**Unformatted text preview: **MA1102R CALCULUS
Review
Wang Fei
[email protected] Department of Mathematics
Office: S17-06-16
Tel: 6516-2937 Chapter 0:
Chapter 1:
Chapter 2:
Chapter 3:
Chapter 4:
Chapter 5:
Chapter 6:
Chapter 7:
Chapter 8: Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Applications of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Transcendental Functions and Techniques of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
Applications of Integration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 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 (Chapter 4).
2 / 63 Chapter 1:
lim f (x) = L. Limit of function.
x → a (x 6= 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 L
f (x)
=
(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 / 63 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: f (x)g(x) : (Refer to Chapter 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 / 63 Limits How to evaluate the limit of function? (Continued)
Left- and right-hand 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 for details.) 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 / 63 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 / 63 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 / 63 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 / 63 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 / 63 Examples
Find lim (1 + sin 2x)1/x .
x→0 (15/16 Semester 1, Q4(ii)). lim (1 + sin 2x) x→0 Exercise. 1/x
1
= lim exp
ln(1 + sin 2x)
x→0
x
ln(1 + sin 2x)
= exp lim
x→0
x
2 cos 2x/(1 + sin 2x)
= exp lim
x→0
1
2
= exp(2) = e . (1 + sin 2x)1/x − e2
. (15/16 S1, Q4(iii)).
x→0
x
lim 10 / 63 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 ! 2 + e1/x sin x
+
|x|
1 + e4/x +1
− x42 e4/x
1
= lim
+ 1 = 0 + 1 = 1.
+
x→0 4e3/x
! = lim x→0+ = lim x→0+ lim x→0− = lim x→0− =
Therefore, lim x→0 2 + e1/x sin x
+
1 + e4/x
|x| sin x
2 + e1/x
+
lim
1 + e4/x x→0+ x
− x12 e1/x 2 + e1/x
sin x
+ lim
4/x
−
−x
1+e
x→0 2+0
+ (−1) = 1.
1+0 exists and equals 1.
11 / 63 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 / 63 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 / 63 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 S1 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 / 63 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 / 63 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 Chapter 6)
dy
from the obtained equation.
dx d
f is continuous ⇒
dx Z x f (t) dt = f (x). a Logarithmic differentiation. (Refer to Chapter 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 / 63 Examples
Suppose f is continuous on R and differentiable on R \ {a}.
If lim f ′ (x) exists, show that 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 dy Evaluate
if y = x(sin x)cos x . (10/11 S2 Q3(a)) dx x=π/2
f ′ (a) = lim 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 1
02
π
− 1 · ln 1 +
=
· = 1. dx x=π/2
π/2
1
2 9 17 / 63 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 / 63 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 / 63 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 / 63 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 RHS exists or is ±∞.
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 / 63 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 interval I . How to maximize or minimize y = f (x) on I ?
If I is a finite closed interval [a, b],
then use the Closed Interval Method.
If I 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 I .
22 / 63 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 / 63 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 / 63 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
for x > 0.
Define f (x) =
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 a
ln b
⇒
>
a
b
⇒ b ln a > a ln b
⇒ eb ln a > ea ln b
⇒ ab > ba . 25 / 63 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 / 63 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 / 63 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 / 63 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 P (a, a2 )
x O Max A(a) = 32 (a + 2 − a2 ),
′ 3
(1
2 B b b b −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 is the largest.
A (a) = 29 / 63 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 / 63 Examples
√
r = 5θ/π , h = 5 4π 2 − θ2 /π .
125 2 √ 2
1
θ 4π − θ2 , 0 ≤ θ ≤ 2π .
Maximize V (θ) = πr 2 h =
3
3π 2
125θ(3θ2 − 8π 2 )
√
.
V ′ (θ) = · · · =
3π 2 4π 2 − θ2
√
2
6
π.
Let V ′ (θ) = 0 on (0, 2π). Then θ =
3 √
2 6
Compare V (0) = 0, V (2π) = 0 and V (
π) > 0.
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 2nd derivative test in optimization problem.
(Cont’d) 31 / 63 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 / 63 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 / 63 17 Chapter 5:
Integrals Definite Integral.
Riemann sum: Z b f (x) dx = lim n→∞ a n
X f (x∗i )∆x. i=1 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
Za 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 / 63 Chapter 5:
Integrals
Let f be continuous on [a, b]. Fundamental Theorem of Calculus: Part I: g(x) = Z x f (t) dt is continuous on [a, b], differentiable on (a, b), and g′ (x) = f (x) on (a, b).
a Part II: If F is continuous on [a, b], differentiable on (a, b) and F ′ (x) = f (x) on (a, b), then Z a b f (x) dx = F (b) − F (a).
f (x) = Indefinite Integral. Methods of Integration. F ′ (x) ⇔ Z f (x) dx = F (x) + C . Fundamental Theorem
Z of Calculus Part II.
′ Substitution Rule I: Substitution Rule II:
Z 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. Integration by parts, Integration of rational functions. 35 / 63 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 / 63 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 √
Z 4
√ 5 x=4 422
2
(1 + x)4
√
=
dx = (1 + x) .
x
5
5
1
x=1 19 37 / 63 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+ Z 1 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 / 63 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) > 0.
39 / 63 20 E...

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