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5. Calculus
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Maths AA
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Topics
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5. Calculus
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5. Calculus
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If
f
(
x
)
f(x)
f
(
x
)
is as close as we like to some real number
A
A
A
for all
x
x
x
sufficiently close to (but not equal to)
a
a
a
, then we say that
f
(
x
)
f(x)
f
(
x
)
has a
limit
of
A
A
A
as
x
x
x
approaches
a
a
a
, and we write
lim
x
→
a
f
(
x
)
=
A
\lim\limits_{x \to a} f(x) = A
x
→
a
lim
f
(
x
)
=
A
.
In this case,
f
(
x
)
f(x)
f
(
x
)
is said to
converge to
A
A
A
as
x
x
x
approaches
a
a
a
. Otherwise, we say a limit is
divergent
.
Figure 5.1.1
Convergence of limits for continuous (left) and discontinuous function (right)
We define continuity using limits. Consider a real function
f
f
f
defined on an open interval containing the value
a
a
a
. We say
f
f
f
is continuous at
x
=
a
x = a
x
=
a
if
lim
x
→
a
f
(
x
)
=
f
(
a
)
\lim\limits_{x \to a} f(x) = f(a)
x
→
a
lim
f
(
x
)
=
f
(
a
)
.
If
f
f
f
is continuous at
x
=
a
x = a
x
=
a
for all
a
ϵ
R
a \epsilon ℝ
a
ϵ
R
, we say
f
f
f
is continuous on
R
ℝ
R
.
5.1 Introduction to Differential Calculus
Derivatives can describe a function in multiple ways.
Figure 5.2.1
Stationary points of a curve
Suppose
S
S
S
is an interval in the domain of
f
(
x
)
f(x)
f
(
x
)
, so
f
(
x
)
f(x)
f
(
x
)
is defined for all
x
x
x
in
S
S
S
. Then, we have:
1.
f
(
x
)
f(x)
f
(
x
)
is increasing on
S
S
S
if
f
′
(
x
)
>
0
f'(x) > 0
f
′
(
x
)
>
0
for all
x
x
x
in
S
S
S
2.
f
(
x
)
f(x)
f
(
x
)
is decreasing on
S
S
S
if
f
′
(
x
)
<
0
f'(x) < 0
f
′
(
x
)
<
0
for all
x
x
x
in
S
S
S
3.
f
(
x
)
f(x)
f
(
x
)
has a stationary point when
f
′
(
x
)
=
0
f'(x) = 0
f
′
(
x
)
=
0
:
a.
Local maximum (increases then decreases: derivative sign changes from + to -)
b.
Local minimum (decreases then increases: derivative sign changes from - to +)
c.
Stationary inflection (derivative sign does not change)
5.2 Properties of Curves
There are different applications you need to be aware of.
5.3 Application of Differential Calculus
Integration is an inverse operation of differentiation.
Thus, we call
F
(
x
)
=
f
(
x
)
d
x
F(x) = f(x) dx
F
(
x
)
=
f
(
x
)
d
x
as the
antiderivative
of
f
(
x
)
f(x)
f
(
x
)
, and
d
d
x
F
(
x
)
=
f
(
x
)
\frac d{dx}F(x) = f(x)
d
x
d
F
(
x
)
=
f
(
x
)
.
Indefinite integral
denotes an integration without the bounds, and thus has a constant of integration
C
C
C
.
Figure 5.5.1
Definite integral represented as the area under the curve for a specific range
Definite integral
denotes an integration with the bounds, and represents the area below
f
f
f
from
a
a
a
to
b
b
b
.
5.4 Integration
There are 3 different applications you need to be aware of.
5.5 Application of Integration
Another application is
differential equation
, an equation involving a derivative of a function. (HL)
There are two ways to solve a differential equation,
numerical method
, and
analytic method
. A general solution of a differential equation is a function with a constant
c
c
c
that satisfies the differential equation.
Below regards the analytic method.
Form
Steps
d
y
d
x
=
f
(
x
)
\frac{dy}{dx}=f(x)
d
x
d
y
=
f
(
x
)
1. If
f
(
x
)
f(x)
f
(
x
)
is integrable, then simply,
y
=
f
(
x
)
d
x
y = f(x) dx
y
=
f
(
x
)
d
x
. 2. Calculate the constant of integration using the
particular value
.
d
y
d
x
=
f
(
x
)
g
(
y
)
\frac{dy}{dx}=f(x)g(y)
d
x
d
y
=
f
(
x
)
g
(
y
)
Separable
1. Separate the given equation to
g
(
y
)
d
y
=
f
(
x
)
d
x
g(y) dy = f(x) dx
g
(
y
)
d
y
=
f
(
x
)
d
x
. 2. Integrate both side to obtain
G
(
y
)
=
F
(
x
)
+
G(y) =F(x) +
G
(
y
)
=
F
(
x
)
+
C . Isolate
y
y
y
if possible. 3. Calculate the constant of integration using the particular value.
d
y
d
x
=
f
(
y
x
)
\frac{dy}{dx}=f(\frac yx)
d
x
d
y
=
f
(
x
y
)
Dimensionally Homogeneous
1. Substitute
y
=
v
x
y=vx
y
=
vx
. Note that
d
y
d
x
=
v
+
x
d
v
d
x
\frac{dy}{dx}=v+x\frac{dv}{dx}
d
x
d
y
=
v
+
x
d
x
d
v
2. Solve the now separable equation in terms of
v
v
v
. Replace
v
v
v
back in terms of
y
y
y
. 3. Calculate the constant of integration using the particular value.
d
y
d
x
+
p
(
x
)
y
=
q
(
x
)
\frac{dy}{dx}+p(x)y=q(x)
d
x
d
y
+
p
(
x
)
y
=
q
(
x
)
Inhomogeneous
1. Calculate the integrating factor
I
(
x
)
=
e
p
(
x
)
d
x
I(x)=e^{p(x) dx}
I
(
x
)
=
e
p
(
x
)
d
x
2. Multiply both LHS and RHS by the
I
I
I
to obtain the form:
d
d
y
[
I
(
x
)
y
]
=
I
(
x
)
q
(
x
)
\frac d{dy}[I(x)y]=I(x)q(x)
d
y
d
[
I
(
x
)
y
]
=
I
(
x
)
q
(
x
)
3. Solve the given equation using one of the methods for the above forms. 4. Calculate the constant of integration using the particular value.
5.6 Differential Equations
