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---
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author: Alvie Rahman
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date: \today
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title: MMME1026 // Calculus
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tags:
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- uni
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- nottingham
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- mechanical
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- engineering
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- mmme1026
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- maths
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- calculus
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uuid: 126b21f8-e188-48f6-9151-5407f2b2b644
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---
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# Calculus of One Variable Functions
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## Key Terms
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<details>
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<summary>
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### Function
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A function is a rule that assigns a **unique** value $f(x)$ to each value $x$ in a given *domain*.
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</summary>
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The set of value taken by $f(x)$ when $x$ takes all possible value in the domain is the *range* of
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$f(x)$.
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</details>
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<details>
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<summary>
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### Rational Functions
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A function of the type
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$$ \frac{f(x)}{g(x)} $$
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</summary>
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where $f$ and $g$ are polynomials, is called a rational function.
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Its range has to exclude all those values of $x$ where $g(x) = 0$.
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</details>
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<details>
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<summary>
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### Inverse Functions
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Consider the function $f(x) = y$.
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If $f$ is such that for each $y$ in the range there is exactly one $x$ in the domain,
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we can define the inverse $f^{-1}$ as:
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$$f^{-1}(y) = f^{-1}(f(x)) = x$$
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</summary>
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</details>
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<details>
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<summary>
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### Limits
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Consider the following:
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$$f(x) = \frac{\sin x}{x}$$
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The value of the function can be easily calculated when $x \neq 0$, but when $x=0$, we get the
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expression $\frac{\sin 0 }{0}$.
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However, when we evaluate $f(x)$ for values that approach 0, those values of $f(x)$ approach 1.
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This suggests defining the limit of a function
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$$\lim_{x \rightarrow a} f(x)$$
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to be the limiting value, if it exists, of $f(x)$ as $x$ gets approaches $a$.
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</summary>
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#### Limits from Above and Below
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Sometimes approaching 0 with small positive values of $x$ gives you a different limit from
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approaching with small negative values of $x$.
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The limit you get from approaching 0 with positive values is known as the limit from above:
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$$\lim_{x \rightarrow a^+} f(x)$$
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and with negative values is known as the limit from below:
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$$\lim_{x \rightarrow a^-} f(x)$$
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If the two limits are equal, we simply refer to the *limit*.
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</details>
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## Important Functions
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<details>
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<summary>
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### Exponential Functions
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$$f(x) = e^x = \exp x$$
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</summary>
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It can also be written as an infinite series:
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$$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
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The two important limits to know are:
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- as $x \rightarrow + \infty$, $\exp x \rightarrow +\infty$ ($e^x \rightarrow +\infty$)
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- as $x \rightarrow -\infty$, $\exp x \rightarrow 0$ ($e^x \rightarrow 0$)
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Note that $e^x > 0$ for all real values of $x$.
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</details>
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<details>
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<summary>
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### Hyperbolic Functions (sinh and cosh)
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The hyperbolic sine ($\sinh$) and hyperbolic cosine function ($\cosh$) are defined by:
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$$\sinh x = \frac 1 2 (e^x - e^{-x}) \text{ and } \cosh x = \frac 1 2 (e^x + e^{-x})$$
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$$\tanh = \frac{\sinh x}{\cosh x}$$
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</summary>
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](./images/Sinh_cosh_tanh.svg)
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Some key facts about these functions:
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- $\cosh$ has even symmetry and $\sinh$ and $\tanh$ have odd symmetry
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- as $x \rightarrow + \infty$, $\cosh x \rightarrow +\infty$ and $\sinh x \rightarrow +\infty$
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- $\cosh^2x - \sinh^2x = 1$
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- $\tanh$'s limits are -1 and +1
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- Derivatives:
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- $\frac{\mathrm{d}}{\mathrm{d}x} \sinh x = \cosh x$
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- $\frac{\mathrm{d}}{\mathrm{d}x} \cosh x = \sinh x$
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- $\frac{\mathrm{d}}{\mathrm{d}x} \tanh x = \frac{1}{\cosh^2x}$
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</details>
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<details>
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<summary>
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### Natural Logarithm
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$$\ln{e^y} = \ln{\exp y} = y$$
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</summary>
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Since the exponential of any real number is positive, the domain of $\ln$ is $x > 0$.
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</details>
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<details>
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<summary>
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### Implicit Functions
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An implicit function takes the form
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$$f(x, y) = 0$$
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</summary>
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To draw the curve of an implicit function you have to rewrite it in the form $y = f(x)$.
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There may be more than one $y$ value for each $x$ value.
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</details>
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# Differentiation
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The derivative of the function $f(x)$ is denoted by:
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$$f'(x) \text{ or } \frac{\mathrm{d}}{\mathrm dx} f(x)$$
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Geometrically, the derivative is the gradient of the curve $y = f(x)$.
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It is a measure of the rate of change of $f(x)$ as $x$ varies.
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For example, velocity, $v$, is the rate of change of displacement, $s$, with respect to time, $t$,
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or:
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$$v = \frac{\mathrm ds}{dt}$$
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<details>
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<summary>
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#### Formal Definition
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</summary>
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As $h\rightarrow 0$, the clospe of the cord $\rightarrow$ slope of the tangent, or:
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$$f'(x_0) = \lim_{h\rightarrow0}\frac{f(x_0+h) - f(x_0)}{h}$$
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whenever this limit exists.
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</details>
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## Rules for Differentiation
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### Powers
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$$\frac{\mathrm d}{\mathrm dx} x^n = nx^{-1}$$
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### Trigonometric Functions
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$$\frac{\mathrm d}{\mathrm dx} \sin x = \cos x$$
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$$\frac{\mathrm d}{\mathrm dx} \cos x = \sin x$$
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### Exponential Functions
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$$\frac{\mathrm d}{\mathrm dx} e^{kx} = ke^{kx}$$
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$$\frac{\mathrm d}{\mathrm dx} \ln kx^n = \frac n x$$
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where $n$ and $k$ are constant.
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### Linearity
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$$\frac{\mathrm d}{\mathrm dx} (f + g) = \frac{\mathrm d}{\mathrm dx} f + \frac{\mathrm d}{\mathrm dx} g$$
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### Product Rule
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$$\frac{\mathrm d}{\mathrm dx} (fg) = \frac{\mathrm df}{\mathrm dx}g + \frac{\mathrm dg}{\mathrm dx}f$$
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### Quotient Rule
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$$ \frac{\mathrm d}{\mathrm dx} \frac f g = \frac 1 {g^2} \left( \frac{\mathrm df}{\mathrm dx} g - f \frac{\mathrm dg}{\mathrm dx} \right) $$
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$$ \left( \frac f g \right)' = \frac 1 {g^2} (gf' - fg')$$
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### Chain Rule
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Let
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$$f(x) = F(u(x))$$
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$$ \frac{\mathrm df}{\mathrm dx} = \frac{\mathrm{d}F}{\mathrm du} \frac{\mathrm du}{\mathrm dx} $$
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<details>
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<summary>
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#### Example 1
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Differentiate $f(x) = \cos{x^2}$.
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</summary>
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Let $u(x) = x^2$, $F(u) = \cos u$
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$$ \frac{\mathrm df}{\mathrm dx} = -\sin u \cdot 2x = 2x\sin{x^2} $$
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</details>
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## L'Hôpital's Rule
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l'Hôpital's rule provides a systematic way of dealing with limits of functions like
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$\frac{\sin x} x$.
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Suppose
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$$\lim_{x\rightarrow{a}} f(x) = 0$$
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and
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$$\lim_{x\rightarrow{a}} g(x) = 0$$
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and we want $\lim_{x\rightarrow{a}} \frac{f(x)}{g(x)}$.
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If
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$$\lim_{x\rightarrow{a}} \frac{f'(x)}{g'(x)} = L $$
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where any $L$ is any real number or $\pm \infty$, then
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$$\lim_{x\rightarrow{a}} \frac{f(x)}{g(x)} = L$$
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You can keep applying the rule until you get a sensible answer.
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# Graphs
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## Stationary Points
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An important application of calculus is to find where a function is a maximum or minimum.
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when these occur the gradient of the tangent to the curve, $f'(x) = 0$.
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The condition $f'(x) = 0$ alone however does not guarantee a minimum or maximum.
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It only means that point is a *stationary point*.
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There are three main types of stationary points:
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- maximum
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- minimum
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- point of inflection
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### Local Maximum
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The point $x = a$ is a local maximum if:
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$$f'(a) = 0 \text{ and } f''(a) < 0$$
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This is because $f'(x)$ is a decreasing function of $x$ near $x=a$.
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### Local Minimum
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The point $x = a$ is a local minimum if:
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$$f'(a) = 0 \text{ and } f''(a) > 0$$
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This is because $f'(x)$ is a increasing function of $x$ near $x=a$.
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### Point of Inflection
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$$f'(a) = 0 \text{ and } f''(a) = 0 \text { and } f'''(a) \ne 0$$
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#### $f'''(a) > 0$
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#### $f'''(a) < 0$
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# Approximating with the Taylor series
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The expansion
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$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$
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is an example of a *Taylor series*.
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These enable us to approximate a given function f(x) using a series which is often easier to
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calculate.
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Among other uses, they help us:
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- calculate complicated function using simple arithmetic operations
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- find useful analytical approximations which work for $x$ near a given value
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(e.g. $e^x \approx 1 + x$ for $x$ near 0)
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- Understand the behaviour of a function near a stationary point
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## Strategy
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Suppose we know information about $f(x)$ only at the point $x=0$.
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How can we find out about $f$ for other values of $x$?
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We could approximate the function by successive polynomials,
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each time matching more derivatives at $x=0$.
|
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\begin{align*}
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g(x) = a_0 &\text{ using } f(0) \\
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g(x) = a_0 + a_1x &\text{ using } f(0), f'(0) \\
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g(x) = a_0 + a_1x + a_2x^2 &\text{ using } f(0), f'(0), f''(0) \\
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&\text{and so on...}
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\end{align*}
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<details>
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<summary>
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#### Example 1
|
||||
|
||||
For $x$ near 0, approximate $f(x) = \cos x$ by a quadratic.
|
||||
|
||||
</summary>
|
||||
|
||||
1. Set $f(0) = g(0$:
|
||||
|
||||
$$f(0) = 1 \rightarrow g(0) = a_0 = 1$$
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2. Set $f'(0) = g'(0$:
|
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|
||||
$$f'(0) = -\sin0 = 0 \rightarrow g'(0) = a_1 = 0$$
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3. Set $f''(0) = g''(0$:
|
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$$f''(0) = -\cos = -1 \rightarrow g''(0) = 2a_2 = -1 \rightarrow a_2 = -0.5$$
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|
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So for $x$ near 0,
|
||||
|
||||
$$\cos x \approx 1 - \frac 1 2 x^2$$
|
||||
|
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Check:
|
||||
|
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$x$ | $\cos x$ | $1 - 0.5x^2$
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--- | -------- | ------------
|
||||
0.4 | 0.921061 | 0.920
|
||||
0.2 | 0.960066 | 0.980
|
||||
0.1 | 0.995004 | 0.995
|
||||
|
||||
</details>
|
||||
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||||
## 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.
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### 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*}
|
||||
|
||||
</summary>
|
||||
|
||||
|
||||
|
||||
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*}
|
||||
|
||||
</details>
|
||||
|
||||
## 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.
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
Find
|
||||
|
||||
$$I = \int\!(5x - 1)^3 \mathrm dx$$
|
||||
|
||||
</summary>
|
||||
|
||||
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
|
||||
|
||||
$$I = \frac{(5x-1)^4}{20} + c$$
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 2
|
||||
|
||||
Find
|
||||
|
||||
$$I = \int\! \cos x \sqrt{\sin x + 1} \mathrm dx$$
|
||||
|
||||
</summary>
|
||||
|
||||
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:
|
||||
|
||||
$$I = \frac23 (\sin x + 1)^{\frac32} + c$$
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 3
|
||||
|
||||
Find
|
||||
|
||||
$$I = \int^{\frac\pi2}_0\! \cos x \sqrt{\sin x + 1} \,\mathrm dx$$
|
||||
|
||||
</summary>
|
||||
|
||||
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
|
||||
|
||||
$$I = \left[ \frac23w^{\frac32} \right]^2_1 = \frac23 (2^{\frac23} = 1)$$
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 4
|
||||
|
||||
Find
|
||||
|
||||
$$I = \int^1_0\! \sqrt{1 - x^2} \,\mathrm dx$$
|
||||
|
||||
</summary>
|
||||
|
||||
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.
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
Find
|
||||
|
||||
$$I = \int\! \ln x \,\mathrm dx$$
|
||||
|
||||
</summary>
|
||||
|
||||
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*}
|
||||
|
||||
</details>
|
||||
|
||||
# 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:
|
||||
|
||||
$$\frac1{g(y} \cdot \frac{\mathrm dy}{\mathrm dx} = f(x)$$
|
||||
|
||||
$$\int\! \frac1{g(y)} \,\mathrm dy = \int\! f(x) \,\mathrm dx + c$$
|
||||
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
#### Example 1
|
||||
|
||||
Find $y$ such that
|
||||
|
||||
$$\frac{\mathrm dy}{\mathrm dx} = ky$$
|
||||
|
||||
where $k$ is a constant.
|
||||
|
||||
</summary>
|
||||
|
||||
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.
|
||||
|
||||
</details>
|
||||
Reference in New Issue
Block a user