---
author: Alvie Rahman
date: \today
title: MMME1026 // Calculus
tags:
- uni
- nottingham
- mechanical
- engineering
- mmme1026
- maths
- calculus
uuid: 126b21f8-e188-48f6-9151-5407f2b2b644
---
# Calculus of One Variable Functions
## Key Terms
### Function
A function is a rule that assigns a **unique** value $f(x)$ to each value $x$ in a given *domain*.
The set of value taken by $f(x)$ when $x$ takes all possible value in the domain is the *range* of
$f(x)$.
### Rational Functions
A function of the type
$$ \frac{f(x)}{g(x)} $$
where $f$ and $g$ are polynomials, is called a rational function.
Its range has to exclude all those values of $x$ where $g(x) = 0$.
### Inverse Functions
Consider the function $f(x) = y$.
If $f$ is such that for each $y$ in the range there is exactly one $x$ in the domain,
we can define the inverse $f^{-1}$ as:
$$f^{-1}(y) = f^{-1}(f(x)) = x$$
### Limits
Consider the following:
$$f(x) = \frac{\sin x}{x}$$
The value of the function can be easily calculated when $x \neq 0$, but when $x=0$, we get the
expression $\frac{\sin 0 }{0}$.
However, when we evaluate $f(x)$ for values that approach 0, those values of $f(x)$ approach 1.
This suggests defining the limit of a function
$$\lim_{x \rightarrow a} f(x)$$
to be the limiting value, if it exists, of $f(x)$ as $x$ gets approaches $a$.
#### Limits from Above and Below
Sometimes approaching 0 with small positive values of $x$ gives you a different limit from
approaching with small negative values of $x$.
The limit you get from approaching 0 with positive values is known as the limit from above:
$$\lim_{x \rightarrow a^+} f(x)$$
and with negative values is known as the limit from below:
$$\lim_{x \rightarrow a^-} f(x)$$
If the two limits are equal, we simply refer to the *limit*.
## Important Functions
### Exponential Functions
$$f(x) = e^x = \exp x$$
It can also be written as an infinite series:
$$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
The two important limits to know are:
- as $x \rightarrow + \infty$, $\exp x \rightarrow +\infty$ ($e^x \rightarrow +\infty$)
- as $x \rightarrow -\infty$, $\exp x \rightarrow 0$ ($e^x \rightarrow 0$)
Note that $e^x > 0$ for all real values of $x$.
### Hyperbolic Functions (sinh and cosh)
The hyperbolic sine ($\sinh$) and hyperbolic cosine function ($\cosh$) are defined by:
$$\sinh x = \frac 1 2 (e^x - e^{-x}) \text{ and } \cosh x = \frac 1 2 (e^x + e^{-x})$$
$$\tanh = \frac{\sinh x}{\cosh x}$$
](./images/Sinh_cosh_tanh.svg)
Some key facts about these functions:
- $\cosh$ has even symmetry and $\sinh$ and $\tanh$ have odd symmetry
- as $x \rightarrow + \infty$, $\cosh x \rightarrow +\infty$ and $\sinh x \rightarrow +\infty$
- $\cosh^2x - \sinh^2x = 1$
- $\tanh$'s limits are -1 and +1
- Derivatives:
- $\frac{\mathrm{d}}{\mathrm{d}x} \sinh x = \cosh x$
- $\frac{\mathrm{d}}{\mathrm{d}x} \cosh x = \sinh x$
- $\frac{\mathrm{d}}{\mathrm{d}x} \tanh x = \frac{1}{\cosh^2x}$
### Natural Logarithm
$$\ln{e^y} = \ln{\exp y} = y$$
Since the exponential of any real number is positive, the domain of $\ln$ is $x > 0$.
### Implicit Functions
An implicit function takes the form
$$f(x, y) = 0$$
To draw the curve of an implicit function you have to rewrite it in the form $y = f(x)$.
There may be more than one $y$ value for each $x$ value.
# Differentiation
The derivative of the function $f(x)$ is denoted by:
$$f'(x) \text{ or } \frac{\mathrm{d}}{\mathrm dx} f(x)$$
Geometrically, the derivative is the gradient of the curve $y = f(x)$.
It is a measure of the rate of change of $f(x)$ as $x$ varies.
For example, velocity, $v$, is the rate of change of displacement, $s$, with respect to time, $t$,
or:
$$v = \frac{\mathrm ds}{dt}$$
#### Formal Definition
As $h\rightarrow 0$, the clospe of the cord $\rightarrow$ slope of the tangent, or:
$$f'(x_0) = \lim_{h\rightarrow0}\frac{f(x_0+h) - f(x_0)}{h}$$
whenever this limit exists.
## Rules for Differentiation
### Powers
$$\frac{\mathrm d}{\mathrm dx} x^n = nx^{-1}$$
### Trigonometric Functions
$$\frac{\mathrm d}{\mathrm dx} \sin x = \cos x$$
$$\frac{\mathrm d}{\mathrm dx} \cos x = \sin x$$
### Exponential Functions
$$\frac{\mathrm d}{\mathrm dx} e^{kx} = ke^{kx}$$
$$\frac{\mathrm d}{\mathrm dx} \ln kx^n = \frac n x$$
where $n$ and $k$ are constant.
### Linearity
$$\frac{\mathrm d}{\mathrm dx} (f + g) = \frac{\mathrm d}{\mathrm dx} f + \frac{\mathrm d}{\mathrm dx} g$$
### Product Rule
$$\frac{\mathrm d}{\mathrm dx} (fg) = \frac{\mathrm df}{\mathrm dx}g + \frac{\mathrm dg}{\mathrm dx}f$$
### Quotient Rule
$$ \frac{\mathrm d}{\mathrm dx} \frac f g = \frac 1 {g^2} \left( \frac{\mathrm df}{\mathrm dx} g - f \frac{\mathrm dg}{\mathrm dx} \right) $$
$$ \left( \frac f g \right)' = \frac 1 {g^2} (gf' - fg')$$
### Chain Rule
Let
$$f(x) = F(u(x))$$
$$ \frac{\mathrm df}{\mathrm dx} = \frac{\mathrm{d}F}{\mathrm du} \frac{\mathrm du}{\mathrm dx} $$
#### Example 1
Differentiate $f(x) = \cos{x^2}$.
Let $u(x) = x^2$, $F(u) = \cos u$
$$ \frac{\mathrm df}{\mathrm dx} = -\sin u \cdot 2x = 2x\sin{x^2} $$
## L'Hôpital's Rule
l'Hôpital's rule provides a systematic way of dealing with limits of functions like
$\frac{\sin x} x$.
Suppose
$$\lim_{x\rightarrow{a}} f(x) = 0$$
and
$$\lim_{x\rightarrow{a}} g(x) = 0$$
and we want $\lim_{x\rightarrow{a}} \frac{f(x)}{g(x)}$.
If
$$\lim_{x\rightarrow{a}} \frac{f'(x)}{g'(x)} = L $$
where any $L$ is any real number or $\pm \infty$, then
$$\lim_{x\rightarrow{a}} \frac{f(x)}{g(x)} = L$$
You can keep applying the rule until you get a sensible answer.
# Graphs
## Stationary Points
An important application of calculus is to find where a function is a maximum or minimum.
when these occur the gradient of the tangent to the curve, $f'(x) = 0$.
The condition $f'(x) = 0$ alone however does not guarantee a minimum or maximum.
It only means that point is a *stationary point*.
There are three main types of stationary points:
- maximum
- minimum
- point of inflection
### Local Maximum
The point $x = a$ is a local maximum if:
$$f'(a) = 0 \text{ and } f''(a) < 0$$
This is because $f'(x)$ is a decreasing function of $x$ near $x=a$.
### Local Minimum
The point $x = a$ is a local minimum if:
$$f'(a) = 0 \text{ and } f''(a) > 0$$
This is because $f'(x)$ is a increasing function of $x$ near $x=a$.
### Point of Inflection
$$f'(a) = 0 \text{ and } f''(a) = 0 \text { and } f'''(a) \ne 0$$
#### $f'''(a) > 0$
#### $f'''(a) < 0$
# Approximating with the Taylor series
The expansion
$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$
is an example of a *Taylor series*.
These enable us to approximate a given function f(x) using a series which is often easier to
calculate.
Among other uses, they help us:
- calculate complicated function using simple arithmetic operations
- find useful analytical approximations which work for $x$ near a given value
(e.g. $e^x \approx 1 + x$ for $x$ near 0)
- Understand the behaviour of a function near a stationary point
## Strategy
Suppose we know information about $f(x)$ only at the point $x=0$.
How can we find out about $f$ for other values of $x$?
We could approximate the function by successive polynomials,
each time matching more derivatives at $x=0$.
\begin{align*}
g(x) = a_0 &\text{ using } f(0) \\
g(x) = a_0 + a_1x &\text{ using } f(0), f'(0) \\
g(x) = a_0 + a_1x + a_2x^2 &\text{ using } f(0), f'(0), f''(0) \\
&\text{and so on...}
\end{align*}
#### Example 1
For $x$ near 0, approximate $f(x) = \cos x$ by a quadratic.
1. Set $f(0) = g(0$:
$$f(0) = 1 \rightarrow g(0) = a_0 = 1$$
2. Set $f'(0) = g'(0$:
$$f'(0) = -\sin0 = 0 \rightarrow g'(0) = a_1 = 0$$
3. Set $f''(0) = g''(0$:
$$f''(0) = -\cos = -1 \rightarrow g''(0) = 2a_2 = -1 \rightarrow a_2 = -0.5$$
So for $x$ near 0,
$$\cos x \approx 1 - \frac 1 2 x^2$$
Check:
$x$ | $\cos x$ | $1 - 0.5x^2$
--- | -------- | ------------
0.4 | 0.921061 | 0.920
0.2 | 0.960066 | 0.980
0.1 | 0.995004 | 0.995
## General Case
### Maclaurin Series
A Maclaurin series is a Taylor series expansion of a function about 0.
Any function $f(x)$ can be written as an infinite *Maclaurin Series*
$$f(x) = a_0 + a_1x + a_2x^2 + a_3x^2 + \cdots$$
where
$$a_0 = f(0) \qquad a_n = \frac 1 {n!} \frac{\mathrm d^nf}{\mathrm dx^n} \bigg|_{x=0}$$
($|_{x=0}$ means evaluated at $x=0$)
### Taylor Series
We may alternatively expand about any point $x=a$ to give a Taylor series:
\begin{align*}
f(x) = &f(a) + (x-a)f'(a) \\
& + \frac 1 {2!}(x-a)^2f''(a) \\
& + \frac 1 {3!}(x-a)^3f'''(a) \\
& + \cdots + \frac 1 {n!}(x-a)^nf^{(n)}(a)
\end{align*}
a generalisation of a Maclaurin series.
An alternative form of Taylor series is given by setting $x = a+h$ where $h$ is small:
$$f(a+h) = f(a) + hf'(a) + \cdots + \frac 1 {n!}h^nf^{(n)}(a) + \cdots$$
## Taylor Series at a Stationary Point
If f(x) has a stationary point at $x=a$, then $f'(a) = 0$ and the Taylor series begins
$$f(x) = f(a) + \frac 1 2 f''(a)(x-a)^2 + \cdots$$
- If $f''(a) > 0$ then the quadratic part makes the function increase going away from $x=a$ and we
have a minimum
- If $f''(a) < 0$ then the quadratic part makes the function decrease going away from $x=a$ and we
have a maximum
- If $f''(a) = 0$ then we must include a higer order terms to determine what happens
have a minimum
# Integration
Integration is the reverse of [differentiation](#differentiation).
Take velocity and displacement as an example:
$$\int\! v \mathrm dt = s + c$$
where $c$ is the constant of integration, which is required for
[indefinite integrals](#indefinite-integrals).A
## Definite Integrals
The definite integral of a function $f(x)$ in the range $a \le x \le b$ is denoted be:
$$\int^b_a \! f(x) \,\mathrm dx$$
If $f(x) = F'(x)$ ($f(x)$ is the derivative of $F(x)$) then
$$\int^b_a \! f(x) \,\mathrm dx = \left[F(x)\right]^b_a = F(b) - F(a)$$
## Area and Integration
Approximate the area under a smooth curve using a large number of narrow rectangles.
Area under curve $\approx \sum_{n} f(x_n)\Delta x_n$.
As the rectangles get more numerous and narrow, the approximation approaches the real area.
The limiting value is denoted
$$\approx \sum_{n} f(x_n)\Delta x_n \rightarrow \int^b_a\! f(x) \mathrm dx$$
This explains the notation used for integrals.
#### Example 1
Calculate the area between these two curves:
\begin{align*}
y &= f_1(x) = 2 - x^2 \\
y &= f_2(x) = x
\end{align*}
1. Find the crossing points $P$ and $Q$
\begin{align*}
f_1(x) &= f_2(x) \\
x &= 2-x^2 \\
x &= 1 \\
x &= -2
\end{align*}
2. Since $f_1(x) \ge f_2(x)$ between $P$ and $Q$
\begin{align*}
A &= \int^1_{-2}\! (f_1(x) - f_2(x)) \mathrm dx \\
&= \int^1_{-2}\! (2 - x^2 - x) \mathrm dx \\
&= \left[ 2x - \frac 13 x^3 - \frac 12 x^2 \right]^1_{-2} \\
&= \left(2 - \frac 13 - \frac 12 \right) - \left( -4 + \frac 83 - \frac 42 \right) \\
&= \frac 92
\end{align*}
## Techniques for Integration
Integration requires multiple techniques and methods to do correctly because it is a PITA.
These are best explained by examples so try to follow those rather than expect and explanation.
### Integration by Substitution
Integration but substitution lets us integrate functions of functions.
#### Example 1
Find
$$I = \int\!(5x - 1)^3 \mathrm dx$$
1. Let $w(x) = 5x - 1$
2.
\begin{align*}
\frac{\mathrm d}{\mathrm dx} w &= 5 \\
\frac 15 \mathrm dw &= \mathrm dx
\end{align*}
3. The integral is then
\begin{align*}
I &= \int\! w^3 \frac 15 \mathrm dw \\
&= \frac 15 \cdot \frac 14 \cdot w^4 + c \\
&= \frac{1}{20}w^4 + c
\end{align*}
4. Finally substitute $w$ out
$$I = \frac{(5x-1)^4}{20} + c$$
#### Example 2
Find
$$I = \int\! \cos x \sqrt{\sin x + 1} \mathrm dx$$
1. Let
$$w(x) = \sin x + 1$$
2. Then
\begin{align*}
\frac{\mathrm d}{\mathrm dx} w = \cos x \\
\mathrm dw = \cos x \mathrm dx \\
\end{align*}
3. The integral is now
\begin{align*}
I &= \int\! \sqrt w \,\mathrm dw \\
&= \int\! w^{\frac12} \,\mathrm dw \\
&= \frac23w^{\frac32} + c
\end{align*}
4. Finally substitute $w$ out to get:
$$I = \frac23 (\sin x + 1)^{\frac32} + c$$
#### Example 3
Find
$$I = \int^{\frac\pi2}_0\! \cos x \sqrt{\sin x + 1} \,\mathrm dx$$
1. Use the previous example to get to
$$I = \int^2_1\! \sqrt w \,\mathrm dw = \frac23w^{\frac32} + c$$
2. Since $w(x) = \sin x + 1$ the limits are:
\begin{align*}
x = 0 &\rightarrow w = 1\\
x = \frac\pi2 &\rightarrow w = 2
\end{align*}
3. This gives us
$$I = \left[ \frac23w^{\frac32} \right]^2_1 = \frac23 (2^{\frac23} = 1)$$
#### Example 4
Find
$$I = \int^1_0\! \sqrt{1 - x^2} \,\mathrm dx$$
1. Try a trigonmetrical substitution:
\begin{align*}
x &= \sin w \\
\\
\frac{\mathrm dx}{\mathrm dw} = \cos w \\
\mathrm dx = \cos 2 \,\mathrm dw \\
\end{align*}
2.
\begin{align*}
x=0 &\rightarrow w=0 \\
x=1 &\rightarrow w=\frac\pi2
\end{align*}
3. Therefore
\begin{align*}
I &= \int^{\frac\pi2}_0\! \sqrt{1 - \sin^2 w} \cos w \,\mathrm dw \\
&= \int^{\frac\pi2}_0\! \cos^w w \,\mathrm dw
\end{align*}
But $\cos(2w) = 2\cos^2w - 1$ so:
$$\cos^2w = \frac12 \cos(2w) + \frac12$$
Hence
\begin{align*}
I &= \int^{\frac\pi2}_0\! \frac12 \cos(2w) + \frac12 \,\mathrm dw \\
&= \left[ \frac14 \sin(2w) + \frac w2 \right]^{\frac\pi2}_0 \\
&= \left( \frac14 \sin\pi + \frac\pi4 \right) - 0 \\
&= \frac\pi4
\end{align*}
### Integration by Parts
$$uv = \int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx + \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$
or
$$\int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx = uv - \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$
This technique is derived from integrating the product rule.
#### Example 1
Find
$$I = \int\! \ln x \,\mathrm dx$$
1. Use
$$\int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx = uv - \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$
2. Set $u = \ln x$
and $v' = 1$.
3. This means that $u' = \frac1x$ and $v = x$.
4.
\begin{align*}
I &= x\ln x - \int\! x\cdot\frac1x \,\mathrm dx + c \\
&= x\ln x - \int\! \,\mathrm dx + c \\
&= x\ln x - x + c \\
\end{align*}
# Application of Integration
## Differential Equations
Consider the equation
$$\frac{\mathrm dy}{\mathrm dx} = y^2$$
To find $y$, is not a straightforward integration:
$$y = \int\!y^2 \,\mathrm dx$$
The equation above does not solve for $y$ as we can't integrate the right until we know $y$...
which is what we're trying to find.
This is an example of a first order differential equation.
The general form is:
$$\frac{\mathrm dy}{\mathrm dx} = F(x, y)$$
### Separable Differential Equations
A first order diferential equation is called *separable* if it is of the form
$$\frac{\mathrm dy}{\mathrm dx} = f(x)g(y)$$
We can solve these by rearranging:
$$\frac1{g(y} \cdot \frac{\mathrm dy}{\mathrm dx} = f(x)$$
$$\int\! \frac1{g(y)} \,\mathrm dy = \int\! f(x) \,\mathrm dx + c$$
#### Example 1
Find $y$ such that
$$\frac{\mathrm dy}{\mathrm dx} = ky$$
where $k$ is a constant.
Rearrange to get
\begin{align*}
\int\! \frac1y \,\mathrm dy &= \int\! k \mathrm dx + c \\
\ln y &= kx + c \\
y &= e^{kx + c} = e^ce^{kx} \\
&= Ae^{kx}
\end{align*}
where $A = e^c$ is an arbitrary constant.