0 < < 2 there is no value of δ > 0 such that | sin(1/x1 ) − sin(1/x2 )| < Thus sin(1/x) is not uniformly continuous in the open interval (0, 1) for all x1 , x2 ∈ (0, 1) and |x1 − x2 | < δ Solution 3.6 √ √ √ First consider the function x Note that the function x + δ − x is a decreasing function of x and an increasing √ √ √ function of δ for positive x and δ Thus for any fixed δ, the maximum value of x + δ − x is bounded by δ √ √ √ Therefore on the interval (0, 1), a sufficient condition for | x − ξ| < is |x − ξ| < 2 The function x is uniformly continuous on the interval (0, 1) Consider any positive δ and Note that 1 1 − > x x+δ for 1 4δ −δ x< δ2 + 2 Thus there is no value of δ such that for all |x − ξ| < δ The function 1 x 1 1 − < x ξ is not uniformly continuous on the interval (0, 1) Solution 3.7 Let the function f (x) be continuous on a closed interval Consider the function e(x, δ) = sup |f (ξ) − f (x)| |ξ−x| 0, ∃ N s.t n > N ⇒ |an − L| < We want to show that ∀ δ > 0, ∃ M s.t m > M ⇒ |a2 − L2 | < δ n Suppose that |an − L| < We obtain an upper bound on |a2 − L2 | n |a2 − L2 | = |an − L||an + L| < (|2L| + ) n Now we choose a value of such that |a2 − L2 | < δ n (|2L| + ) = δ √ = L2 + δ − |L| Consider any fixed δ > 0 We see that since √ for = L2 + δ − |L|, ∃ N s.t n > N ⇒ |an − L| < implies that n > N ⇒ |a2 − L2 | < δ n Therefore ∀ δ > 0, ∃ M s.t m > M ⇒ |a2 − L2 | < δ n We conclude that limn→∞ a2 = L2 n 2 limn→∞ a2 = L2 does not imply that limn→∞ an = L Consider an = −1 In this case limn→∞ a2 = 1 and n n limn→∞ an = −1 95 3 If an > 0 for all n > 200, and limn→∞ an = L, then L is not necessarily positive Consider an = 1/n, which satisfies the two constraints 1 lim = 0 n→∞ n 4 The statement limx→∞ f (x) = L is equivalent to ∀ > 0, ∃ X s.t x > X ⇒ |f (x) − L| < This implies that for n > X , |f (n) − L| < ∀ > 0, ∃ N s.t n > N ⇒ |f (n) − L| < lim f (n) = L n→∞ 5 If f : R → R is continuous and limn→∞ f (n) = L, then for x ∈ R, it is not necessarily true that limx→∞ f (x) = L Consider f (x) = sin(πx) lim sin(πn) = lim 0 = 0 n→∞ n→∞ limx→∞ sin(πx) does not exist Solution 3.9 a d n (x + ∆x)n − xn (x ) = lim ∆x→0 dx ∆x  xn + nxn−1 ∆x + = lim  ∆x→0 = lim ∆x→0 nxn−1 + n(n−1) n−2 x ∆x2 2 + · · · + ∆xn − xn ∆x n(n − 1) n−2 x ∆x + · · · + ∆xn−1 2 = nxn−1 96   d n (x ) = nxn−1 dx b f (x + ∆x)g(x + ∆x) − f (x)g(x) d (f (x)g(x)) = lim ∆x→0 dx ∆x [f (x + ∆x)g(x + ∆x) − f (x)g(x + ∆x)] + [f (x)g(x + ∆x) − f (x)g(x)] = lim ∆x→0 ∆x f (x + ∆x) − f (x) g(x + ∆x) − g(x) = lim [g(x + ∆x)] lim + f (x) lim ∆x→0 ∆x→0 ∆x→0 ∆x ∆x = g(x)f (x) + f (x)g (x) d (f (x)g(x)) = f (x)g (x) + f (x)g(x) dx c Consider a right triangle with hypotenuse of length 1 in the first quadrant of the plane Label the vertices A, B, C, in clockwise order, starting with the vertex at the origin The angle of A is θ The length of a circular arc of radius cos θ that connects C to the hypotenuse is θ cos θ The length of the side BC is sin θ The length of a circular arc of radius 1 that connects B to the x axis is θ (See Figure 3.20.) Considering the length of these three curves gives us the inequality: θ cos θ ≤ sin θ ≤ θ Dividing by θ, cos θ ≤ sin θ ≤ 1 θ Taking the limit as θ → 0 gives us sin θ = 1 θ→0 θ lim 97 B θ θ cos θ A sin θ θ C Figure 3.20: One more little tidbit we’ll need to know is cos θ − 1 cos θ − 1 cos θ + 1 lim = lim θ→0 θ→0 θ θ cos θ + 1 2 cos θ − 1 = lim θ→0 θ(cos θ + 1) − sin2 θ = lim θ→0 θ(cos θ + 1) − sin θ sin θ = lim lim θ→0 θ→0 (cos θ + 1) θ 0 = (−1) 2 = 0 98 Now we’re ready to find the derivative of sin x sin(x + ∆x) − sin x d (sin x) = lim ∆x→0 dx ∆x cos x sin ∆x + sin x cos ∆x − sin x = lim ∆x→0 ∆x sin ∆x cos ∆x − 1 = cos x lim + sin x lim ∆x→0 ∆x→0 ∆x ∆x = cos x d (sin x) = cos x dx d Let u = g(x) Consider a nonzero increment ∆x, which induces the increments ∆u and ∆f By definition, ∆f = f (u + ∆u) − f (u), ∆u = g(x + ∆x) − g(x), and ∆f, ∆u → 0 as ∆x → 0 If ∆u = 0 then we have = ∆f df − → 0 as ∆u → 0 ∆u du If ∆u = 0 for some values of ∆x then ∆f also vanishes and we define = 0 for theses values In either case, ∆y = df ∆u + ∆u du 99 We divide this equation by ∆x and take the limit as ∆x → 0 df ∆f = lim dx ∆x→0 ∆x df ∆u ∆u = lim + ∆x→0 du ∆x ∆x df ∆f = lim + ∆x→0 ∆x du df du du = + (0) du dx dx df du = du dx lim ∆x→0 Thus we see that d (f (g(x))) = f (g(x))g (x) dx Solution 3.10 1 f (0) = lim → 0 | |−0 = lim → 0| | =0 The function is differentiable at x = 0 100 lim ∆x→0 ∆u ∆x 2 1+| |−1 f (0) = lim → 0 1 = lim → 0 2 (1 + | |)−1/2 sign( ) 1 1 = lim → 0 sign( ) 2 Since the limit does not exist, the function is not differentiable at x = 0 Solution 3.11 a d d d [x sin(cos x)] = [x] sin(cos x) + x [sin(cos x)] dx dx dx d = sin(cos x) + x cos(cos x) [cos x] dx = sin(cos x) − x cos(cos x) sin x d [x sin(cos x)] = sin(cos x) − x cos(cos x) sin x dx b d d [f (cos(g(x)))] = f (cos(g(x))) [cos(g(x))] dx dx d = −f (cos(g(x))) sin(g(x)) [g(x)] dx = −f (cos(g(x))) sin(g(x))g (x) d [f (cos(g(x)))] = −f (cos(g(x))) sin(g(x))g (x) dx 101 c d [f (ln x)] d 1 = − dx dx f (ln x) [f (ln x)]2 d f (ln x) dx [ln x] =− [f (ln x)]2 f (ln x) =− x[f (ln x)]2 d 1 f (ln x) =− dx f (ln x) x[f (ln x)]2 d First we write the expression in terms exponentials and logarithms, x xx = xexp(x ln x) = exp(exp(x ln x) ln x) Then we differentiate using the chain rule and the product rule d d exp(exp(x ln x) ln x) = exp(exp(x ln x) ln x) (exp(x ln x) ln x) dx dx d 1 x = xx exp(x ln x) (x ln x) ln x + exp(x ln x) dx x 1 x = xx xx (ln x + x ) ln x + x−1 exp(x ln x) x xx x = x x (ln x + 1) ln x + x−1 xx = xx x +x x−1 + ln x + ln2 x d xx x x = xx +x x−1 + ln x + ln2 x dx 102 Figure 4.1: Plot of ln |x| and 1/x Note the absolute value signs This is because this d dx ln |x| = 1 x for x = 0 In Figure 4.1 is a plot of ln |x| and Example 4.1.1 Consider I= (x2 x dx + 1)2 We evaluate the integral by choosing u = x2 + 1, du = 2x dx 1 2x dx 2 + 1)2 2 (x 1 du = 2 u2 1 −1 = 2 u 1 =− 2 + 1) 2(x I= Example 4.1.2 Consider I= tan x dx = 118 sin x dx cos x 1 x to reinforce By choosing f (x) = cos x, f (x) = − sin x, we see that the integral is I=− − sin x dx = − ln | cos x| + c cos x Change of Variable The differential of a function g(x) is dg = g (x) dx Thus one might suspect that for ξ = g(x), f (ξ) dξ = f (g(x))g (x) dx, (4.1) since dξ = dg = g (x) dx This turns out to be true To prove it we will appeal to the the chain rule for differentiation Let ξ be a function of x The chain rule is d f (ξ) = f (ξ)ξ (x), dx df dξ d f (ξ) = dx dξ dx We can also write this as df dx df = , dξ dξ dx or in operator notation, d dx d = dξ dξ dx Now we’re ready to start The derivative of the left side of Equation 4.1 is d dξ f (ξ) dξ = f (ξ) 119 Next we differentiate the right side, d dξ dx d f (g(x))g (x) dx dξ dx dx = f (g(x))g (x) dξ dx dg = f (g(x)) dg dx = f (g(x)) = f (ξ) f (g(x))g (x) dx = to see that it is in fact an identity for ξ = g(x) Example 4.1.3 Consider x sin(x2 ) dx We choose ξ = x2 , dξ = 2xdx to evaluate the integral 1 sin(x2 )2x dx 2 1 = sin ξ dξ 2 1 = (− cos ξ) + c 2 1 = − cos(x2 ) + c 2 x sin(x2 ) dx = 120 Integration by Parts The product rule for differentiation gives us an identity called integration by parts We start with the product rule and then integrate both sides of the equation d (u(x)v(x)) = u (x)v(x) + u(x)v (x) dx (u (x)v(x) + u(x)v (x)) dx = u(x)v(x) + c u (x)v(x) dx + u(x)v (x)) dx = u(x)v(x) u(x)v (x)) dx = u(x)v(x) − v(x)u (x) dx The theorem is most often written in the form u dv = uv − v du So what is the usefulness of this? Well, it may happen for some integrals and a good choice of u and v that the integral on the right is easier to evaluate than the integral on the left Example 4.1.4 Consider x ex dx If we choose u = x, dv = ex dx then integration by parts yields x ex dx = x ex − ex dx = (x − 1) ex Now notice what happens when we choose u = ex , dv = x dx 1 x ex dx = x2 ex − 2 1 2 x x e dx 2 The integral gets harder instead of easier When applying integration by parts, one must choose u and dv wisely As general rules of thumb: 121 • Pick u so that u is simpler than u • Pick dv so that v is not more complicated, (hopefully simpler), than dv Also note that you may have to apply integration by parts several times to evaluate some integrals 4.2 4.2.1 The Definite Integral Definition The area bounded by the x axis, the vertical lines x = a and x = b and the function f (x) is denoted with a definite integral, b f (x) dx a The area is signed, that is, if f (x) is negative, then the area is negative We measure the area with a divide-and-conquer strategy First partition the interval (a, b) with a = x0 < x1 < · · · < xn−1 < xn = b Note that the area under the curve on the subinterval is approximately the area of a rectangle of base ∆xi = xi+1 − xi and height f (ξi ), where ξi ∈ [xi , xi+1 ] If we add up the areas of the rectangles, we get an approximation of the area under the curve See Figure 4.2 n−1 b f (x) dx ≈ a f (ξi )∆xi i=0 As the ∆xi ’s get smaller, we expect the approximation of the area to get better Let ∆x = max0≤i≤n−1 ∆xi We define the definite integral as the sum of the areas of the rectangles in the limit that ∆x → 0 n−1 b f (x) dx = lim a ∆x→0 f (ξi )∆xi i=0 The integral is defined when the limit exists This is known as the Riemann integral of f (x) f (x) is called the integrand 122 f(ξ1 ) a x1 x2 x3 ∆ x n-2 x n-1 b xi Figure 4.2: Divide-and-Conquer Strategy for Approximating a Definite Integral 4.2.2 Properties Linearity and the Basics Because summation is a linear operator, that is n−1 n−1 (cfi + dgi ) = c i=0 n−1 fi + d i=0 gi , i=0 definite integrals are linear, b b (cf (x) + dg(x)) dx = c a b f (x) dx + d a g(x) dx a One can also divide the range of integration b c f (x) dx = a b f (x) dx + a f (x) dx c 123 We assume that each of the above integrals exist If a ≤ b, and we integrate from b to a, then each of the ∆xi will be negative From this observation, it is clear that b a f (x) dx = − a f (x) dx b If we integrate any function from a point a to that same point a, then all the ∆xi are zero and a f (x) dx = 0 a Bounding the Integral Recall that if fi ≤ gi , then n−1 n−1 fi ≤ i=0 gi i=0 Let m = minx∈[a,b] f (x) and M = maxx∈[a,b] f (x) Then n−1 (b − a)m = n−1 m∆xi ≤ i=0 n−1 f (ξi )∆xi ≤ i=0 M ∆xi = (b − a)M i=0 implies that b (b − a)m ≤ f (x) dx ≤ (b − a)M a Since n−1 n−1 fi ≤ i=0 we have |fi |, i=0 b b f (x) dx ≤ a |f (x)| dx a 124 Mean Value Theorem of Integral Calculus Let f (x) be continuous We know from above that b (b − a)m ≤ f (x) dx ≤ (b − a)M a Therefore there exists a constant c ∈ [m, M ] satisfying b f (x) dx = (b − a)c a Since f (x) is continuous, there is a point ξ ∈ [a, b] such that f (ξ) = c Thus we see that b f (x) dx = (b − a)f (ξ), a for some ξ ∈ [a, b] 4.3 The Fundamental Theorem of Integral Calculus Definite Integrals with Variable Limits of Integration Consider a to be a constant and x variable, then the function F (x) defined by x F (x) = f (t) dt a 125 (4.2) is an anti-derivative of f (x), that is F (x) = f (x) To show this we apply the definition of differentiation and the integral mean value theorem F (x + ∆x) − F (x) ∆x x+∆x x f (t) dt − a f (t) dt a = lim ∆x→0 ∆x x+∆x f (t) dt = lim x ∆x→0 ∆x f (ξ)∆x = lim , ξ ∈ [x, x + ∆x] ∆x→0 ∆x = f (x) F (x) = lim ∆x→0 The Fundamental Theorem of Integral Calculus Let F (x) be any anti-derivative of f (x) Noting that all anti-derivatives of f (x) differ by a constant and replacing x by b in Equation 4.2, we see that there exists a constant c such that b f (x) dx = F (b) + c a Now to find the constant By plugging in b = a, a f (x) dx = F (a) + c = 0, a we see that c = −F (a) This gives us a result known as the Fundamental Theorem of Integral Calculus b f (x) dx = F (b) − F (a) a We introduce the notation [F (x)]b ≡ F (b) − F (a) a 126 Example 4.3.1 π sin x dx = [− cos x]π = − cos(π) + cos(0) = 2 0 0 4.4 4.4.1 Techniques of Integration Partial Fractions A proper rational function p(x) p(x) = q(x) (x − a)n r(x) Can be written in the form p(x) = (x − α)n r(x) a0 a1 an−1 + + ··· + n n−1 (x − α) (x − α) x−α + (· · · ) where the ak ’s are constants and the last ellipses represents the partial fractions expansion of the roots of r(x) The coefficients are 1 dk p(x) ak = k! dxk r(x) x=α Example 4.4.1 Consider the partial fraction expansion of 1 + x + x2 (x − 1)3 The expansion has the form a0 a1 a2 + + 3 2 (x − 1) (x − 1) x−1 127 The coefficients are 1 (1 + x + x2 )|x=1 = 3, 0! 1 d a1 = (1 + x + x2 )|x=1 = (1 + 2x)|x=1 = 3, 1! dx 1 d2 1 a2 = (1 + x + x2 )|x=1 = (2)|x=1 = 1 2 2! dx 2 a0 = Thus we have 1 + x + x2 3 3 1 = + + 3 3 2 (x − 1) (x − 1) (x − 1) x−1 Example 4.4.2 Suppose we want to evaluate 1 + x + x2 dx (x − 1)3 If we expand the integrand in a partial fraction expansion, then the integral becomes easy 1 + x + x2 dx = (x − 1)3 3 3 1 + + dx (x − 1)3 (x − 1)2 x − 1 3 3 =− − + ln(x − 1) 2 2(x − 1) (x − 1) Example 4.4.3 Consider the partial fraction expansion of 1 + x + x2 x2 (x − 1)2 The expansion has the form a0 a1 b0 b1 + + + 2 2 x x (x − 1) x−1 128 The coefficients are 1 1 + x + x2 = 1, 0! (x − 1)2 x=0 1 d 1 + x + x2 = a1 = 1! dx (x − 1)2 x=0 1 1 + x + x2 = 3, b0 = 0! x2 x=1 1 d 1 + x + x2 b1 = = 1! dx x2 x=1 a0 = Thus we have 2(1 + x + x2 ) 1 + 2x − (x − 1)2 (x − 1)3 1 + 2x 2(1 + x + x2 ) − x2 x3 = 3, x=0 = −3, x=1 1 + x + x2 1 3 3 3 = 2+ + − 2 (x − 1)2 2 x x x (x − 1) x−1 If the rational function has real coefficients and the denominator has complex roots, then you can reduce the work in finding the partial fraction expansion with the following trick: Let α and α be complex conjugate pairs of roots of the denominator p(x) = (x − α)n (x − α)n r(x) a0 a1 an−1 + + ··· + (x − α)n (x − α)n−1 x−α a0 a1 an−1 + + + ··· + n n−1 (x − α) (x − α) x−α + (· · · ) Thus we don’t have to calculate the coefficients for the root at α We just take the complex conjugate of the coefficients for α Example 4.4.4 Consider the partial fraction expansion of 1+x x2 + 1 129 The expansion has the form a0 a0 + x−i x+i The coefficients are 1 1 1+x = (1 − i), 0! x + i x=i 2 1 1 a0 = (1 − i) = (1 + i) 2 2 a0 = Thus we have 1+x 1−i 1+i = + 2+1 x 2(x − i) 2(x + i) 4.5 Improper Integrals If the range of integration is infinite or f (x) is discontinuous at some points then integral b a f (x) dx is called an improper Discontinuous Functions If f (x) is continuous on the interval a ≤ x ≤ b except at the point x = c where a < c < b then b a c−δ f (x) dx = lim + δ→0 a b f (x) dx + lim + →0 f (x) dx c+ provided that both limits exist Example 4.5.1 Consider the integral of ln x on the interval [0, 1] Since the logarithm has a singularity at x = 0, this 130 is an improper integral We write the integral in terms of a limit and evaluate the limit with L’Hospital’s rule 1 1 ln x dx = lim δ→0 0 ln x dx δ = lim[x ln x − x]1 δ δ→0 = 1 ln(1) − 1 − lim(δ ln δ − δ) δ→0 = −1 − lim(δ ln δ) δ→0 = −1 − lim δ→0 = −1 − lim δ→0 ln δ 1/δ 1/δ −1/δ 2 = −1 Example 4.5.2 Consider the integral of xa on the range [0, 1] If a < 0 then there is a singularity at x = 0 First assume that a = −1 1 xa dx = lim + δ→0 0 xa+1 a+1 1 δ 1 δ a+1 = − lim a + 1 δ→0+ a + 1 This limit exists only for a > −1 Now consider the case that a = −1 1 x−1 dx = lim [ln x]1 δ + δ→0 0 = ln(0) − lim ln δ + δ→0 This limit does not exist We obtain the result, 1 xa dx = 0 1 , a+1 131 for a > −1 Infinite Limits of Integration If the range of integration is infinite, say [a, ∞) then we define the integral as ∞ α f (x) dx, f (x) dx = lim α→∞ a a provided that the limit exists If the range of integration is (−∞, ∞) then ∞ a f (x) dx = lim α→−∞ −∞ β f (x) dx + lim β→+∞ α f (x) dx a Example 4.5.3 ∞ 1 ∞ ln x dx = x2 ln x 1 −1 = ln x x = lim x→+∞ = lim x→+∞ d −1 dx x ∞ dx ∞ −1 1 dx x x 1 1 ∞ ln x 1 − − x x 1 1/x 1 − − lim + 1 x→∞ x 1 − =1 Example 4.5.4 Consider the integral of xa on [1, ∞) First assume that a = −1 ∞ β xa+1 x dx = lim β→+∞ a + 1 1 β a+1 1 = lim − β→+∞ a + 1 a+1 a 1 132 ... approximate sin (1) , 13 15 1? ?? + ≈ 0.8 41 6 67 12 0 10 7 To see that this has the required accuracy, sin (1) ≈ 0.8 41 4 71 Solution 3 .19 Expanding the terms in the approximation in Taylor series, ∆x3 ∆x4 ∆x2 f... Example 4. 4 .1 Consider the partial fraction expansion of + x + x2 (x − 1) 3 The expansion has the form a0 a1 a2 + + (x − 1) (x − 1) x? ?1 127 The coefficients are (1 + x + x2 )|x =1 = 3, 0! d a1 = (1 +... − x→0 10 9 x =0 c ln lim x→+∞ 1+ x x = lim x→+∞ = lim x→+∞ = lim x→+∞ = lim ln + 1/ x 1+ x→+∞ 1+ =1 Thus we have lim x→+∞ 1+ 11 0 x x ? ?1 x − x2 ? ?1/ x2 x→+∞ = lim x ln 1+ x x ln + x x = e x ? ?1 d It