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Chapter 5 Integration 1 5.1 Estimating with Finite Sums 2 Riemann Sums Approximating area bounded by the graph between [a,b] 3 Area is approximately given by f(c1)Dx1 + f(c2)Dx2+ f(c3)Dx3+ … + f(cn)Dxn Partition of [a,b] is the set of P = {x0, x1, x2, … xn-1, xn} a = x0< x1< x2 …< xn-1 < xn=b cn[xn-1, xn] ||P|| = norm of P = the largest of all subinterval width 4 Riemann sum for f on [a,b] Rn = f(c1)Dx1 + f(c2)Dx2+ f(c3)Dx3+ … +f(cn)Dxn 5 Figure 5.4 Let the true value of the area is R Two approximations to R: cn= xn corresponds to case (a). This under estimates the true value of the area R if n is finite. cn= xn-1 corresponds to case (b). This over estimates the true value of the area S if n is finite. go back 6 Limits of finite sums Example 5 The limit of finite approximation to an area Find the limiting value of lower sum approximation to the area of the region R below the graphs f(x) = 1 - x2 on the interval [0,1] based on Figure 5.4(a) 7 Solution Dxk = (1 - 0)/n= 1/n ≡Dx; k = 1,2,…n Partition on the x-axis: [0,1/n], [1/n, 2/n],…, [(n-1)/n,1]. ck = xk = kDx = k/n The sum of the stripes is Rn = Dx1 f(c1) + Dx2 f(c2) + Dx3 f(c3) + …+ Dxn f(cn) = Dx f(1/n) + Dx f(2/n) + Dx f(3/n) + …+ Dxn f(1) = ∑k=1n Dx f(kDx) = Dx ∑k=1n f (k/n) = (1/ n) ∑k=1n [1 - (k/n)2] = ∑k=1n 1/ n - k2/n3 = 1 – (∑k=1n k2)/ n3 = 1 – [(n) (n+1) (2n+1)/6]/ n3 = 1 – [2 n3 + 3 n2+n]/(6n3) ∑k=1n k2 = (n) (n+1) (2n+1)/6 8 Taking the limit of n → ∞ 2n3 + 3n2 + n lim Rn = R = 1 = 1 2 / 6 = 2 / 3 3 n 6n The same limit is also obtained if cn = xn-1 is chosen instead. For all choice of cn [xn-1,xn] and partition of P, the same limit for S is obtained when n ∞ 9 5.3 The Definite Integral 10 11 “The integral from a to b of f of x with respect to x” 12 The limit of the Riemann sums of f on [a,b] converges to the finite integral I n lim || P|| 0 f (c )Dx k =1 k k b = I = f ( x)dx a We say f is integrable over [a,b] Can also write the definite integral as b b b a a a I = f ( x)dx = f (t )dt = f (u )du b = f (what ever) d (what ever) a The variable of integration is what we call a ‘dummy variable’ 13 Question: is a non continuous function integrable? 14 Integral and nonintegrable functions Example 1 A nonintegrable function on [0,1] 1, if x is rational f ( x) = 0, if x is irrational Not integrable 15 Properties of definite integrals 16 17 18 Example 3 Finding bounds for an integral Show that the value of is less than 3/2 1 0 1 + cos xdx Solution Use rule 6 Max-Min Inequality 19 Area under the graphs of a nonnegative function 20 Example 4 Area under the line y = x b Compute 0 xdx (the Riemann sum) and find the area A under y = x over the interval [0,b], b>0 21 By geometrical consideration: Solution A=(1/2)highwidth= (1/2)bb= b2/2 Choose partition of n subinterval with equal width: 0 = x0 , x1, x2 , xn = b, Dxk = xk xk 1 = Dx = b / n Riemann sum: n n lim Dx f (ck ) = lim Dx f ( xk ) n n k =1 n k =1 n = lim Dx xk = lim Dx k Dx n n k =1 k =1 b = lim ( Dx ) k = lim n n n k =1 2 n 2 n k k =1 b n ( n + 1) b n ( n + 1) = lim = lim n n n n 2 2 b2 1 b2 = lim 1 + = n 2 n 2 2 2 22 Using the additivity rule for definite integration: b a b 0 0 a xdx = xdx + xdx b b a b2 a 2 xdx = xdx xdx = , a b 2 2 a 0 0 Using geometry, the area is the area of a trapezium A= (1/2)(b-a)(b+a) = b2/2 - a2/2 Both approaches to evaluate the area agree 23 One can prove the following Riemannian sum of the functions f(x)=c and f(x)= x2: 24 Average value of a continuous function revisited Average value of nonnegative continuous function f over an interval [a,b] is f (c1 ) + f (c2 ) + n f (cn ) 1 n = f (ck ) n k =1 Dx n 1 n = f (ck ) = Dxf (ck ) b a k =1 b a k =1 In the limit of n ∞, the average = b 1 f ( x)dx ba a 25 26 27 Example 5 Finding average value Find the average value of f ( x) = 4 x 2 over [-2,2] 28 5.4 The Fundamental Theorem of Calculus 29 Mean value theorem for definite integrals 30 31 32 Example 1 Applying the mean value theorem for integrals Find the average value of f(x)=4-x on [0,3] and where f actually takes on this value as some point in the given domain. Solution Average = 5/2 Happens at x=3/2 33 Fundamental theorem Part 1 x Define a function F(x): F ( x ) = f (t )dt a x,a I, an interval over which f(t) > 0 is integrable. The function F(x) is the area under the graph of f(t) over [a,x], x > a 34 35 Fundamental theorem Part 1 (cont.) F ( x + h) F ( x) = x +h f ( t ) dt x F ( x + h ) F ( x ) 1 x +h = f ( t ) dt = f ( c ) ; x c x + h h h x mean value theorem F ( x + h) F ( x) lim = F ( x ) = f ( x ) h 0 h The above result holds true even if f is not positive definite over [a,b] 36 37 Note: Convince yourself that (i) F(x) is an antiderivative of f(x) (ii)f(x) is a derivative of F(x) d/dx f(x) is a derivative of F(x) because F'(x)= f(x) x F ( x ) = f (t )dt a ( f(x) = F'(x) )dx F(x) is an antiderivative x of f(x) because F ( x) = a f (t )dt 38 The main use of theorem 4 is … d It tells us that dx ( x a ) f (t )dt = f ( x ) In pragmatic terms, if a function is expressed in terms of an integral of the form x F ( x ) = f (t )dt a d then the derivative of F(x), F ( x ) , is simply dx f(x) Example 3 Applying the fundamental theorem Use the fundamental theorem to find x x d (a ) cos tdt dx a d 1 (b) dt 2 dx a 1 + t 5 x2 dy (c ) if y = 3t sin tdt dx x dy (d ) if y = cos tdt dx 1 40 Solution for (d): you have to invoke chain rule x 2 d F ( x ), where F ( x ) cos tdt dx 1 Chain rule says if F(x)= (f◦u)(x)= f [u(x)], d d d d d F ( x) = ( f u ) ( x) = f [u( x )] = f (u ) u( x ) dx dx dx du dx 41 Solution for (d): you have to invoke chain rule x2 F ( x ) = cos t dt is a composite function of the form F(x)=f [u(x)] 1 F ( x ) = f [u ( x )], where u f (u ) = cos t dt , u ( x ) = x 2 so that 1 u d 2 d d d d F ( x) = f (u ) u ( x ) = cos t dt ( x ) dx du dx du 1 dx u d 2 = cos t dt 2 x = cos u 2 x = 2 x cos x ( ) du 1 42 Example 4 Constructing a function with a given derivative and value Find a function y = f(x) on the domain (-p /2, p/2) with derivative dy/dx = tan x that satisfies f(3)=5. The strategy: Use the fundamental theorem of calculus. Think along this line: find a function F(x) of the form x F ( x ) = q(t )dt such that a d F ( x ) = q( x ), with q( x ) = tan x dx Example 4 (Cont. 1) Solution x dF = tan x. Stage 1: If F ( x ) = tan tdt , then a dx Stage 2: construct the function f(x) using F(x), and then try to make f(x) so constructed fulfills the condition of f(3)=5. The way to construct f(x) from F(x) is obviously dy y = f ( x ) = F ( x ) + constant (so that = tan x ) dx x = tan tdt + constant a 44 Example 4 (Cont. 2) x f ( x ) = tan tdt + constant a Find the values of a and constant so that f(3)=5 This can be done by choosing a = 3, constant =5. Verify this: x =3 f (3) = tan tdt + 5=0+5=5 a =3 So, finally, the function we are seeking is x f ( x ) = tan tdt + 5 3 45 Fundamental theorem, part 2 (The evaluation theorem) 46 To calculate the definite integral of f over [a,b], do the following 1. Find an antiderivative F of f, and 2. Calculate the number b f ( x)dx = F (b) F (a) = a 47 To summarise x d dF ( x) f (t )dt = = f ( x) dx a dx dF (t ) a dt dt = a f (t )dt = F ( x) F (a) x x 48 Example 5 Evaluating integrals p ( a ) cos xdx 0 0 ( b) p sec x tan xdx /4 4 4 3 x 2 dx (c) x 2 1 49 Example 7 Canceling areas Compute (a) the definite integral of f(x) over [0,2p] (b) the area between the graph of f(x) and the x-axis over [0,2p] 50 Example 8 Finding area using antiderivative Find the area of the region between the xaxis and the graph of f(x) = x3 - x2 – 2x, -1 ≤ x ≤ 2. Solution First find the zeros of f. f(x) = x(x+1) (x-2) 51 52 5.5 Indefinite Integrals and the Substitution Rule 53 Note The indefinite integral of f with respect to x, f ( x)dx is a function plus an arbitrary constant b A definite integral f ( x)dx is a number. a 54 Antiderivative and indefinite integral in terms of variable x If F(x) is an antiderivative of f(x), d F ( x) = f ( x) dx the indefinite integral of f(x) is f ( x )dx = F ( x ) + C 55 A useful mnemonic d ( dx Example: + constant ) = dx = + constant d (tan x + constant) = sec 2 x dx 2 sec x dx = tan x + constant 56 Antiderivative and indefinite integral with chain rule d F ( x ) = f ( x ) , i.e., F ( x ) antiderivative of f ( x ), dx d F [u ] = f ( u ) , where u = u ( x ) . du d Applying chain rule to F u ] : dx du ( x ) dF ( u ) du d d du F u ] = = f (u ) F u ] = f (u ) dx dx du dx dx dx du In other words, F u ] is an antiderivative of f ( u ) , so that we can write dx du dx f ( u ) dx = F u ] + C 57 The power rule in integral form u n +1 d u n +1 n du n du u =u +C dx = dx n + 1 dx dx n +1 n du n du n u dx = u dx = u dx dx du du differential of u ( x), du is du = dx dx 58 Example 1 Using the power rule The strategy is to convert the integral into the form 1 + y 2 y dy = ? 2 du Let u = 1 + y , du = dy = 2 ydy. dy 2 1 + y 2 2 y dy = u du = ... 59 Example 2 Adjusting the integrand by a constant 4t 1 dt = ? Let u = 4t 1, du = 4dt , 1 1 4t 1 dt = u dt = u 4dt = u du = ... 4 4 60 Substitution: Running the chain rule backwards du dx = f (u )du dx Used to find the integration with the integrand in the form of the product of f [ g ( x )] g '( x ) let u = g ( x); f [ g ( x)] g ( x)dx = f (u ) f [ g ( x)] g '( x)dx = f (u)du f (u ) du 61 Example 3 Using substitution cos(7 x + 5) dx = cos u u 1 du 7 du 1 1 = sin u + C = sin ( 7 x + 5 ) + C 7 7 7 62 Example 4 Using substitution ? = dx x sin x 3 2 u = x ; du = 3x dx 2 3 sin x 3 u 1 1 x dx = sin u du = cos u + C 3 3 1 2 3 du 1 3 = cos x + C 3 63 Example 5 Using Identities and substitution 1 2 2 dx = sec 2 x dx = sec = cos2 2 x 2ux dx 1 2 du 1 1 1 2 sec u du = tan u + C = tan 2 x + C 2 d 2 2 du tan u 64 Example 6 Using different substitutions 2z 3 z2 +1 dz = ( z + 1) 2 u 1/ 3 1/ 3 2 zdz = u 1/ 3du =... du 65 The integrals of 2 sin x and 2 cos x Example 7 1 sin x dx = 2 1 cos 2 x dx x 1 = cos 2 x dx 2 2 u 1 2 2 du x 1 = cos udu =... 2 4 66 The integrals of 2 sin x and 2 cos x Example 7(b) 1 cos x dx = 2 cos 2 x + 1 dx = ... 2 67 Example 8 Area beneath the curve y=sin2 x For Figure 5.24, find (a) the definite integral of y(x) over [0,2p]. (b) the area between the graph and the xaxis over [0,2p]. 68 5.6 Substitution and Area Between Curves 69 Substitution formula x =b let u = g ( x); x =a f [ g ( x)] g ( x)dx = x =b x =a u = g (b ) du f [u ] dx = f (u )du dx u=g (a) 70 Example 1 Substitution Evaluate 1 2 3 3 x x + 1 dx 1 u ( x =1) x =1 x =1 x 3 + 1 3 x 2 dx = u1/ 2 du u1/ 2 du =... u ( x =1) 71 Example 2 Using the substitution formula x =p / 2 p cot x csc 2 xdx = ? x= / 4 u2 cot x csc xdx = cotu x csc duxdx = udu = 2 + c 2 2 cot 2 x = +c 2 1 cot x cot x 1 2 2 = = cot (p / 4 ) cot (p / 2 ) = p / 4 cot x csc xdx = 2 2 p /2 2 2 p /4 1 0 p /2 2 p /2 2 p /4 2 72 Definite integrals of symmetric functions 73 74 Example 3 Integral of an even function 2 Evaluate 4 2 x 4 x + 6 ) dx ( 2 Solution: f ( x) = x 4 4 x 2 + 6; f ( x) = ( x ) 4 ( x ) + 6 = x 4 4 x 2 + 6 = f ( x) 4 2 even function How about integration of the same function from x=-1 to x=2 75 Area between curves 76 n n k =1 k =1 A DAk = Dxk ( f (ck ) g (ck ) A = lim || P|| 0 n Dx ( f (c ) g (c ) = f ( x) g ( x)] dx k =1 b k k k a 77 78 Example 4 Area between intersecting curves Find the area of the region enclosed by the parabola y = 2 – x2 and the line y = -x. DA = ( f ( x) g ( x) ) Dx n A = lim DAk = dA; n A= A 0 k =1 b2 a 1 [ f ( x) g ( x)]dx 2 x2 = 2 1 (2 x 2 x x ) dx = ... 79 Example 5 Changing the integral to match a boundary change Find the area of the shaded region Area = A + B A= 2 B= 4 0 2 xdx; x ( x 2)dx 80 n n k =1 k =1 A DAk = Dyk ( f (ck ) g (ck ) A = lim || P|| 0 n Dy ( f (c ) g (c ) = f ( y) g ( y)] dy k =1 d k k k c DAk 81 Example 6 Find the area of the region in Example 5 by integrating with respect to y DA = ( f ( y) g ( y)) Dy n A = lim DAk = n k =1 y =4 y =0 [ f ( y ) g ( y )]dy = ( y + 2 ) ( y 2 ) dy = ... 2 0 82