CALCULUS REFERENCE 3/18/03 10:49 AM Page 1
SPARKCHARTSTM
CALCULUS REFERENCE
SPARK
CHARTS
TM
THEORY
SPARKCHARTS
T...
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CALCULUS REFERENCE 3/18/03 10:49 AM Page 1
SPARKCHARTSTM
CALCULUS REFERENCE
SPARK
CHARTS
TM
THEORY
SPARKCHARTS
TM
DERIVATIVES AND DIFFERENTIATION f (x+h)−f (x) h h→0
Definition: f � (x) = lim
is continuous and differentiable on the interval and F � (x) = f (x).
d f (x) ± g(x) = f � (x) ± g � (x) 1. Sum and Difference: dx � � d cf (x) = cf � (x) 2. Scalar Multiple: dx � � d f (x)g(x) = f � (x)g(x) + f (x)g � (x) 3. Product: dx
�
Mnemonic: If f is “hi” and g is “ho,” then the product rule is “ho d hi plus hi d ho.” � � � (x)g � (x) f (x) d = f (x)g(x)−f (g(x)) 2 g(x) dx
4. Quotient:
Mnemonic: “Ho d hi minus hi d ho over ho ho.”
5. The Chain Rule
• First formulation: (f ◦ g)� (x) = f � (g(x)) g � (x) dy du dy = du • Second formulation: dx dx
dy dx
9 781586 638962 $5.95 CAN $3.95
1. Left-hand rectangle approximation: n−1 � Ln = ∆x f (xk )
dx dx
y) dy = 3 dx − 2y dx cos y + x d(cos , dx dx
and then cos y − x(sin y)y � − 2yy � = 3. Finally, solve for y � =
cos y−3 . x sin y+2y
Mn = ∆x
=0
2. Linear:
d (mx dx
+ b) = m
3. Powers:
d (xn ) dx
= nx n−1 (true for all real n �= 0)
4. Polynomials:
d (an xn dx
+ · · · + a2 x + a1 x + a0 ) = an nx
6. Logarithmic • Base e:
d (ln x) dx
d (sin dx
x
1 x
x) = cos x
d (tan dx
d (sec dx
=
x) = sec 2 x
x) = sec x tan x
n−1
• Arbitrary base:
d (ax ) dx
5. Simpson’s Rule: Sn =
f (x0 )+4f (x1 )+2f (x2 )+· · ·+2f (xn−2 )+4f (xn−1 )+f (xn )
� ∆x 3
�
�
�
�
• Arbitrary base:
d (log a dx
• Cosine: • Cotangent: • Cosecant:
d (cos dx
d (cot dx
d (csc dx
• Definite integrals: concatenation:
= a ln a
x) =
x) = − sin x
x) = − csc 2 x
x) = − csc x cot x
• Arccosine:
d (cos −1 dx
1 1+x2
• Arccotangent:
1 x) = − √1−x 2
x) =
d (cot −1 dx
• Arcsecant:
d (sec −1 dx
x) =
√1 x x2 −1
• Arccosecant:
1 x) = − 1+x 2
d (csc −1 dx
x) = − x√x12 −1
INTEGRALS AND INTEGRATION
�
�
b
f (x) dx = −
a
p
f (x) dx + a
• Definite integrals: comparison: If f (x) ≤ g(x) on the interval [a, b], then
1 x ln a
d (tan −1 dx
√ 1 1−x2
• Definite integrals: reversing the limits:
x
d (sin −1 dx
�
�
a
f (x) dx b
b
f (x) dx = p
b a
�
f (x) dx ≤
�
�
b
f (x) dx a
b
g(x) dx. a
� � � 2. Substitution Rule—a.k.a. u-substitutions: f g(x) g � (x) dx = f (u) du � � � � f g(x) g � (x) dx = F (g(x)) + C if f (x) dx = F (x) + C . • �
3. Integration by Parts Best used to integrate a product when one factor (u = f (x)) has a simple derivative and the other factor (dv = g � (x) dx) is easy to integrate.
� �• Indefinite Integrals: � � f (x)g � (x) dx = f (x)g(x) − f � (x)g(x) dx or u dv = uv − v du
• Definite Integrals:
�
b
b
f (x)g � (x) dx = f (x)g(x)]a −
a
�
b
f � (x)g(x) dx a
4. Trigonometric Substitutions: Used to integrate expressions of the form
DEFINITE INTEGRAL The definite integral
f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )
∆x 2
+ · · · + 2a2 x + a1
• Arctangent:
�
�
4. Trapezoidal Rule: Tn =
�
• Arcsine:
x) =
f
k=0
xk + xk+1 2
• Constant multiples: cf (x) dx = c f (x) dx 2
=e
k=1
�
� � 1. Properties of Integrals � � � • Sums and differences: f (x) ± g(x) dx = f (x) dx ± g(x) dx
1. Constants:
d (c) dx
d (ex ) dx
n−1 �
TECHNIQUES OF INTEGRATION
COMMON DERIVATIVES
5. Exponential • Base e:
2. Right-hand rectangle approximation: n � f (xk )
Rn = ∆x
k=0
= y � and solve for y � .
f (t) dt
a
of x. Differentiate both sides of the equation with respect to x. Use the chain rule
Ex: x cos y − y 2 = 3x. Differentiate to first obtain
x a
APPROXIMATING DEFINITE INTEGRALS
6. Implicit differentiation: Used for curves when it is difficult to express y as a function carefully whenever y appears. Then, rewrite
�
Part 2: If f (x) is continuous on the interval [a, b] and F (x) is any antiderivative of f (x), � b f (x) dx = F (b) − F (a) . then
3. Midpoint Rule:
8. Inverse Trigonometric
Printed the USA
f (x) dx = F (x) + C if F � (x) = f (x).
Part 1: If f (x) is continuous on the interval [a, b], then the area function F (x) =
�
• Tangent: • Secant:
50395 ISBN 1-58663-896-3
�
FUNDAMENTAL THEOREM OF CALCULUS
DERIVATIVE RULES
7. Trigonometric • Sine:
Copyright © 2003 by SparkNotes LLC. All rights reserved. SparkCharts is a registered trademark of SparkNotes LLC. A Barnes & Noble Publication 10 9 8 7 6 5 4 3 2 1
antiderivatives:
b
√
±a2 ± x2 .
f (x) dx is the signed area between the function y = f (x) and the a
x-axis from x = a to x = b.
• Formal definition: Let n be an integer and ∆x =
For each k = 0, 2, . . . , n − 1, n−1 � f (x∗k ) pick point x∗k in the interval [a + k∆x, a + (k + 1)∆x]. The expression ∆x k=0 � b n−1 � f (x) dx is defined as lim ∆x f (x∗k ). is a Riemann sum. The definite integral b−a . n
n→∞
a
k=0
Expression
Trig substitution
Expression becomes
Range of θ
Pythagorean identity used
�
a2 − x2
x = a sin θ dx = a cos θ dθ
a cos θ
− π2 ≤ θ ≤ π2 (−a ≤ x ≤ a)
1 − sin 2 θ = cos 2 θ
�
x2 − a2
x = a sec θ dx = a sec θ tan θ dθ
a tan θ
0≤θ< π≤θ
0, this is exponential growth; if r < 0, exponential decay.
> 0.
INTEREST σ
COST
dP dt
∂g ∂p
• P (t): the amount after t years. • P0 = P (0): the original amount invested (the principal). • r: the yearly interest rate (the yearly percentage is 100r%).
MICROECONOMICS
In all the following models • P (t): size of the population at time t; • P0 = P (0), the size of the population at time t = 0; • r: coefficient of rate of growth.
x
1
FINANCE
95% 68%
• P (µ − 2σ ≤ X ≤ µ + 2σ) = 95.5%
2. χ-square distribution: with mean ν and variance 2ν: x ν 1 f (x) = ν � ν � x 2 −1 e− 2 22Γ 2 � • Gamma function: Γ(x) = 0∞ tx−1 e−t dt
BIOLOGY
0
SUBSTITUTE AND COMPLEMENTATRY COMMODITIES
COMMON DISTRIBUTIONS
C (x) x
y = L(x) completely equitable distribution
X and Y are two commodities with unit price p and q, respectively. • The amount of X demanded is given by f (p, q). • The amount of Y demanded is given by g(p, q). > 0 and 1. X and Y are substitute commodites (Ex: pet mice and pet rats) if ∂f ∂q 2. X and Y are complementary commodities (Ex: pet mice and mouse feed) ∂g ∂f if ∂q < 0 and ∂p < 0.
Cov(X, Y ) σ Correlation: ρ(X, Y ) = X Y = � σX σY Var(X)Var(Y )
for x in
y 1
The quantity L is between 0 and 1. The closer L is to 1, the more equitable the income distribution.
f (y) dy
g(x, y) dy . �∞ �∞ �
• Covariance: Cov(X, Y ) = σX Y =
(¯ p − S(x)) dx
0
• Joint probability density function g(x, y) chronicles distribution of X and Y . Then �∞
x
x
dP dt
P
P0 r0
0
= r (A − P )
A
0
$5.95 CAN
−∞
0
p = D(x)
$3.95
�x
S(x) dx =
� x¯
The Lorentz Curve L(x) is the fraction of income received by the poorest x fraction of the population. 1. Domain and range of L(x) is the interval [0, 1]. 2. Endpoints: L(0) = 0 and L(1) = 1 3. Curve is nondecreasing: L� (x) ≥ 0 for all x 4. L(x) ≤ x for all x • Coefficient of Inequality (a.k.a. Gini Index): � 1 � � L=2 x − f (x) dx .
• Probability density function f (x) of the random variable X satisfies:
F (x) = P (X ≤ x) =
0
LORENTZ CURVE
X and Y are random variables. 1. f (x) ≥ 0 for all x; �∞ 2. −∞ f (x) dx = 1.
� x¯
( x, p)
0
• Producer surplus:
f (x) dx .
consumer surplus producer surplus
3
1 b−a
p
x) = S (¯ x).) (So p¯ = D (¯ Consumer � x¯surplus: � x¯ ¯ = 0 (D(x) − p¯) dx CS = 0 D(x) dx − p¯x
20593 38963
=
p = S(x)
• Supply function: p = S(x) gives price per unit
–
0
• Average value of f (x) between a and b is f
p
(p) when x units demanded.
+
ACCELERATION
3. Acceleration a(t) vs. time t graph: • The (signed) area under the graph gives the change � t in velocity: a(τ ) dτ v(t) − v(0) =
TM
CONSUMER AND PRODUCER SURPLUS
–
v(τ ) dτ
0
7
t
• Formula relating elasticity and revenue: R� (p) = x(p) 1 − E(p)
SPARKCHARTS
+
Writer: Anna Medvedovsky Design: Dan O. Williams Illustration: Matt Daniels, Dan O. Williams Series Editor: Sarah Friedberg
2. Velocity v(t) vs. time t graph: • The slope of the graph is the accleration: v � (t) = a(t). • The (signed) area under the graph gives the displacement�(change in position): s(t) − s(0) =
x (p) • Price elasticity of demand: E(p) = − px(p) • Demand is elastic if E(p) > 1. Percentage change in p leads to larger percentage
POSITION
1. Position s(t) vs. time t graph: • The slope of the graph is the velocity: s� (t) = v(t). • The concavity of the graph is the acceleration: s�� (t) = a(t).
� = rP 1 −
P K
• K : the carrying
P0 > A
P
�
k
capacity
P0 < A
• Solution: t
P (t) =
0
P0 > k
k < P0 < k 2
P0