = Z 8 0 y 2 3 dy = apple 3 5 y 5 3 8 0 = 96 5.
Other volumes Example The region R enclosed by the curves y = x and y = x 2 is rotated about the x-axis. Find the volume V of the solid region.
Example The region R enclosed by the curves y = x and y = x 2 is rotated about the x-axis. Find the volume V of the solid region. Solution: The curves y = x and y = x 2 intersect at the points (0, 0) and (, ). Cross section of rotated surface has the shape of a washer (annular ring). So the cross sectional area is: A(x) = x 2 (x 2 ) 2 = (x 2 x 4 ). So, V = R 0 A(x) dx = R h i 0 (x 2 x 4 x ) dx = 3 x 5 3 5 = 2 5.
Example The region R enclosed by the curves y = x and y = x 2 is rotated about the line y = 2. Find the volume V.
Example The region R enclosed by the curves y = x and y = x 2 is rotated about the line y = 2. Find the volume V. Solution: The cross section is again a washer. The cross sectional area is: A(x) = (2 x 2 ) 2 (2 x) 2 = (x 4 5x 2 +4x). So, V = R 0 A(x) dx = R 0 (x 4 5x 2 +4x) dx = h i x 5 5 5 x 3 3 +4x 2 2 = 8 5. 0
Integration by parts We now return to integration methods. Recall our first method was substitution which came from the chain rule. Integration by parts comes from the product rule: d dx [f (x)g(x)] = f (x)g 0 (x)+g(x)f 0 (x). So, f (x)g 0 (x) =(f (x)g(x)) 0 g(x)f 0 (x). Thus, after taking antiderivatives, we get Z Z f (x)g 0 (x) dx = f (x)g(x) g(x)f 0 (x) dx.
Integration by parts Z f (x)g 0 (x) dx = f (x)g(x) Z f 0 (x)g(x) dx The above is called the formula for integration by parts. If we let u = f (x) andv = g(x), then du = f 0 (x) dx and dv = g 0 (x) dx. So the formula becomes: Z udv = uv Z vdu.
Strategy for using integration by parts Recall the integration by parts formula: Z Z udv = uv vdu. To apply this formula we must choose dv so that we can integrate it! Frequently, we choose u so that the derivative of u is simpler than u. If both properties hold, then you have made the correct choice.
Examples using strategy: R udv = uv R vdu Z xe x dx : Choose u = x and dv = e x dx Z t 2 e t dt : Choose u = t 2 and dv = e t dt Z ln xdx: Choose u =lnx and dv = dx Z x sin xdx: Choose u = x and dv =sinxdx Z x 2 sin 2x dx: Choose u = x 2 and dv =sin2x dx
R udv = uv R vdu Example Find R xe x dx. Solution: Let u = x dv = e x dx. Then du = dx v = e x. Integrating by parts gives Z xe x dx = xe x Z e x dx = xe x e x + C.
Example Evaluate R ln xdx. Solution: Let u =lnx dv = dx. Then du = dx v = x. x Integrating by parts, we get Z ln xdx= x ln x = x ln x Z Z x dx x dx = x ln x x + C.
Example Find R t 2 e t dt. Solution: Let u = t 2 dv = e t dt Then du = 2tdt v = e t. Integration by parts gives R t 2 e t dt = t 2 e t 2 R te t dt () Using integration by parts a second time, this time with u = t dv = e t dt. Then and du = dt, v = e t, R R te t dt = te t e t dt = te t e t + C. Putting this in Equation (), we get Z Z t 2 e t dt = t 2 e t 2 te t dt = t 2 e t 2(te t e t + C) = t 2 e t 2te t +2e t + C wherec = 2C.
Example Evaluate R e x sin xdx. Solution: Integrate by parts twice. Choose u = e x and dv =sinxdx. Then du = e x dx and v = cos x, so integration by parts gives R e x sin xdx= e x cos x + R e x cos xdx. (2) Next use u = e x and dv = cos x dx. Thendu = e x dx, v =sinx, and R R e x cos xdx= e x sin x e x sin xdx. (3). Put Equation (3) into Equation (2) and we get R R e x sin xdx= e x cos x + e x sin x e x sin xdx. This can be regarded as an equation to be solved for the unknown integral. Adding R e x sin xdx to both sides, we obtain 2 R e x sin xdx= e x cos x + e x sin x. Dividing by 2 and adding the constant of integration, we get R e x sin xdx= 2 ex (sin x cos x)+c.
Trigonometric integrals Trigonometric integrals are integrals of functions f (x) that can be expressed as a product of functions from trigonometry. For example; f (x) = cos 3 x f (x) =sin 5 x cos 2 x f (x) =sin 2 x. Integrating such functions involve several techniques and strategies which we will describe today.
Aside from the most basic relations such as tan( ) = sin( ) cos( ) and sec( ) =, you should know the following trig identities: cos( ) cos 2 ( )+sin 2 ( ) =. sec 2 ( ) tan 2 ( ) =. cos 2 ( ) = + cos(2 ) 2 sin 2 ( ) = cos(2 ) 2 sin(2 ) =2sin( ) cos( ) cos(2 ) = cos 2 ( ) sin 2 ( ) = 2 cos 2 ( ) = 2 sin 2 ( )
Example Find R 0 sin2 xdx. Solution: We will use the double angle formula: sin 2 xdx= 2 ( cos 2x). Z 0 sin 2 xdx= 2 Z 0 ( cos 2x) dx = 2 x = 2 ( 2 sin 2 ) 2 (0 2 sin 0) = 2. sin 2x 2 0 Here we mentally made the substitution u = 2x when integrating cos 2x.
Strategy for Evaluating R sin m x cos n xdx (a) If the power of cosine is odd (n =2k + ), save one cosine factor and use cos 2 x = sin 2 x to express the remaining factors in terms of sine: Z Z sin m x cos 2k+ xdx= sin m x(cos 2 x) k cos xdx Z = sin m x( sin 2 x) k cos xdx Then substitute u =sinx.
Strategy for Evaluating R sin m x cos n xdx (b) If the power of sine is odd (m =2k + ), save one sine factor and use sin 2 x = cos 2 x to express the remaining factors in terms of cosine: Z Z sin 2k+ x cos n xdx= (sin 2 x) k cos n x sin xdx Z = ( cos 2 x) k cos n x sin xdx. Then substitute u = cos x.
Strategy for Evaluating R sin m x cos n xdx (c) If the powers of both sine and cosine are even, use the half-angle identities sin 2 x = 2 ( cos 2x) cos2 x = 2 ( + cos 2x) It is sometimes helpful to use the identity sin x cos x = 2 sin 2x
Strategy for Evaluating R tan m x sec n xdx (a) If the power of secant is even (n =2k, k 2), save a factor of sec 2 x and use sec 2 =+tan 2 x to express the remaining factors in terms of tan x: Z Z tan m x sec 2k xdx= tan m x(sec 2 x) k sec 2 xdx Then substitute u =tanx. Z = tan m x( + tan 2 x) k sec 2 xdx
Strategy for Evaluating R tan m x sec n xdx (b) If the power of tangent is odd (m =2k + ), save a factor of sec x tan x and use tan 2 x =sec 2 x to express the remaining factors in terms of sec x: Z Z tan 2k+ x sec n xdx= (tan 2 x) k sec n x sec x tan xdx Then substitute u =secx. Z = (sec 2 x ) k sec n x sec x tan xdx
Two other useful formulas Recall that we proved the following formula in class using integration by parts. Z tan xdx=ln sec x + C. The next formula can be checked by di erentiating the right hand side. Z sec xdx=ln sec x +tanx + C. Also, don t forget that d dx tan x =sec2 x and d dx sec x =secxtan x.
Example Find R tan 3 xdx. Solution: Here only tan x occurs, so we use tan 2 x =sec 2 x to rewrite a tan 2 x factor in terms of sec 2 x: R tan 3 xdx= R tan x tan 2 xdx= R tan x (sec 2 x ) dx = R R tan x sec 2 xdx tan xdx= tan 2 x 2 ln sec x + C. In the first integral we mentally substituted u =tanx so that du =sec 2 x dx
To evaluate the integrals (a) R sin mx cos nx dx, (b) R sin mx sin n; dx, or (c) R cos mx cos nx dx, use the corresponding identity: (a) sin A cos B = 2 [sin(a B)+sin(A + B)] (b) sin A sin B = 2 [cos(a B) cos(a + B)] (c) cos A cos B = 2 [cos(a B) + cos(a + B)] Example Evaluation R sin 4x cos 5x dx. Solution: Z = Z 2 sin 4x cos 5x dx= Z [sin( x)+sin9x] dx 2 ( sin x +sin9x) dx = 2 (cos x cos 9x)+C. 9
Table of Trigonometric Substitution Expression p Substitution Identity a 2 x 2 p x = a sin, 2 apple apple 2 sin 2 = cos 2 a 2 + x 2 p x = a tan, 2 < < 2 + tan 2 =sec 2 x 2 a 2 x = a sec, 0 apple < 2 sec 2 = tan 2 or apple < 3 2
p a 2 x 2, x = a sin, sin 2 =cos 2 Example Find the area enclosed by the ellipse x 2 a + y 2 b =.
p a 2 x 2, x = a sin, sin 2 =cos 2 Example Find the area enclosed by the ellipse x 2 a 2 + y 2 b 2 =. Solution: Solving for y gives y = b a p a 2 x 2 and A = 4b a R a 0 p a 2 x 2 dx Substitute x = a sin, dx = a cos d and use p a 2 x 2 = a cos. R p a 2 x 2 dx = R a cos a cos d = a R 2 cos 2 d = a R 2 2 ( + cos 2 ) d = 2 a2 ( + 2 sin 2 ). R A = 4b a p a 0 a 2 x 2 =2ab + 2 sin 2 0 = ab.
p a 2 x 2, x = a sin, sin 2 =cos Example Evaluate R p 9 x 2 x 2 dx.
p a 2 x 2, x = a sin, sin 2 =cos Example Evaluate R p 9 x 2 x 2 dx. Solution: Let x = 3 sin, dx = 3 cos d. p 9 x 2 = p 9 9 sin 2 = p 9 cos 2 = 3 cos Z p Z 9 x 2 3 cos x 2 dx = 9 sin 2 3 cos d Z cos 2 Z = sin 2 d = cot 2 d Z p 9 x = (csc 2 2 ) d = cot + C = x
p a2 + x 2, x = a tan, +tan 2 =sec 2 Example Evaluate R x 2p x 2 +4 dx.
p a2 + x 2, x = a tan, +tan 2 =sec 2 Example Evaluate R x 2p x 2 +4 dx. Solution: Let x = 2 tan, dx = 2 sec 2 d and p x 2 +4=2sec. Thus, R dx x 2p = R 2sec 2 dt x 4 +4 4tan 2 2sec = R sec 4 tan 2 d Put everything in terms of sin, cos : R dx x 2p = R x 4 +4 4 cos cos2 sin 2 d = R cos 4 sin 2 d = 4 ( u )+C = 4sin + C = csc 4 + C = p x 2 +4 4x + C.
Integration by partial fractions Definition A function f (x) is called a rational function if it can be expressed as the ratio of two polynomials. Example f (x) = 2 x So, R x+5 x 2 x 2 dx = R x 2(x+2) (x ) x+2 = (x )(x+2) = x+5 x 2 +x 2, is a rational function. x+2 dx =2ln x ln x +2 + C.
Proper and improper rational functions Recall that the degree polynomial P(x) =a n x n + a n x n +...+ a x + a 0 is n if a n 6= 0. We write this as deg(p) =n. A rational function f (x) = P(x) Q(x) is called proper if the degree of P(x) is less than the degree of Q(x); otherwise it is called improper. If deg(p) deg(q), then after long division, f (x) = P(x) R(x) Q(x) = S(x)+ Q(x),whereR(x) istheremainder and deg(r) < deg(q).
Example Find R x 3 +x x dx. Solution: After long division, wefind x 3 + x x = x 2 + x +2+ 2 x. So, Z x 3 Z + x x dx = x 2 + x +2+ 2 x = x 3 3 + x 2 +2x +2ln x + C. 2 dx
Partial Fraction decomposition First step, factor Q(x) andexpress R(x) Q(x) using partial fractions. Any polynomial Q(x) can be factored into a product of linear factors (ax + b) andquadraticfactors(ax 2 + bx + c), where b 2 4ac < 0. For example, suppose Q(x) =(x 2 4)(x 2 + 4) = (x 2)(x + 2)(x 2 + 4). Next express R(x) Q(x) as a sum of partial fractions of the form. A (ax + b) i or Ax + B (ax 2 + bx + c) j
Case Q(x) is a product of linear factors Example Evaluate R x 2 +2x 2x 3 +3x 2 2x dx. Solution: First factor Q(x). Q(x) =2x 3 +3x 2 2x = x(2x 2 +3x 2) = x(2x )(x + 2). So, x 2 +2x x(2x )(x + 2) = A x + B 2x + C x +2. Now multiply both sides by Q(x): x 2 +2x =A(2x )(x + 2) + Bx(x + 2) + Cx(2x ). Next simplify: x 2 +2x =(2A + B +2C)x 2 +(3A +2B + C)x 2A.
Continuation: Next solve the linear equations: 2A + B +2C = 3A +2B C =2 2A = Solving we get A = 2, B = 5, C = 0. Thus, = = 2 Z x 2 +2x 2x 3 +3x 2 2x dx Z 2 x + 5 2x 0 dx x +2 ln x + 0 ln 2x ln x +2 + C. 0
Q(x) isaproductofrepeatedlinearfactors Example Find R Solution: 4x x 3 x 2 x+ dx. The first step is to factor the denominator Q(x) =x 3 x 2 x +. Since Q() = 0, we know that x is a factor. So, x 3 x 2 x +=(x )(x 2 ) = (x )(x )(x + ) = (x ) 2 (x + ). Since the linear factor x occurs twice, the partial fraction decomposition is 4x (x ) 2 (x + ) = A x + B (x ) 2 + C x +. Now solve for A, B, andc and integrate.
Q(x) has irreducible repeated quadratic factors Example Write out the form of the partial fraction decomposition of the function x 3 + x 2 + x(x )(x 2 + x + )(x 2 + ) 3. Solution: The form of the partial fraction decomposition is x 3 + x 2 + x(x )(x 2 + x + )(x 2 + ) 3 = A x + B x + Cx + D x 2 + x + + Ex + F x 2 + + Gx + H (x 2 + ) 2 + Ix + J (x 2 + ) 3.
Q(x) contains irreducible quadratic factors Example Evaluate R 2x 2 x+4 x 3 +4x ; dx. Solution: Since x 3 +4x = x(x 2 + 4) can t be factored further, we write 2x 2 x +4 x(x 2 + 4) = A x + Bx + C x 2 +4 Multiplying by x(x 2 + 4), we have 2x 2 x +4=A(x 2 + 4) + (Bx + C)x =(A + B)x 2 + Cx +4A Equating coe cients, we obtain A + B =2 C = 4A =4 Thus A =, B =, and C = and so Z 2x 2 Z x +4 x 3 dx = +4x x + x x 2 dx +4 In order to integrate the second term we split it into two parts:
Continuation: Z Z x x 2 +4 dx = x x 2 +4 dx Z x 2 +4 dx We make the substitution u = x 2 + 4 in the first of these integrals so that du = 2x dx. Integrating we obtain: Z 2x 2 x +4 x(x 2 dx = + 4) Z Z x dx + =ln x + 2 ln(x 2 + 4) Z x x 2 +4 dx x 2 +4 dx 2 tan ( x 2 )+K
Example Z p x +4 Evaluate dx. x Solution: Let u = p x + 4, u 2 = x + 4, x = u 2 Therefore, Z =2 Z p Z x +4 u dx = Z x u 2 =2 ( + 4 u 2 ) du =2 4 Z du +8 4 u 2 4, dx = 2u du. Z 4 2udu=2 Z + u 2 u 2 4 (u 2)(u + 3) 4 du du =2u +2ln u 2 2ln u +2 + C u +2
Strategies: Simplify the integrand if possible Z Z Z tan sin sec 2 d = cos cos2 d = sin cos d = Z sin 2 d. 2 Z Z (sin x + cos x) 2 dx = sin 2 x +2sinx cos x + cos 2 xdx Z = + 2 sin x cos xdx.
Strategies: Look for an obvious substitution Z x x 2 dx = Z 2 Here, u = x 2, du = 2x dx. 2x x 2 dx = 2 ln x 2.
Strategies: Classify integrand according to form Trigonometric functions. Rational functions. Integration by parts. transcendental function. Radicals. Use recommended substitutions. Use partial fractions. Use if f (x) is a product of x n and a Particular substitutions are recommended. (a) If p ±x 2 ± a 2 occurs, use trig substitutions. (b) If p ax + b occurs, use the rationalizing substitution u = p ax + b.
Strategies: Try again Try substitution. Try parts. Manipulate the integrand. For example, R dx cos x = R cos x +cos x +cos x dx = R +cos 2 x dx sin 2 x Relate R the problem to previous problems. For example, tan 2 x sec xdx= R R R sec 3 xdx sec xdxandyou know sec 3 xdxby previous work. Use several methods.
Methods for approximate integration. We have already considered several methods for estimating left endpoint approximation right endpoint approximation midpoint approximation Z b a f (x) dx.
Example Use the Midpoint Rule with n = 4 to approximate the integral R 8 0 ex2 dx. Solution: Since a = 0, b = 8, and n = 4, the Midpoint Rule gives R 8 0 ex2 dx x[f () + f (3) + f (5) + f (7)] =2 e + e 9 + e 25 + e 49.
Trapezoidal Rule Z b a x f (x) dx T n = 2 [f (x 0)+2f (x )+2f(x 2 )+ +2f(x n )+f(x n )] (b a) where x = n and x i = a + i x. Example Use the Trapezoidal Rule to approximate the integral Solution: Z 2 x dx. (2 ) With n = 5, a =, and b = 2, we have x = 5 = 0.2, and so the Trapezoidal Rule gives Z 2 x dx T 5 = 0.2 [f ()+2f (.2)+2f (.4)+2f (.6)+2f (.8)+f (2)] = 2 0. + 2.2 + 2.4 + 2.6 + 2.8 +. 2
Error Bounds Theorem (Error Bounds) Suppose f 00 (x) apple K for a apple x apple b. IfE T and E M are the errors in the Trapezoidal and Midpoint Rules, then and K(b a)3 E T apple 2n 2 K(b a)3 E M apple 24n 2
Example Use the Midpoint Rule with n = 0 to approximate the integral Z e x 2 0 Solution: dx. Since a = 0, b =, and n = 0, the Midpoint Rule gives Z 0 e x 2 dx [f (0.05) + f (0.5) + + f (0.85) + f (0.95)] =0. e 0.0025 + e 0.0225 + e 0.0625 +...+ e 0.7225 + e 0.9025.460393
Example Suppose f 00 (x) apple K for a apple x apple b. If E T and E M are the errors in the Trapezoidal and Midpoint Rules, then K(b a)3 K(b a)3 E T apple 2n and E 2 M apple 24n 2 Give an upper bound for the error involved in this approximation. Solution: Since f (x) =e x 2,wehavef 0 (x) =2xe x 2 and f 00 (x) =(2+4x 2 )e x 2. Also, since 0 apple x apple, we have x 2 apple and so 0 apple f 00 (x) =(2+4x 2 )e x 2 apple 6e Taking K =6e, a = 0, b =, and n = 0 in the error estimate, we see that an upper bound for the error is 6e() 3 24(0) 2 = e 400 0.007
Simpson s Rule Simpson s Rule generalizes the method in the Trapezoidal Rule by using parabola approximations. It gives a much better approximation. Simpson s Rule: R b a f (x) dx S n = x 3 [f (x 0)+4f (x )+2(x 2 )+4f (x 3 )+ +2f (x n 2 )+4f (x n )+f (x n )], where n is even and x = (b a) n. Error Bound for Simpson s Rule Suppose that f (4) (x) apple K for a apple x apple b. IfE S is the error involved in using Simpson s Rule, then E S apple K(b a)5 80n 4.
Example How large should we take n in order to guarantee that the Simpson s Rule approximation for R 2 ( x ) dx is accurate to within 0.000? Solution: If f (x) = x,thenf (4) (x) = 24 x 5. Since x, we have x apple and so f (4) (x) = 24 x apple 24. 5 Therefore, we can take K = 24. Thus, for an error less than 0.000 we should choose n so that 24() 5 80n 4 < 0.000 This gives n 4 > 24 80(0.000) or n > 4p 0.00075 6.04 Therefore, n =8(n must be even) gives the desired accuracy.
Definition of an Improper Integral of Type Definition (Improper Integral of Type ) (a) If R t f (x) dx exists for every number t a, then R a R t f (x) dx =lim a t! f (x) dx provided this limit exists (as a a finite number). (b) If R b f (x) dx exists for every number t apple b, then t R b f (x) dx =lim R b t! f (x) dx t provided this limit exists (as a finite number). The improper integrals R f (x) dx and R b f (x) dx are called a convergent if the corresponding limit exists and divergent if the limit does not exist. (c) If both R f (x) dx and R a f (x) dx are convergent, then we define a R f (x) dx = R a f (x) dx + R f (x) dx. a In part (c) any real number a can be used.
Example Find the area A under the graph of f (x) = x 2 from x =tox =. Solution: A = Z dx = lim x 2 t! This means that the area is equal to also. x t 0 =.
Example Determine whether the integral Solution: Z x dx is convergent or divergent. According to part (a) of Definition above, we have Z x dx = lim t! Z t x dx = lim ln x t = lim (ln t ln ) t! t! = lim ln t =. t! The limit does not exist as a finite number and so the improper integral R ( x ) dx is divergent.
Example Evaluate Solution: Z 0 e x dx. Using part (b) of Definition above, Z 0 e x dx = Z 0 lim t! t e x dx 0 = lim t! ex = lim t! t e t =.
Example Evaluate R Solution: R +x 2 dx. +x 2 dx = R 0 +x dx + R 2 0 +x 2 dx Since f (x) = +x 2,thenf (x) =f ( x) andso R 0 Then R 0 +x dx + R 2 0 +x dx =2 R 2 0 +x 2 dx. R t dx +x dx =lim 2 t! 0 +x = lim 2 t! tan x t = 0 lim t! (tan t tan 0) = lim t! tan t = 2. So, R +x dx =2 2 2 =.
Example Evaluate R 0 xex dx. Solution: Using part (b) of Definition above, we have R 0 xex dx =lim t! R 0 t xex dx We integrate by parts with u = x, dv = e x dx so that du = dx, v = e x : R 0 t xex dx = xe x 0 R 0 t t ex dx = te t +e t. We know that e t! 0ast! have, and by Hospital s Rule we lim t! te t t =lim t! e =lim t t! e =lim t t! ( e t )= 0 Therefore R 0 xex dx =lim t! ( te t +e t )= 0 +0=
Definition of an Improper Integral of Type 2 Definition (Improper Integral of Type 2) (a) If f is continuous on [a, b) and is discontinuous at b, then Z b Z t f (x) dx = lim f (x) dx a t!b a if this limit exists (as a finite number). (b) If f is continuous on (a, b] and is discontinuous at a, then Z b f (x) dx = lim a t!a + number). Z b t f (x) dx if this limit exists (as a finite
Definition of an Improper Integral of Type 2 Definition (Improper Integral of Type 2) The improper integral Z b a f (x) dx is called convergent if the corresponding limit exists and divergent if the limit does not exist. (c) If f has a discontinuity at c, wherea < c < b, and both and Z b Z b c f (x) dx are convergent, then we define f (x) dx = Z c f (x) dx + Z b a a c f (x) dx Z c a f (x) dx Example Z Z 0 x 2 = Z x 2 dx + 0 x 2 dx.
Example Find Z 5 2 p x 2 dx. Solution: We note first that the given integral is improper because f (x) = p x 2 has the vertical asymptote x = 2. Since the infinite discontinuity occurs at the left endpoint of [2, 5], we use (b): Z 5 Z dx 5 dx p = lim p 2 x 2 t!2 + t x 2 = lim t!2 + 2p x 2 5 t = lim t!2 + 2(p 3 p t 2) = 2 p 3
Example Evaluate R 0 ln xdx=lim R t!0 + ln xdx. t Solution: We know that the function f (x) =lnx has a vertical asymptote at 0, since lim x!0 + ln x =. Thus, the given integral is improper and we have R 0 ln xdx=lim R t!0 + ln xdx t Now we integrate by parts with u =lnx, dv = dx, du = dx x, and v = x: R t ln xdx= x ln x ] t R t dx =ln t ln t ( t) = t ln t +t To find the limit of the first term we use Hospital s Rule: lim t!0 + t ln t =lim t!0 + ln t t =lim t!0 + t =lim t!0 +( t) =0 t 2 Therefore R 0 ln xdx=lim t!0 +(lim t!0 +( t ln t +t) = 0 +0=.
Example For what values of p is the integral Z x p dx convergent? Solution: If p =, then we know integral is divergent. Assume now p 6=. R x x dx =lim p+ p t! p+ i x=t x= =lim t! p h t p If p >, then as t!, t p!and t p! 0. So, R 0 x dx = p p if p >. But if p <, t p = t p!as t!and so the integral is divergent. i
Example Find Z 5 2 p x 2 dx. Solution: We note first that the given integral is improper because f (x) = p x 2 has the vertical asymptote x = 2. Since the infinite discontinuity occurs at the left endpoint of [2, 5], we use : Z 5 2 dx p x = lim t!2 + 2p x 2 5 t Z 5 dx = lim p 2 t!2 + t x 2 = lim t!2 + 2(p 3 p t 2) = 2 p 3
Comparison Theorem Theorem (Comparison Theorem) Suppose that f and g are continuous functions with f (x) g(x) 0for x a. (a) If (b) If Z a Z f (x) dx is convergent, then g(x) dx is divergent, then Z a Z a a g(x) dx is convergent. f (x) dx is divergent.
Example Show that R 0 e x 2 dx is convergent. Proof. First write Z 0 e x 2 = Z Since on [, ), e x 2 apple e x and Z e x dx = lim t! 0 e x 2 dx + e x t Z e x 2 dx. = e e t = e, the integral is convergent by the Comparison Theorem.
Example The integral Z +e x is divergent by the Comparison Theorem because x dx and Z x dx is divergent. +e x x > x
Definition Asequence{a n } has the limit L and we write lim a n = L or a n! L as n! n! if for every " > 0 there is a corresponding integer N such that if n > N, then a n L < ".
Definition Asequencea n has the limit L and we write lim a n = L or a n! L as n! n! if we can make the terms a n as close to L as we like by taking n su ciently large. If lim n! a n exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent). Example Let f (x) = x. Consider the sequence a n = f (n) = n Then lim a n = lim n! n! n =0. for n an integer.
Definition Asequence{a n } is called increasing if a n < a n+ for all n, that is, a < a 2 < a 3 <. It is called decreasing if a n > a n+ for all n. It is called monotonic if it is either increasing or decreasing. Example n o 3 3 The sequence n+5 is decreasing because n+5 > 3 (n+)+5 = 3 n+6 a n > a n+ for all n. and so
Definition Asequence{a n } is bounded above if there is a number M such that a n apple M for all n. It is bounded below if there is a number m such that m apple a n for all n. If it is bounded above and below, then {a n } is a bounded sequence. Theorem (Monotonic Sequence Theorem) Every bounded, monotonic sequence is convergent.
Limit Laws for sequences If {a n } and {b} n are convergent sequences and c if a constant, then lim (a n + b n )= lim a n + lim b n n! n! n! lim (a n b n )= lim a n lim b n n! n! n! lim ca n = c lim a n lim c = c n! n! n! lim n! lim (a nb n )= lim a n lim b n n! n! n! a n = lim n! a n if lim b n lim n! b b n 6= 0 n n! lim n! ap n =[lim n! a n] p if p > 0 and a n > 0
Theorem If lim n! a n = 0, then lim n! a n = 0. Example Calculate lim n! ln n n. Solution: Notice that both numerator and denominator approach infinity as n!. We can apply Hospital s Rule to the related function f (x) = ln x x and obtain ln x lim x! x x = lim x! =0.
Theorem If P a n and P b n are convergent series, then so are the series P c a n (where c is a constant), P (a n + b n ), and P (a n b n ): (i) (ii) (iii) X X c a n = c n= n= a n X X X (a n + b n )= a n + n= n= n= X X (a n b n )= b n X a n b n n= n= n=
Theorem If lim n! a n = L and the function f is continuous at L, then lim n! f (a n )=f (L) Example Discuss the convergence of the sequence a n = n! n n,where n! = 2 3 n. Solution: Write out a few terms: a = a 2 = 2 2 2 a 3 = 2 3 3 3 3 a n = 2 3 n n n n n. Observe from above equation that a n = n 2 3 n n n n. The expression in parentheses is at most because the numerator is less than the denominator. So 0 < a n apple n. We know that n! 0asn!. Therefore a n! 0asn!by the Squeeze Theorem.
Definition Given a series P n= a n = a + a 2 + a 3 +,lets n denote its n-th partial sum nx s n = a i = a + a 2 + + a n. i= If the sequence {s n } is convergent and lim n! s n = s exists as a real number, then the series P a n is called convergent. When it is convergent, we write a + a 2 + + a n + = s or P n= a n = s. The number s is called the sum of the series. Otherwise, the series is called divergent.
Big Question! What is the di erence between a sequence and a series? A sequence is just a listing of numbers, usually given by a formula. For example, {a =2, a 2 =4,...,a n =2n, a n+ =2n +2,...} is the sequence of positive integers. P n= 2n =2+4+...+2n +(2n + 2) +... is the sum of the all even integers. Aseries P n= a n creates the sequence which is its sequence n-th partial sums s n = P n i= a i = a + a 2 + + a n, and we say that the series converges if is sequence of n-th partial sums has a limit = L = lim n! s n.
An important example of an infinite series is the geometric series a + ar + ar 2 + ar 3 + + ar n + = X ar n. Each term is obtained from the preceding one by multiplying it by the common ratio r. If r 6=, we have the n-th partial sum n= s n = a + ar + ar 2 + + + ar n and rs n = ar + ar 2 + + ar n + ar n. Subtracting these equations, we get s n rs n = a ar n. Thus, s n = a( rn ). r
The geometric series a + ar + ar 2 + ar 3 + + ar n + = has the n-th partial sum s n = a( rn ). r X ar n, n= If < r <, we know that r n! 0asn!,so a( r n lim s ) n = lim = a a n! n! r r r rn = a r a r lim n! rn = a r.
Theorem The geometric series P n i= arn = a + ar + ar 2 + is convergent if r < and its sum is X X ar n = ar n = a r <. r n=0 n= If r, the geometric series is divergent. Example Find the sum of the geometric series 0 5 3 + 20 40 9 27 + = 5 + 5 2 3 + 5 ( 2 3 )2 Solution: The first term is a = 5 and the common ratio is r = 2 3. Since r = 2 3 <, the series is convergent and its sum is 0 5 3 + 20 40 9 27 + = 5 = 5 2 ( 3 ) 5 = 3. 3
Example Find the exact value of the repeating decimal 0.56 = 0.565656... as a rational number. Solution: Note that 0.56 = X n=0 n 56. 000 000 This is a geometric series with a = 56 000 and r = 000 and has the value: 56 000 000 = 56 000 999 000 = 56 999.
Example Is the series P n= 22n 3 n convergent or divergent? Solution: Let s rewrite the n-th term of the series in the form ar n : X X 2 2n 3 n = (2 2 ) n (n ) 3 n= = X n= 4 n 3 n = X n= n= 4 n 4 3. We recognize this series as a geometric series with a =4andr = 4 3. Since r >, the series diverge.
Example Show that the series P n= Solution: n(n+) is convergent, and find its sum. Compute the partial sums s n = P n i= i(i+) = 2 + 2 3 + 3 4 + + n(n+). Simplify expression using the partial fractions Thus, s n = P n i= 2 + 2 i(i+) = P n i= 3 + 3 So lim n! s n =lim n! n+ Therefore, i(i + ) = i i +. i i+ = 4 + + n n+ = 0 =. X n= n(n + ) =. = n+.
Theorem If the series P n= a n is convergent, then lim n! a n = 0. The converse of the above theorem is false. If lim n! a n = 0, we cannot conclude that P a n is convergent. For the harmonic series P n,wehavea n = n! 0 as n!,but is divergent! P n This same example shows that the converse to the next theorem is not true! Theorem (Test For Divergence) If lim n! a n does not exist or if lim n! a n 6= 0, thentheseries P n= a n is divergent.
Example Show that the series P n 2 n= 5n 2 +4 diverges. Solution: lim a n = lim n! n! n 2 5n 2 + 4 = lim n! So the series diverges by the Test of Divergence. 5 + 4 = 6= 0. 5 n 2
Example Consider the series 2 + 2 2 + 3 2 + 4 2 + = X n 2. n= By comparing the areas, we see that X x dx =. So, 2 R n= X n=2 n 2 < +=2. is less than the area n2
Example Consider the series X n= p = p + p + p +. n 2 3 By P comparing the areas, we see that p n= n > R p x dx =, so the series fails to converge.
Theorem (Integral Test) Suppose f is a continuous, positive, decreasing function on [, ) andleta n = f (n). X Then the series a n is convergent if and only if the improper integral Z In other words: (i) If (ii) If Z Z n= f(x) dx is convergent. f(x) dx is convergent, then f(x) dx is divergent, then X a n is convergent. n= n= X a n is divergent.
Example Test the series P n= Solution: n 2 + The function f (x) = (x 2 +) on [, ). for convergence or divergence. By the Integral Test: Z Z t x 2 dx = lim + t! x 2 + dx = = lim tan t = t! 4 2 4 = 4. Thus Z P n= n 2 + is continuous, positive and decreasing lim t! tan x t (x 2 dx is a convergent integral and the series + ) is convergent.
Theorem (p-series Test) A series of the form X n= n p is convergent if p > and divergent if p apple. Proof. This Z follows from the Integral Test and our earlier calculation that dx is convergent if and only if p >. x p
Example (a) The series X n= n 3 = 3 + 2 3 + 3 3 + 4 3 + is convergent because it is a p-series with p =3>. (b) The series X n= n 3 = X n= 3p n = + 3p 2 + 3p 3 + 3p 4 + is divergent because it is a p-series with p = 3 <.
Suppose f (x) is a positive decreasing function of x. Comparing the areas of the rectangles with the area under y = f (x) for x > n in figure above, we see that R n = a n+ + a n+2 + apple Z n f(x) dx.
Similarly, we see from figure below R n = a n+ + a n+2 + Z n+ So we have proved the following error estimate. f(x) dx
Theorem (Remainder Estimate For The Integral Test) Suppose f (k) =a k,wheref is a continuous, positive, decreasing function for x n and P a n is convergent. If R n = s s n,then Z n+ f(x) dx apple R n apple Z n f(x) dx.
Example It can be shown that 2 6 = X n 2. Estimate how close is R k = P k n= n 2 = + 4 + 9 + + k 2 to 2 6? n= Solution: R n apple Z k dx =lim x2 t!0 x t k = k.
Theorem (Comparison Test) Suppose that P a n and P b n are series with positive terms. (i) If P b n is convergent and a n apple b n for all n, then P a n is also convergent. (ii) If P b n is divergent and a n divergent. Solution to part (i): b n for all n, then P a n is also Let s n = nx a i t n = i= nx X b i t = b n. i= n= The sequences {s n } and {t n } are increasing (s n+ = s n + a n+ s n ). Also t n! t, sot n apple t for all n. Since a i apple b i,wehaves n apple t n.thuss n apple t for all n. So {s n } is increasing and bounded above and converges by the Monotonic Sequence Theorem.
Example Determine whether the series P n= Solution: 5 2n 2 +4n+3 5 2n 2 +4n +3 < 5 2n 2 because left side has a bigger denominator. X 5 2n 2 = 5 X 2 n 2 n= n= is convergent because it s a constant times a p p =2>. Therefore X n= 5 2n 2 + 4n + 3 converges or diverges. is convergent by part (i) of the Comparison Test. series with
Example Test the series Solution: X n= ln n n Observe that ln n > forn for convergence or divergence. ln n n 3 and so > n n 3. We know that X is divergent (p series with p = ). n Thus the given series is divergent by the Comparison Test.
Theorem (Limit Comparison Test) Suppose that P a n and P b n are series with positive terms. If a n lim = c n! b n where c is a finite number and c > 0, then either both series converge or both diverge.
Example Test the series P n= Solution: 2 n for convergence or divergence. We use the Limit Comparison Test with a n = 2 n b n = 2 n We obtain = lim n! a n lim = lim n! b n n! 2 n 2 n = lim n! 2 n 2 n 2 n = > 0 Since this limit exists and P 2 n is a convergent geometric series, the given series converges by the Limit Comparison Test.
Theorem (The Alternating Series Test.) If the alternating series P n= ( )n b n = b b 2 + b 3 b 4 + b 5 b 6 + for b n > 0 satisfies: (i) b n+ apple b n for all n, (ii) lim b n =0, n! then the series is convergent. The proof is contained in the next figure.
Example The alternating harmonic series 2 + 3 4 + = X n= ( ) n n satisfies (i) b n+ < b n because n+ < n (ii) lim b n = lim n! n! n =0. So the series is convergent by the Alternating Series Test.
Theorem (Alternating Series Estimation Theorem) If s = P ( ) n b n is the sum of an alternating series that satisfies then R n = s s n apple b n+. (i) 0 apple b n+ apple b n and (ii) lim n! b n =0, Proof. We know from the proof of the Alternating Series Test that s lies between any two consecutive partial sums s n and s n+. It follows that s n s n apple s n+ s n = b n+.
Definition Aseries P a n is called absolutely convergent if the series of absolute values P a n is convergent. Definition Aseries P a n is called conditionally convergent if it is convergent but not absolutely convergent. Example ( ) n The series P n= = + + is absolutely n 2 2 2 3 2 4 2 convergent because P ( ) n n= = P n 2 n= = + + + + is a convergent n 2 2 2 3 2 4 2 p-series (p = 2).
Theorem If a series P a n is absolutely convergent, thenitisconvergent. Proof. Observe the inequality 0 apple a n + a n apple 2 a n. If P a n is absolutely convergent, then P a n is convergent, so P 2 an is convergent. By the Comparison Test, P (a n + a n ) is convergent, then X an = X X (a n + a n ) an is the di erence of two convergent series and is convergent.
Theorem (Ratio Test) a n+ X (i) If lim = L <, then the series a n is absolutely n! a n n= convergent (and therefore convergent). a n+ a n+ X (ii) If lim = L > or lim =, thentheseries n! a n n! a n n= is divergent. a (iii) If lim n+ n! a n =, the Ratio Test is inconclusive; no conclusion can be drawn about the convergence or divergence of X a n. n= a n
Proof of the Ratio Test. Assume the series P n i= a n satisfies a j+ a j! L < r < as j!. Then for a large integer i and n i, a n+ < a n r. Thus, for any k > 0, a n+ a n < r or a i+k < a i+k r < a i+k 2 r 2...< a i r k. Hence, for k > 0, a i+k < a i r k. Since P k=0 a i r k is a convergent geometric series, the Comparison Test implies P k=0 a i+k converges and so P n converges absolutely. Thus, P n=0 a n converges absolutely. i= a n
Example Test the series P n= n3 3 n for absolute convergence. Solution: We use the Ratio Test with a n = n3 3 n : = 3 n + n a n+ a n = (n+) 3 3 (n + n+ )3 = n 3 3 n+ 3n n 3 3 n 3 = + 3! < as n!. 3 n 3 Thus, by the Ratio Test, the given series is absolutely convergent and therefore convergent.
Theorem (Root Test) p n (i) If lim an = L <, then the series P n! n= a n is absolutely convergent (and therefore convergent). p p n n (ii) If lim an = L > or lim an =, thentheseries P n! is divergent. n! (iii) If lim n! n p a n =, the Root Test is inconclusive. n= a n
Example Test the convergence of the series P n= 2n+3 3n+2 n. Solution: a n = n 2n +3 3n +2 p n 2n +3 an = 3n +2 = 2+ 3 n 3+ 2 n! 2 < as n!. 3 Thus the given series converges by the Root Test.
Example P n= n 2n+ Solution: Since a n! 2 6=0asn!, it fails to converge by the Test for Divergence. Example P n= Solution: p n 3 + 3n 3 +4n 2 +2 Since a n is an algebraic function of n, we compare the given series with a p-series. The comparison series for the Limit Comparison Test is P b n, where p b n = n 3 3n = n 3 2 3 3n =. 3 3n 3 2
Example P n2 n= ne Solution: Since the integral R xe x 2 dx is easily evaluated, we use the Integral Test. The Ratio Test also works. Example P n3 n= ( )n n 4 + Solution: Since the series is alternating, we use the Alternating Series Test.
Example P k= 2k k! Solution: Since the series involves k!, we use the Ratio Test. Example P n= 2+3 n Solution: Since the series is closely related to the geometric series P 3 n,weuse the Comparison Test.
A power series is a series of the form X c n x n = c 0 + c x + c 2 x 2 + c 3 x 3 + () n=0 where x is a variable and the c n s are constants called the coe cients of the series. For each fixed x, theseries() is a series of constants that we can test for convergence or divergence. A power series may converge for some values of x and diverge for other values of x. The sum of the series is a function f(x) =c 0 + c x + c 2 x 2 + + c n x n + whose domain is the set of all x for which the series converges.
Definition A series of the form X c n (x a) n = c 0 + c (x a)+c 2 (x a) 2 + n=0 is called any of the following: a power series in (x a) a power series centered at a a power series about a.
Example For what values of x does the series P (x 3) n n= n Solution: Let a n = Then (x 3)n n. an+ a n = (x 3) n+ n+ (x 3) n n = (x 3)n+ n + converge? n = n x 3 (x 3) n n + = x 3! x 3 as n!. + n By the RatioTest, thegivenseriesisabsolutelyconvergent,when x 3 < anddivergentwhen x 3 >. So the series converges when 2 < x < 4anddivergeswhenx < 2or x > 4. For x =4,theseriesistheharmonic series P, which is divergent by n p-test. For x =2,theseriesis P ( ) n, which converges by the Alternating n Series Test. Thus the given power series converges for 2 apple x < 4.
Theorem For a given power series P n=0 c n(x a) n, there are only three possibilities: (i) The series converges only when x = a. (ii) The series converges for all x. (iii) There is a positive number R such that the series converges if x a < R and diverges if x a > R. The number R in case (iii) is called the radius of convergence of the power series. By convention, the radius of convergence is R = 0 in case (i) and R = in case (ii). The interval of convergence of a power series is the interval that consists of all values of x for which the series converges.
Example Find the radius of convergence and the interval of convergence for P x n n=. n Solution: Let a n = xn n. Then a n+ a n = x n+ n+ x n n = xn+ n + n = n x! x as n!. x n n + By the Ratio Test, this series converges for x < and diverges for x >. So the radius of convergence is R = and the series converges on the interval (, ). We now check end points. For x =,wehavetheharmonicseries P n= which diverges by the n Integral Test or by the p-test. For x =, we have the alternating harmonic series P ( ) n n= which n converges by the Alternating Series Test. So the interval of convergence is [, ).
Example Express +x 2 as the sum of a power series and find the interval of convergence. Solution: Recall x = + x + x2 + x 3 + = P n=0 xn x <. Thus, + x 2 = ( x 2 ) = X ( x 2 ) n = X ( ) n x 2n n=0 n=0 = x 2 + x 4 x 6 + x 8. Because this is a geometric series, it converges precisely when x 2 <, that is x 2 <, or x <. Therefore the interval of convergence is (, ).
Theorem If the power series P n=0 c n(x a) n has radius of convergence R > 0, then the function f defined by X f(x) =c 0 + c (x a)+c 2 (x a) 2 + = c n (x a) n is di erentiable on the interval (a R, a + R) and n=0 (i) f 0 (x) =c + 2c 2 (x a)+3c 3 (x a) 2 + = P n= nc n(x a) n. (ii) R (x a) f(x) dx = C + c 0 (x a)+c 2 (x a) 2 + c 3 2 3 + = C + P n= c (x a)n+ n n+. The radii of convergence of the power series in Equations (i) and (ii) are both R.
Example Express ( x) 2 as a power series by di erentiating ( x) = P n=0 xn. What is the radius of convergence? Solution: Di erentiate each side of the equation ( x) = + x + x2 + x 3 + = P n=0 xn. We get ( x) 2 = + 2x + 3x2 + = Replace n by n +andwritetheansweras ( x) 2 = X (n + )x n. n=0 X nx n. By the theorem the radius of convergence of the di erentiated series is the same, so, R =. n=
Example Find a power series representation for ln( convergence. x) and its radius of Solution: By FTC and the geometric series, ln( x) = R x dx = R ( + x + x 2 + ) dx = x + x2 2 + x3 3 + + C = P n= xn n + C x <. To determine the value of C we put x = 0 is this equation and obtain ln( 0) = C. Thus C =0and ln( x) = x x 2 2 x 3 3 = P n= xn n x <. The radius of convergence is the same as for the original series: R =.
Example Find a power series representation for f (x) =tan x. Solution: Since f 0 (x) = (+x 2 ), find the required series by integrating the geometric series (+x 2 ) = ( ( x 2 )) = P n=0 ( x2 ) n = P n=0 ( )n x 2n. Z Z tan x = + x 2 dx = ( x 2 + x 4 x 6 + ) dx = C + x x 3 3 + x5 5 x 7 7 +. To find C we put x = 0 and obtain C =tan 0 = 0. Therefore tan x = x x 3 3 + x5 5 x 7 7 + = X ( ) n x2n+ 2n +. n=0
Theorem If f has a power series representation (expansion) at a, thatis, if X f(x) = c n (x a) n x a < R, n=0 then its coe cients are given by the formula c n = f(n) (a). n!
Theorem If f has a power series expansion at a, then = f(a)+ f0 (a)! f(x) = X n=0 (x a)+ f00 (a) 2! f (n) (a) (x a) n n! (x a) 2 + f000 (a) (x a) 3 +. 3! This power series is called the Taylor series of the function f(x) at a (or centered at a). If a=0, then the Taylor series of f(x) is called the Maclaurin series of f(x).
Example Find the Maclaurin series of the function f (x) =e x and its radius of convergence. Solution: If f (x) =e x,thenf (n) (x) =e x, so f (n) (0) = e 0 =foralln. Therefore, the Taylor series for f at 0 (that is, the Maclaurin series) is: X X n=0 f (n) (0) x n = n! n=0 x n n! = + x! + x2 2! + x3 3! +. To find the radius of convergence we let a n = x n n!. Then a n+ a n = x n+ (n+)! x n n! = xn+ (n + )! n! x n = x n +! 0 <, So, by the Ratio Test, the series converges for all x and the radius of convergence is R =.
Definition The partial sum nx T n (x) = i=0 f (i)(a) i! (x a) i = f(a)+ f0 (a) (x a)+ f00 (a) (x a) 2 + + f(n) (a) (x a) n! 2! n! is called the n-th-degree Taylor polynomial of f.
Theorem If f (x) =T n (x)+r n (x), where T n is the n-th-degree Taylor polynomial of f at a and lim R n(x) =0 n! for x a < R, thenf is equal to the sum of its Taylor series on the interval x a < R. R n is called the remainder. Theorem (Taylor s Inequality) If f (n+) (x) apple M for x a apple d, thenthereminderr n (x) of the Taylor series satisfies the inequality M R n (x) apple (n + )! x a n+ for x a apple d. In applying Taylor s inequality, the following fact is useful: x n lim = 0 for every real number x n! n!
Example Prove that e x is equal to the sum of its Maclaurin series. Solution: If f (x) =e x,thenf (n+) (x) =e x,thenf (n+) (x) =e x. If d is any positive number and x apple d, then f (n+) (x) = e x apple e d. So Taylor s Inequality, with a =0andM = e d,saysthat e d R n (x) apple (n + )! x n+ for x apple d. The same constant M = e d works for every value of n. So we have lim n! e d (n + )! x n+ = e d lim n! x n+ (n + )! =0. It follows that lim n! R n (x) =0andthereforelim n! R n (x) =0 for all values of x. Hence, e x = X n=0 x n n! for all x.
Example Find the Taylor series for f (x) =e x at a = 2. Solution: We have f (n) (2) = e 2 and so, putting a = 2 in the definition of a Taylor series, we get X n=0 f (n) (2) (x 2) n = n! X n=0 e 2 n! (x 2)n. Again it can be verified, that the radius of convergence is R =. As in the previous example we can verify that lim n! R n (x) = 0, so X e x e 2 = n! (x 2)n for all x n=0
Example Find the Maclaurin series for sin x and prove that it represents sin x for all x. Solution: Calculating we get: f (x) =sinx f 0 (x) =cosx f 00 (x) = sin x f 000 (x) = cos x f (4) (x) =sinx; f (0) = 0 f 0 (0) = f 00 (0) = 0 f 000 (0) = f (4) (0) = 0. Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as: f(0)+ f0 (0) x + f00 (0) x 2 + f000 (0) x 3 +! 2! 3! x 3 = x 3! + x5 x 7 5! 7! + = X ( ) n x 2n+ (2n + )! Since f (n+) (x) is± sin x or ± cos x, weknowthat f (n+) (x) apple for all x. So we can take M = in Taylor s Inequality: R n(x) apple M (n +) x n+ = x n+! 0 as n!. (n +)! n=0 So sin x is equal to the sum of its Maclaurin series.
Example Find the Maclaurin series for cos x. Solution: We could proceed directly as in the previous example but it s easier to di erentiate the Maclaurin series for sin x: cos x = d dx (sin x) = d x 3 x dx 3! + x5 x 7 5! 7! + = 3x 2 3! + 5x4 5! 7x 6 7! + = x 2 2! + x4 4! x 6 6! + Since the Maclaurin series for sin x converges for all x, the di erentiated series for cos x also converges for all x.
Example Find the Maclaurin series for the function f (x) = x cos x. Solution: Instead of computing derivatives, it s easier to multiply the Maclaurin series for cos x by x: x cos x = x X ( ) n x2n (2n)! = X ( ) n x2n+ (2n)! n=0 n=0
n= Table of Maclaurin Series x = X x n = + x + x 2 + x 3 + (, ) n=0 X e x x n = n! = + x! + x2 2! + x3 + (, ) 3! n=0 X sin x = ( ) n x 2n+ (2n + )! = x x 3 3! + x5 x 7 + (, ) 5! 7! n=0 X cos x = ( ) n x2n (2n)! = x 2 2! + x4 x 6 + (, ) 4! 6! n=0 X tan x = ( ) n x2n+ 2n + = x x 3 3 + x5 x 7 + [, ] 5 7 n=0 X x n ln( x) = n = x x 2 x 3 (0, 2) 2 3
Example Evaluate R e x 2 dx as an infinite series. Solution: Find the Maclaurin series for f (x) =e x 2. Replace x with x 2 in the series for e x given in the table of Maclaurin series. Thus, for all values of x, X ( x 2 ) n X e x2 = = ( ) n x2n n! n! = x 2! + x4 2! n=0 n=0 Now integrate term by term: Z Z x 2 e x2 dx =! + x4 2! = C + x x 3 3! + x5 5 2! x 6 3! + x 6 x2n + +( )n 3! n! + dx x 7 7 3! + +( x 2n+ )n (2n + )n! + This series converges for all x because the original series for e x 2 converges for all x.
Example Evaluate R 0 e x 2 dx correct to within an error of 0.00. Solution: Apply the Maclaurin series for e x : Z = 0 apple x = e x2 dx = Z 0 x 3 3! + x5 5 2! 3 + 0 x 2! + x4 2! x 7 7 3! + x9 9 4! 42 + 26 x 6 + dx 3! 0.7475 The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than 5! = 320 < 0.00 0
Example Approximate the function f (x) = 3 p x by a Taylor polynomial of degree 2 at a = 8. Solution: p f (x) = 3 x = x 3 f 0 (x) = 3 x 2 3 f 00 (x) = 2 9 x 5 3 f 000 (x) = 0 27 x 8 3 f (8) = 2 f 0 (8) = 2 f 00 (8) = 44 Thus, the second-degree Taylor polynomial is T 2 (x) =f (8) + f 0 (8)! (x 8) + f 00 (8) (x 8) 2 2! =2+ 2 (x 8) 288 (x 8)2. The desired approximation is 3p x T2 (x) =2+ 2 (x 8) 288 (x 8)2.
Example What is the maximum error possible in using the approximation x 3 sin x x 3! + x 5 5! when -0.3 apple x apple 0.3? Solution: Notice that the Maclaurin series x 3 sin x = x 3! + x 5 x 7 5! 7! + is alternating for all nonzero values of x, andaredecreasinginsize because x <, so we can use the Alternating Series Estimation Theorem. The error in approximating sin x by the first three terms of its Maclaurin series is at most x 7 7! = x 7 5040. If -0.3 apple x apple 0.3, then x apple 0.3, so the error is smaller than (0.3) 7 5040 4.3 0 8.
Example For what values of x is this approximation accurate to within 0.00005? Solution: The error will be smaller than 0.00005 if x 7 5040 < 0.00005 item Solving this inequality for x, we get x 7 < 0.252 or x < (0.252) 7 0.82 So the given approximation is accurate to within 0.00005 when x < 0.82.
Polar Coordinates (r, ) in the plane are described by r = distance from the origin and 2 [0, 2 ) is the counter-clockwise angle. We make the convention ( r, ) =(r, + )..
Example Plot the points whose polar coordinates are given. (a), 5 (b) (2, 3 ) (c) 2, 2 4 3 (d) 3, 3 4. Solution: In part (d) the point 3, 3 4 is located three units from the pole in the fourth quadrant because the angle 3 4 is in the second quadrant and r = 3 is negative.
Coordinate conversion - Polar/Cartesian x = r cos y = r sin r 2 = x 2 + y 2 tan = y x
Example Convert the point 2, 3 from polar to Cartesian coordinates. Solution: Since r =2and = 3, x = r cos = 2 cos 3 =2 2 = y = r sin =2sin 3 =2 p 3 2 = p 3 Therefore, the point is (, p 3) in Cartesian coordinates.
Example Represent the point with Cartesian coordinates (, coordinates. ) in terms of polar Solution: If we choose r to be positive, then r = p x 2 + y 2 = p 2 +( ) 2 = p 2 tan = y x =. Since the point (, ) lies in the fourth quadrant, we choose = 4 or =7 4. p Thus, one possible answer is 2, 4 ; another is p 2, 7 4.
Polar Coordinates The coordinates of a point (x, y) 2 R 3 can be described by the equations: x = r cos( ) y = r sin( ), () where r = p x 2 + y 2 is the distance from the origin and ( x r, y r )is (cos( ), sin( )) on the unit circle. Note that r 0and can be taken to lie in the interval [0, 2 ). To find r and when x and y are known, we use the equations: r 2 = x 2 + y 2 tan( ) = y x. (2)
Graph of a polar equation Definition The graph of a polar equation r = f ( ), or more generally F(r, ) = 0, consists of all points P that have at least one polar representation (r, ) whose coordinates satisfy the equation.
Example What curve is represented by the polar equation r = 2? Solution: Curve consists of points (r, ) with r = 2. Since r represents the distance from the point to the origin O, the curve r =2representsthecirclewithcenterO and radius 2. The equation r = a represents a circle with center O and radius a.
Example Sketch the curve with polar equation r = 2 cos. Solution: Plotting points we find what seems to be a circle:
Example Find the Cartesian coordinates for r = 2 cos. Proof. Solution Since x = r cos, the equation r = 2 cos becomes r = 2x r or 2x = r 2 = x 2 + y 2 or or x 2 2x + y 2 =0 (x ) 2 + y 2 =. This is the equation of a circle of radius centered at (, 0).
Cardioid Example Sketch the curve r =+sin. Thiscurveiscalledacardioid. Solution
Example Sketch the curve r = cos 2. This curve is called a four-leaved rose. Solution
Tangents to Polar Curves To find a tangent line to a polar curve r = f( ) we regard as a parameter and write its parametric equations as x = r cos = f( ) cos y = r sin = f( )sin Then, using the method for finding slopes of parametric curves and the Product Rule, we have dy dy dx = d dx d = dr d sin + r cos dr d cos r sin We locate horizontal tangents by finding the points where dy d = 0 (provided that dx d 6= 0.) Likewise, we locate vertical tangents at the points where dx d = 0 (provided that dy d 6= 0).
Example For the cardioid r = +sin find the slope of the tangent line when = 3. Solution: dr dy dx = d sin + r cos cos sin +( +sin ) cos dr d cos = r sin cos cos ( +sin )sin cos ( + 2 sin ) = 2 sin 2 sin = cos ( + 2 sin ) ( +sin )( 2 sin ) The slope of the tangent at the point where = 3 is = dy dx = 3 2 ( + p 3) ( + p 3 2 )( p = 3) = cos 3 ( + 2 sin 3 ) ( +sin 3 )( 2 sin 3 ) + p 3 (2 + p 3)( p = + p 3 p = 3) 3
The area of a region under a polar function r = f( ) isdescribed by either of the following formulas. These formulas arise from the fact that the area of a apple apple 2 portion of a circle of radius r is given by 2 ( 2 )r 2. A = R b a A = R b a 2 [f( )]2 d, 2 r2 d,