MA1102R-Review.pdf - MA1102R CALCULUS Review Wang Fei...

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Unformatted text preview: MA1102R CALCULUS Review Wang Fei [email protected] Department of Mathematics Office: S17-06-16 Tel: 6516-2937 Chapter 0: Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Chapter 1: Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Chapter 2: Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Chapter 3: Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Chapter 4: Applications of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Chapter 5: Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Chapter 6: Transcendental Functions and Techniques of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 Chapter 7: Applications of Integration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Chapter 8: Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Remarks to the Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 1 Chapter 0:  What is a function f : A → B ?  Each a ∈ A is assigned to a unique b ∈ B , f (a) = b.    Domain, codomain, range, graph. Algebras of functions, composite of functions. Classes of functions.    Functions Polynomials, rational functions, root functions, trigonometric functions. Exponential functions, logarithmic functions. Special Classes of Functions.   Even function, odd function. Increasing function: a < b ⇒ f (a) < f (b); Decreasing function: a < b ⇒ f (a) > f (b).  Increasing/Decreasng Test. 2 / 65 Chapter 1:  lim f (x) = L. Limit of Function.   x → a ⇒ f (x) → L. For any ǫ > 0, there is a δ > 0 s.t. 0 < |x − a| < δ ⇒ |f (x) − L| < ǫ. Limit Laws:  x→a Intuitive definition: Precise definition:   Limits Suppose lim f (x) = L and lim g(x) = M . x→a lim cf (x) = cL, lim (f (x) ± g(x)) = L ± M , x→a  x→a lim f (x)g(x) = LM , x→a  x→a lim x→a f (x) L = (M 6= 0). g(x) M Other Limits:    Infinite Limit: Limit at Infinity: lim f (x) = ±∞. x→a One-sided Limit: lim f (x) = L or ± ∞. x→±∞ lim f (x), lim− f (x) = L or ± ∞. x→a+ x→a 3 / 65 2 Chapter 1:  Limits How to Evaluate the Limit of Function?  Direct Substitution:   (Refer to Chapter 3 for continuity) f is continuous at a ⇔ lim f (x) = f (a). x→a f (x) = g(x) for all x near a ⇒ lim f (x) = lim g(x). x→a    x→a (x − 1)(x − 2) . (x − √ 1)(x + 2) x+1−1 Rationalization: lim . x→0 x lim Cancelation: x→1   f is continuous ⇒ lim f (g(x)) = f lim g(x) . x→a  x→a Indeterminate Forms: 0 ∞ 0 0 , 1 , ∞ : (Refer to Chap. 6) lim f (x)g(x) = lim eg(x)·ln f (x) x→a   = exp lim (g(x) · ln f (x)) . x→a x→a Chapter 1:  4 / 65 Limits How to Evaluate the Limit of Function? (Continued)  Left- and right-limit.   lim f (x) = lim− f (x) = L ⇔ lim f (x) = L. x→a+ Squeeze Theorem.  f (x) ≤ g(x) ≤ h(x) and lim f (x) = lim h(x) = L x→a ⇒ lim g(x) = L x→a  x→a x→a l’Hôpital’s rule  x→a (L can be a number or ±∞). (Refer to Chapter 4) Suppose lim f (x) = 0 and lim g(x) = 0 x→a   x→a Or lim |f (x)| = ∞ and lim |g(x)| = ∞. x→a x→0 f ′ (x) f (x) = lim ′ if RHS exists or equals ±∞. lim x→a g (x) x→a g(x) (Refer to Chapter 3 for derivative) 5 / 65 3 Examples   4 5 − . (11/12 Semester 1 Q1(a))  lim x→1 1 − x5 1 − x4   4 5(1 − x4 ) − 4(1 − x5 ) 5 − = lim lim x→1 x→1 1 − x5 1 − x4 (1 − x5 )(1 − x4 ) −20x3 + 20x4 = lim x→1 −4x3 − 5x4 + 9x8 −60x2 + 80x3 = lim x→1 −12x2 − 20x3 + 72x7 −60 + 80 = −12 − 20 + 72 20 1 = = . 40 2 6 / 65 Examples √ 2x2 + 5  lim √ . (2014/2015 Semester 1 Q2(a)). x→∞ 3 x3 − 1 √ √ √ 2x2 + 5 2x2 + 5/ x2 √ lim √ = lim √ x→∞ 3 x2 − 1 x→∞ 3 x3 − 1/ 3 x3 q 2 + x52 = lim q x→∞ 3 1 − x13 √ 2 √ = √ = 2. 3 1 √ 2x2 + 5 Exercise. Find lim √ . 3 x→−∞ x3 − 1 7 / 65 4 Examples  sec2 x − 2 tan x . (09/10 Semester 1 Q1(a)) x→π/4 1 + cos 4x lim 2 sec2 x tan x − 2 sec2 x sec2 x − 2 tan x = lim lim x→π/4 x→π/4 1 + cos 4x −4 sin 4x tan x − 1 = − lim sec2 x · lim x→0 2 sin 4x x→π/4 √ sec2 x = −( 2)2 · lim x→π/4 8 cos 4x √ 2 ( 2) 1 = −2 · = . 8 · (−1) 2 8 / 65 Examples √ π 3 + x2 + x sin . (11/12 Semester 2, Q2(a))  lim x x→0+ x π  For all x > 0, −1 ≤ sin ≤ 1. Then x √ √ π − x3 + x2 + x ≤ x3 + x2 + x sin x √ 3 2 ≤ x + x + x. √   lim+ − x3 + x2 + x = 0. x→0 √  lim+ x3 + x2 + x = 0. x→0 By Squeeze Theorem,  lim+ x→0 √ x3 + x2 + x sin π exists and equals 0. x 5 9 / 65 Examples 1  lim x ln(x3 +1) . (10/11 Semester 1 Q1(a)). x→∞ lim x 1 ln(x3 +1) x→∞  ln x = lim exp x→∞ ln(x3 + 1)   ln x = exp lim x→∞ ln(x3 + 1) # "  1 1 x 2 x→∞ 3x x3 +1 = exp lim 1 + x−3 = exp lim x→∞ 3   √ 1 = 3 e. = exp 3   10 / 65 Examples   2 + e1/x sin x . (10/11 Semester 1 Q1(b)) +  lim x→0 1 + e4/x |x| lim x→0+ 2 + e1/x sin x + |x| 1 + e4/x ! = lim x→0+ sin x 2 + e1/x + lim 4/x + x x→0 1+e − x12 e1/x +1 − x42 e4/x 1 + 1 = 0 + 1 = 1. = lim x→0+ 4e3/x = lim x→0+ lim x→0− 2 + e1/x sin x + |x| 1 + e4/x ! = lim x→0− =  Therefore, lim x→0  2 + e1/x sin x + 1 + e4/x |x|  2 + e1/x sin x + lim 1 + e4/x x→0− −x 2+0 + (−1) = 1. 1+0 exists and equals 1. 11 / 65 6 Chapter 2:  Continuous Functions Definition of Continuity.  f is continuous at a if lim f (x) = f (a). x→a  Algebra of Continuous Functions.   Examples of Continuous Functions.    f and g continuous ⇒ cf , f ± g, f g, f /g, f ◦ g continuous. Rational function, root function, trigonometric function, Exponential function, logarithmic function. Intermediate Value Theorem.  Let f be continuous on [a, b], and N between f (a), f (b). Then there is a c ∈ (a, b) such that f (c) = N .  Show that there are At Least · · · roots.  For At Most part, use Mean Value Theorem. 12 / 65 Examples  Let f be a nonconstant continuous function on [a, b], (a < b). Prove that the range of f is a finite closed interval. (08/09 S2)  By Extreme Value Theorem,   f attains the maximum M at some u ∈ [a, b]; f attains the minimum m at some v ∈ [a, b]. ∴ m ≤ f (x) ≤ M for all x ∈ [a, b]. f nonconstant ⇒ M 6= m ⇒ u 6= v .  By Intermediate Value Theorem,  For any N with m < N < M ,  there exists c between u and v such that f (c) = N . Note that f (u) = M and f (v) = m. ∴ The range of f is [m, M], a finite closed interval. 13 / 65 7 Examples  Suppose f is continuous and f (f (x)) = x for all x ∈ R. Show that f has at least one fixed point. (08/09 Sem 1 Q8)  Take a ∈ R and let b = f (a). Then f (b) = a.  If a = b, then a is a fixed point of f .  If a 6= b, we may assume that a < b. f has a fixed point ⇔ f (x) = x has a real root ⇔ f (x) − x = 0 has a real root  Define g(x) = f (x) − x. Then g is continuous on [a, b], and   g(a) = f (a) − a = b − a > 0, g(b) = f (b) − b = a − b < 0. By IVT, g(c) = 0 for some c ∈ (a, b). That is, f (c) = c. So c is a fixed point of f . 14 / 65 Chapter 3:  Derivatives Definition of Derivative. f (x) − f (a) f (a + h) − f (a) . = lim h→0 x−a h ′  f is Differentiable at a if f (a) exists.  f ′ (a) = lim x→a   f is differentiable at a ⇒ f is continuous at a. Differentiable Functions.      Power functions: (xa )′ = axa−1 . Trigonometric functions: (sin x)′ = cos x, . . . . Exponential function: (ax )′ = ax ln a, Logarithmic function: (ln x)′ = 1/x. How to Differentiate Functions?   Definition. Differentiation formulas: (f ± g)′ = f ′ ± g ′ , (f g)′ = f ′ g + f g ′ , (f /g)′ = (f ′g − f g ′)/g 2, (f ◦ g)′ = (f ′ ◦ g)(g ′). 15 / 65 8 Chapter 3:  Derivatives How to Differentiate Functions? (Continued)   Implicit Differentiation.  Differentiate f (x, y) = 0 with respect to x.  Then solve Fundamental Theorem of Calculus: (Refer to Chap. 6)   dy from the obtained equation. dx d f is continuous ⇒ dx Z x f (t) dt = f (x). a Logarithmic Differentiation. (Ref: Chap. 7)  y = f1 (x) · · · fk (x):   ln |y| = ln |f1 (x)| + · · · + ln |fk (x)|; y = f (x)g(x) :  ln y = g(x) ln f (x). 16 / 65 Examples  Suppose f is continuous on R and differentiable on R \ {a}. If lim f ′ (x) exists, show f is differentiable at a. (08/09 S2) x→a   f ′ (x) f (x) − f (a) = lim exists. x→a x→a x−a 1 f ′ (a) = lim Evaluate   dy if y = x(sin x)cos x . (10/11 S2 Q3(a)) dx x=π/2 ln y = ln x + cos x ln sin x. 1  cos x  1 dy . = + − sin x ln sin x + cos x  y dx x sin x Let x = π/2. Then y = (π/2)10 = π/2.  dy = dx x=π/2  02 1 − 1 · ln 1 + π/2 1  · π = 1. 2 17 / 65 9 Examples  If F (x) = Z 2 √ " 2 x Z t4 16 √ # 1 + u4 du dt, find F ′′ (1). (11/12 S2) u √ Z 16x2 √ 1 + u4 1 + u4 1 du = √ du u x 16 u 16   Z 16x2 √ Z 16x2 √ 1 + u4 1 + u4 1 d 1 d ′′ √ F (x) = du + √ du dx x u x dx 16 u 16 p Z 16x2 √ 1 + (16x2 )4 1 1 + u4 1 . du + √ (32x) = − x−3/2 2 u x 16x2 16 p 2 √ Z 1 + (16 · 12 )4 1 −3/2 16·1 1 + u4 1 ′′ F (1) = − · 1 du + √ (32) 2 u 16 · 12 1 16 √ √ 65537 = 2 65537. = 0 + 32 · 16 √ F (x) = (2 x)′ ′ Z √ (2 x)4 18 / 65 Chapter 4:  The Extreme Values. (Maximum and Minimum)    Global max (min): f (c) ≥ f (x) (≤) for all x. Local max (min): f (c) ≥ f (x) (≤) for all x near c. Extreme Value Theorem:   Continuous function on finite closed interval attains the global maximum and global minimum. Critical Number and Fermat’s Theorem:  f has a local max or min at c ⇒ c is a critical number,   Applications of Differentiation that is, f ′ (c) does not exist or f ′ (c) = 0. Closed Interval Method: Suppose f is continuous on [a, b]. 1. Differentiate f to find critical numbers of f on (a, b). 2. Evaluate y = f (x) at critical numbers and at end points. 3. Compare these values to find global max/min on [a, b]. 19 / 65 10 Chapter 4:  Applications of Differentiation Mean Value Theorem  Rolle’s Theorem:    If f is continuous on [a, b], differentiable on (a, b), f (a) = f (b) ⇒ there is c ∈ (a, b) s.t. f ′ (c) = 0. Mean Value Theorem:  If f is continuous on [a, b] and differentiable on (a, b). then there is c ∈ (a, b) such that f ′ (c) =  Increasing & Decreasing Test.  f (b) − f (a) . b−a Let f be continuous on [a, b], and differentiable on (a, b).    f ′ (x) > 0 on (a, b) ⇒ f is increasing on [a, b]. f ′ (x) < 0 on (a, b) ⇒ f is decreasing on [a, b]. f ′ (x) = 0 on (a, b) ⇔ f is constant on [a, b]. 20 / 65 Chapter 4:  Applications of Mean Value Theorem.    Prove identities, and inequalities; Determine the number of solutions of an equation. l’Hôpital’s Rule: (x → a+ , x → a− , x → ±∞).    Applications of Differentiation Suppose lim f (x) = lim g(x) = 0 (or ±∞). x→a x→a f (x) f ′ (x) Then lim = lim ′ if the RHS exists or equals ±∞. x→a g(x) x→a g (x) Concavity: f is concave up (resp. concave down) on an interval I if its lies above (resp. below) all the tangent lines on I .  f is concave up ⇔ f ′ is increasing. f is concave down ⇔ f ′ is decreasing.  f ′′ > 0 ⇒ concave up, f ′′ < 0 ⇒ concave down.  21 / 65 11 Chapter 4:  First & Second Derivative Tests:   Applications of Differentiation Determine whether the function has a local maximum or minimum at a critical number. Optimization Problem.  Express the problem as   Finding global max or min of y = f (x) on domain A. How to maximize or minimize y = f (x) on A?   If A is a finite closed interval [a, b], then use the Closed Interval Method. If A is not a finite closed interval, then use Increasing/Decreasing Test to determine the intervals on which f is increasing or decreasing, hence find the global max or min of y = f (x) on A. 22 / 65 Examples  Let u, v, w be functions such that  u′ = v , v ′ = w and w ′ = u for all x ∈ R. If u(0) = 1, v(0) = 0 and w(0) = 0, prove that   u3 + v 3 + w 3 − 3uvw = 1 for all x ∈ R. (08/09 S1). Let f (x) = u3 + v 3 + w 3 − 3uvw . 1. Show that f is constant. f ′ (x) = 3u2 u′ + 3v 2 v ′ + 3w2 w′ − 3(u′ vw + uv ′ w + uvw′ ) = 3(u2 v + 3v 2 w + uw2 ) − 3(v 2 w + uw2 + u2 w) = 0. 2. Determine the constant. For any x ∈ R,  f (x) = f (0) = 13 + 03 + 03 − 3 · 1 · 0 · 0 = 1. ∴ u3 + v 3 + w 3 − 3uvw = 1 for all x ∈ R. 23 / 65 12 Examples  ln(1 + x) ≤ x ≤ − ln(1 − x) on (−1, 1). (06/07 Sem 2)  Let f (x) = ln(1 + x) − x on (−1, 1). Then     f attains the maximum at x = 0.   −x 1 −1= . 1+x 1+x −1 < x < 0 ⇒ f ′ (x) > 0 So f is increasing on (−1, 0]. 0 < x < 1 ⇒ f ′ (x) < 0. So f is decreasing on [0, 1). f ′ (x) = For all x ∈ (−1, 1), f (0) ≥ f (x). That is, 0 ≥ ln(1 + x) − x. Replace x by −x in the first inequality:  ln(1 − x) ≤ −x.  That is, x ≤ − ln(1 − x). 24 / 65 Examples  Show that if 3 < a < b, then ba < ab . (06/07 Sem 1). ba < ab ⇔ ln(ba ) < ln(ab ) ⇔ a ln b < b ln a. ln x  Define f (x) = for x > 0. x 1 · x − ln x · 1 1 − ln x = .  f ′ (x) = x 2 2 x x  x > e ⇒ f ′ (x) < 0.   f is decreasing on [e, ∞). 3 < a < b ⇒ f (a) > f (b) ln b ln a > ⇒ a b ⇒ b ln a > a ln b ⇒ eb ln a > ea ln b ⇒ ab > ba . 25 / 65 13 Examples   Let f, g be twice differentiable functions. Suppose f (a) = f (b) = g(a) = g(b) = 0 and g ′′ (x) 6= 0 on (a, b).  Prove that g(x) 6= 0 on (a, b).  Prove that there exists c ∈ (a, b) such that (09/10 Semester 1, Q8) f ′′ (c) f (c) = ′′ . g(c) g (c) Assume that g(x0 ) = 0 for some x0 ∈ (a, b).  g(a) = g(x0 ) = 0. Apply Rolle’s Theorem to g on [a, x0 ],   g(x0 ) = g(b) = 0. Apply Rolle’s Theorem to g on [x0 , b],   there exists α ∈ (a, x0 ) such that g ′ (α) = 0. there exists β ∈ (x0 , b) such that g ′ (β) = 0. g ′ (α) = g ′(β) = 0. Apply Rolle’s Theorem to g ′ on [α, β].  there exists γ ∈ (α, β) such that (g ′ )′ (γ) = 0. This contradicts the assumption that g ′′ (x) 6= 0 on (a, b). 26 / 65 Examples   Let f, g be twice differentiable functions. Suppose f (a) = f (b) = g(a) = g(b) = 0 and g ′′ (x) 6= 0 on (a, b).  Prove that g(x) 6= 0 on (a, b).  Prove that there exists c ∈ (a, b) such that (09/10 Semester 1, Q8) f ′′ (c) f (c) = ′′ . g(c) g (c) Define h(x) = f (x)g ′ (x) − f ′ (x)g(x).  h(a) = f (a)g ′ (a) − f ′ (a)g(a) = 0.  h(b) = f (b)g ′ (b) − f ′ (b)g(a) = 0. Apply Rolle’s Theorem to h on [a, b],  There exists c ∈ (a, b) such that h′ (c) = 0. h′ (x) = [f ′ (x)g ′ (x) + f (x)g ′′ (x)] − [f ′′ (x)g(x) + f ′ (x)g ′ (x)] = f (x)g ′′ (x) − f ′′ (x)g(x). 0 = h′ (c) = f (c)g ′′ (c) − f ′′ (c)g(c). That is, f (c)/g(c) = f ′′ (c)/g ′′ (c). (g ′′ (c) 6= 0, g(c) 6= 0). 27 / 65 14 Examples  Let A, B be the intersection of y = x2 and y = x + 2, and P (a, a2 ) a point on the parabola between A and B . Find the max area of △P AB . (11/12 Semester 2, Q3) y b y = x+2 y = x2 A Q(a, a + 2) B b b P (a, a2 ) b x O x2 = x + 2 ⇒ x = −1 and x = 2.  Area of △AP Q = 12 (a + 2 − a2 )(a + 1). Area of △BP Q = 21 (a + 2 − a2 )(2 − a).  28 / 65 Examples  Let A, B be the intersection of y = x2 and y = x + 2, and P (a, a2 ) a point on the parabola between A and B . Find the max area of △P AB . (11/12 Semester 2, Q3) y y = x+2 y = x2 A Q(a, a + 2) b B b b b P (a, a2 ) x O  Max A(a) = 32 (a + 2 − a2 ), ′ 3 (1 2 −1 ≤ a ≤ 2. − 2a) = 0 ⇒ a = 1/2.  Compare A(−1) = A(2) = 0 and A(1/2) = 27/8. ∴ P (1/2, 1/4) is the point so that △ABP has the largest area.  A (a) = 29 / 65 15 Examples  A sector AOB is to be cut out from a piece of thin cardboard in the shape of a circle of radius 10 cm. The radii OA and OB are joined together so that the sector forms the curved surface of a cone. B A 10 10 h O θ r Find θ such that the cone has the largest volume. (09/10 S2) ⌢ 10 θ = AB = 2πr ⇒ r = 5θ/π . √ √  h = 102 − r 2 = 5 4π 2 − θ 2 /π .  30 / 65 Examples  √ r = 5θ/π , h = 5 4π 2 − θ2 /π . 1 125 2 √ 2  Max V (θ) = πr 2 h = θ 4π − θ2 , 0 ≤ θ ≤ 2π . 3 3π 2 125θ(8π 2 − 3θ2 ) √ .  V ′ (θ) = · · · = 3π 2 4π 2 − θ2 √ 6 2 π.  Let V ′ (θ) = 0 on (0, 2π). Then θ = 3 √ 2 6 π) > 0.  Compare V (0) = 0, V (2π) = 0 and V ( 3 √ 2 6 π.  The cone has the maximum volume when θ = 3 z Verification of maximality/minimality is necessary. z Do NOT use the 1st/2nd derivative test in optimization problem. (Cont’d) 31 / 65 16 Examples  Let g be a function and c a number in the domain. (11/12 S1)    If g ′ (c) = g ′′ (c) = g ′′′ (c) = 0 and g (4) (c) > 0, show that g has a local minimum at c. Let f = g ′′ . Then f (c) = f ′ (c) = 0 and f ′′ (c) > 0.  By 2nd derivative test, f has a local minimum at c.   For all x near c but x 6= c, f (x) > f (c) = 0. For all x near c but x 6= c, g ′′ (x) > 0.  g ′′(x) = (g ′)′ (x) > 0 on some interval (a, c).   By increasing test, g ′ is increasing on (a, c]. a < x < c ⇒ g ′(x) < g ′(c) = 0. g ′′(x) = (g ′)′ (x) > 0 on some interval (c, b).  By increasing test, g ′ is increasing on [c, b). c < x < b ⇒ g ′(x) > g ′ (c) = 0. 32 / 65 Examples  Let g be a function and c a number in the domain. (11/12 S1)    If g ′ (c) = g ′′ (c) = g ′′′ (c) = 0 and g (4) (c) > 0, show that g has a local minimum at c. Let f = g ′′ . Then f (c) = f ′ (c) = 0 and f ′′ (c) > 0.  By 2nd derivative test, f has a local minimum at c.   For all x near c but x 6= c, g ′′ (x) > 0.    For all x near c but x 6= c, f (x) > f (c) = 0. a < x < c ⇒ g ′(x) < 0. c < x < b ⇒ g ′(x) > 0. By 1st derivative test, g has a local minimum at c. 33 / 65 17 Chapter 5:  Integrals Definite Integral.  Riemann Sum: lim Sn , where Sn = n→∞  n P i=1 f (x∗i )∆x. Let f be a continuous function on [a, b]. Z  b f (x) dx is the net area of the region bounded between the graph of y = f (x) and the a x-axis from a to b.  Properties of Definite Integral. Z b b Z Z b (f (x) ± g(x)) dx = f (x) dx ± g(x) dx. a a a Z b Z c Z c f (x) dx + f (x) dx =  f (x) dx. a b Z b a Z b  f (x) ≤ g(x) on [a, b] ⇒ f (x) dx ≤ g(x) dx.  a a 34 / 65 Chapter 5:  Integrals Fundamental Thm of Calculus:  Part I: g(x) = Let f be continuous on [a, b]. x Z f (t) dt is continuous on [a, b], differentiable on (a, b), and g′ (x) = f (x). a  Part II: If F is continuous on [a, b], differentiable on (a, b) and F ′ (x) = f (x), then Z a b f (x) dx = F (b) − F (a). f (x) =  Indefinite Integral.  Methods of Integration. F ′ (x) ⇔  Fundamental Thm of Z Calculus Part II. ′  Substitution Rule I:  Substitution Rule II:  Z Z f (x) dx = F (x) + C . Z f (g(x)) g (x) dx = f (u) du. Z f (x) dx = f (g(t))g′ (t) dt. Trigonometric substitution; universal trigonometric substitution.  Integral by Parts, Integration of Rational Functions. 35 / 65 18 Chapter 5:  Integrals Improper Integral.  Discontinuous Integrand.    f is continuous on [a, b) ⇒ Z f is continuous on (a, b] ⇒ Z Infinite Intervals.   f (x) dx = lim c→b− a b f (x) dx = lim c→a+ a f is continuous on [a, ∞) ⇒ f continuous on (−∞, b]: b Z Z ∞ −∞ Z b f (x) dx = lim c→−∞ ∞ b f (x) dx c c→∞ a f is continuous on (−∞, ∞): Z ∞ Z a Z  f (x) dx = f (x) dx + −∞ f (x) dx a f (x) dx = lim −∞  c Z Z c f (x) dx a Z b f (x) dx c f (x) dx. a 36 / 65 Examples √ Z 4 (1 + x)4 √ dx. (08/09 Semester 1 Q2(a))  x 1 √ 1 du =√ .  Let u = 1 + x. Then dx x √ Z Z 2 (1 + x)4 √ dx = 2u4 du = u5 + C x 5 √ 5 2 = (1 + x) + C. 5 √ 4 Z 4 √ 5 x=4 422 (1 + x) 2 √ . = dx = (1 + x) 5 5 x 1 x=1 19 37 / 65 Examples Z 1 ln x dx.  (07/08 Semester 1 Q7(a)) 0 Z ln x dx = x ln x − Z ′ x(ln x) dx = x ln x − Z x· 1 dx x = x ln x − x + C. x=1 = −1 − a ln a + a. (a > 0) ln x dx = x ln x − x x=a a Z 1 Z 1 ln x dx = lim+ (−1 − a ln a + a) ln x dx = lim+ 1 Z a→0 0 a→0 a ln a = −1 − lim+ + lim+ a a→0 1/a a→0 1/a +0 = −1 − lim+ a→0 −1/a2 = −1 + lim+ a + 0 = −1. a→0 38 / 65 Examples  Suppose f is continuous and positive on [a, b], where a < b. Z b f (x) dx > 0. (06/07 Semester 1 Q8(a)) Z x  Let F (x) = f (t) dt. Then Show that a a   F is continuous on [a, b] and F ′ (x) = f (x) on (a, b). F ′ (x) > 0 ⇒ F (x) is increasing on [a, b]. In particular, F (b) > F (a). ∴ Z b Z b f (x) dx > a  Z a Z b f (x) dx = 0. a By EVT, f has the minimum m on [a, b]. Then m > 0.  a f (x) dx ≥ a m dx = m(b − a)...
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