# Calculus I

This article was considered for deletion at Wikipedia on November 21 2016. This is a backup of Wikipedia:Calculus_I. All of its AfDs can be found at Wikipedia:Special:PrefixIndex/Wikipedia:Articles_for_deletion/Calculus_I, the first at Wikipedia:Wikipedia:Articles_for_deletion/Calculus_I. Purge

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Calculus I is a math course in undergraduate education worldwide and is nearly identical to what is covered in AP Calculus. It typically covers a well-known subset of the topics of calculus, which is a branch of mathematics that deals with the finding and properties of derivatives and integrals of other functions. Many students in the STEM fields worldwide are required to either get credit via AP Calculus (or the equivalent in high school) or take Calculus I. Many institution also have courses called Calculus II and Calculus III. Some students find Calculus I to be daunting because of its abstract nature due to the rigor and formality that expanded and generalized the subject in the 100 years after it was elegantly defined by Newton and Leibniz.

This is a semi-technical article covering some of Calculus I. It reviews only those concepts and functions that can be visualized. Calculus math notation and nomenclature is introduced as late as practicable. The slope-value-area triple is novel to this article (known by some as "original research"). This article will only consider curves that are a smooth graph of a function$f(x)$ with no discontinuities. Curves are generally discussed in the Cartesian coordinate system and points on the curves are presented as $(x,y)$ coordinates.

The scope of the Calculus I topics presented as sections are:

## Functions, curves and the slope-value-area triple

Consider the curve $y(x)=x^2$ which this article refers to as a "simple parabola". For any given positive value of $x$ called $x_1$, this curve passes through the point $(x_1,x_1^2)$. We call that point $X_1$. At $X_1$ the slope -value-area-triple is:

• the slope of that curve at that point (the instantaneous slope or rate of change). This slope is called the "derivative". Calculating the derivative is known as "differentiation".
• the value of the function (the value of $y$) is $x_1^2$.
• the area under the curve (AUC) at $x_1$. That is the area bounded by the curve from:
• $(0,0)$ to $X_1$
• the vertical line segment from $X_1$ to $(x_1,0)$
• the horizontal line segment $(x_1,0)$ to $(0,0)$. This area is called the "integral".

This article lists these three values in tables for the purpose of clarity.

Calculating the integral is known as "integration". For other functions with negative values (such as $sin(x)$ and $cos(x))$, the area accumulated for negative values of $y$ subtract from the positive area accumulated as $x$ increases.

The derivative and integral are linear functions. Thus the slope of $(f(x)+g(x))$ equals the slope of $f(x)$ plus the slope of $g(x)$. The same applies for integral.

Consider skimming Fundamental theorem of calculus for a hint of a much more rigorous and general treatment of these concepts.

## Undefined values and other special cases

In computing the slope of a straight line, the formula is

$\text{slope} = \frac{\Delta y}{\Delta x}$

but we cannot directly use that formula for the instantaneous slope of a curve because $\frac{0}{0}$ is an undefined value and an expression such as zero-divided-by-zero is an indeterminate form. We will solve this problem by utilizing limits.

In exponentiation, another undefined value is $0^0$. This is, in part, because of these two limits that approach two different values:

$\lim_{x \to 0} 0^x=0$
$\lim_{x \to 0} x^0=1$

These two do not approach the same value. We use a positive base such as $2^x$. If we consider $(-2)^x$, we find that at $x=\frac{1}{2}$ (or any other rational number that can be expressed as an odd integer divided by an even integer), the value involves $\sqrt{-2}$ and thus to the imaginary unit, so we avoid such curves in this article. If you have a graph such as $y=\sqrt{|x|}$ or other fractional exponents, then there is a cusp at (0,0) and the slope at that point is undefined.

## Power rule

The power rule is usually one of the first rule covered in Calculus I.Template:Sfn This section will first take a non-calculus approach.

### A pattern in integer powers of x

The term "curve" includes straight lines. Perhaps the simplest curve is the straight line $y=0$:

Slope Value Area
0 0 0

The next curve is $y=x^0=1$:

Slope Value Area
0 1 $x$

By "$x$" what we mean is that the area under the curve is the rectangle that is 1 by $x_1$ or simply $x$ in area.

The next curve is $y=x^1=x$.

Slope Value Area
1 $x$ $\frac{1}{2}x^2$

The area is a square with side length of $x_1$ and cut in half by the curve, which is a diagonal of the square. In this triple, a pattern emerges in that the exponent of $x$ increases by one as the terms progress left-to-right. The pattern is less obvious in the earlier tables.

Consider again the simple parabola, the curve $y=x^2$:

Slope Value Area
$2x$ $x^2$ $\frac{1}{3}x^3$

The pattern in the exponent of $x$ is even more clear and there is a pattern in the coefficients (2,1,$\frac{1}{3}$).

### Two proofs for the simple parabola

While calculus proofs are not usually expected in Calculus I examinations, the following is a proof for the slope and area of the simple parabola. AP Calculus#AP Calculus BC does not always require proofs, but the concept is introduced.

#### Derivative of the simple parabola

To calculate the slope of the simple parabola, consider two points on the parabola for some positive value of $x_1$ and positive $h$ the two points:

$X_1$ at $(x_1,x_1^2)$
$X_2$ at $(x_1+h,(x_1+h)^2)$

Draw a line (either the line segment $X_1X_2$ which is a chord or the corresponding secant line). Now decrease $h$ until it approaches zero. That chord will have a slope that approaches the slope of the curve at the $X_1$. Evaluate the slope of the line and take the limit (the slope of the tangent line):

\begin{align} \text{slope} & =\frac{\Delta y}{\Delta x} \\[8pt] & =\frac{(x_1+h)^2-x_1^2}{(x_1+h)-x_1} \\[8pt] & =\frac{(x_1^2+2hx_1+h^2)-x_1^2} h \end{align}

(note that all terms vanish in the next few steps except the second-from-the-left term $2hx_1$ )

$=\frac{2hx_1+h^2}{h}$
$=2x_1+h$
$\lim_{h \to 0}(2x_1+h)=2x_1$

Note for the term $(x+h)^2$, it expands to $(x^2+2hx+h^2)$, with numerical coefficients of 1,2,1 . To the power (the exponent) of 3, the pattern is 1,3,3,1 . The value $2x$ corresponds to the\ term that is second-from-the-left in row 2 of Pascal's triangle, with all the term(s) to the right having power(s) of $h$ that vanish as $h$ goes to zero. The curve $y=x^3$ leads to row 3 of the triangle and the slope is $3x^2$. We now have the basis for finding the slope of $y=x^n$ for any integer $n$.

#### Integral of the simple parabola

To calculate the area of the simple parabola, imagine dividing some finite amount of the area from 0 to $x_1$ constructing $n$ vertical equal-width rectangles. Calculate the sum of the area of those rectangles and take the limit as $n$ approaches infinity:

\begin{align} \text{area} & =\lim_{n \to \infty} \sum_{i=1}^n \text{rectangle area }_i \\[8pt] & =\lim_{n \to \infty} \sum_{i=1}^n (\text{height }_i) \text{width} \\[8pt] & =\lim_{n \to \infty} \sum_{i=1}^n f(x_i) \, \Delta_x \\[8pt] & =\lim_{n \to \infty} \sum_{i=1}^n \left( \left(\frac{x_1i} n \right)^2 \frac{x_1} n \right) \\[8pt] & =\lim_{n \to \infty} \frac{x_1^3}{n^3} \sum_{i=1}^n i^2 \end{align}

(note the sum of sequence of squares, for which an identity formula exists)

\begin{align} \qquad & =\lim_{n \to \infty} \left(\frac{x_1^3}{n^3}\right) \frac{n(n+1)(2n+1)}{6} \\[8pt] & =\lim_{n \to \infty} \frac{x_1^3(2n^3+3n^2+n)}{6n^3} \end{align}

(note that all terms except the left-most one with the highest exponent $2n^3$ vanish)

\begin{align} \quad & =\frac{2x_1^3}{6} \\[8pt] & =\frac{1}{3}x_1^3 \end{align}

For the curve $y=x^3$, the the identity formulas for the sum of the cubes is $\frac{n^2(n+1)^2}{4}$, all of the terms except the left-most term with the highest exponent vanish and the results are:

Slope Value Area
$3x^2$ $x^3$ $\frac{1}{4}x^4$

And so on for higher powers, thus we can calculate the area for $y=x^n$ for any integer $n$ except for $n=(-1)$.

### The power rule for differentiation and integration and implications for calculus

For integer $n$ and the curve $y=x^n$, the power rule states:

Derivative of fuction Function Integral of fuction
$nx^{n-1}$ $x^n$ $\frac{1}{n+1}x^{n+1}$

These formulas are true for any integer value of $n$ except for $n=(-1)$ for the integral. That is the curve $y=\frac{1}{x}$. It is a special case that defines the natural logarithm. These formulas can be generalized to any value of $n$ (any real number including non-integer values), but this article avoids covering that topic.

We already know that the slope is the derivative of the value-curve and the area is the integral of the curve. By examining the power rule, we see that derivative and integral are inverse functions of each other. This is true not just for powers of x but for all functions. This is one of the great insights of Newton and Leibniz. To be complete: the integral is the antiderivative plus the constant of integration.Template:Sfn The derivative of the integral of the curve is just the curve again (and the other way around). This implies the surprising results that the value is the integral of the slope and the value is the derivative of the area. The relationship between area and value (and that between value and slope) is bidirectional. This turns out to be true not just for powers of $x$ but for any function curve. Given two adjacent columns in one of the tables, the item to the left is the derivative of the item to the right and the item to the right is the integral of the item to the left. On top of that, the item two columns to the left is the second derivative. More columns can be added to the tables for demonstrating even higher-order derivatives and integrals.

To emphasize this bidirectional relationship, consider the two functions $y=2x$ and $y=x^2$. The area of the first curve is $2(\frac{1}{2}x^2)$ or just $x^2$. The slope of that latter curve is $2x$ and we return to our original formula as we take the derivative of the integral (or the integral of the derivative).

## Physical models of derivative and integral

### Derivatives for 1-,2- and 3-dimensional objects

Imagine that the value of $x$ represented time $t$. Think of $y$ as a time-varying function $y=f(t)$ and examine the differential. $dt$ is an infinitesimal (akin to the limit as $dt$ goes to zero).

For the one-dimensional case, y is length of steadily-growing thread $L$ with one end fixed at (0) and growing at the other end, thus $L(t)=t$ i.e. length grows linearly with time, like, perhaps a thin icicle, thus the growth is:

$dL=dt$
$\frac{d}{dt}L=1$
$\frac{d}{dt}t=1$

This last line is one of the standard mathematical notations for the derivative.

For the two-dimensional case, imagine we have a square thin layer of a ice that is stable on the origin (0,0) grows along the two sides opposite the x-axis and the y-axis, thus the area is $A(t)=t^2$. The growth over a little bit of time is the sume of:

• the growth of the two growing sides of the square
• the little square of growth at corner opposite (0,0):
$dA=(2t)dt+(dt)^2$
$\frac{d}{dt}A=2t+dt$

but $dt$ is an infinitesimal:

$\frac{d}{dt}A=2t$
$\frac{d}{dt}t^2=2t$

For the three-dimensional case, consider a growing ice cube with one corner fixed at (0,0,0) with volume $V(t)=t^3$. The growth of the volume over a short period of time $dV(t)$ is the sum of:

• the growth along the three faces of the cube opposite the three planes of the coordinate system
• the growth along the edges opposite the three axes of the coordinate system
• the little cube opposite the (0,0,0) corner:
$dV=(3t^2)dt+(3t)(dt)^2+(dt)^3$
$\frac{d}{dt}V=3t^2+(3t)dt+(dt)^2$

and since $dt$ is an infinitesimal:

$\frac{d}{dt}V=3t^2$
$\frac{d}{dt}t^3=3t^2$

In all three cases, the derivative (rate of growth) of the dimensional object quantity is the same as for the slopes of the curves of the powers of $x$ and the same as what the power rule finds.

### Physics

Consider in physics a ball thrown horizontally off of a tall building. The horizontal speed is constant and the horizontal distance traveled is horizontal velocity times time. In the vertical, We have the distance-velocity-acceleration (s-v-a) triple where vertical acceleration is constant and the downwards direction is positive for our coordinate system. The ball follows the path of a simple parabola. The middle column in the table below represents $v(t)=at$. In the vertical:

Acceleration Velocity Distance
a $at$ $a\frac{1}{2}t^2$

So we have the familiar formula $s(t)=\frac{1}{2}at^2$. The formula for kinetic energy, $E=\frac{1}{2}mv^2$ has the same form for similar reasons.

### A money triple

Assume that you rent an apartment without a lease and the rent goes up a little every month. The rent is a rate: dollars per month. A graph of the rent each month since you moved in and a smooth curve through the data points would present the rent as a function of time. The triple is:

Slope Value Area
Rate of monthly rise of rent Rent Total money paid

A proper financial planner can use these three values to make better-informed decisions and predictions. Many other practical analogies such as fuel being filled and consumed from a fuel tank (or a water tank, etc.) are possible.

## Non-zero constant of integration

The antiderivatives of the curves that have been considered so far pass through (0,0) and thus have a constant of integration are zero and we have ignored them.

In the next section, the antiderivatiive of sine is negative cosine. The cosine function (positive or negative) does not pass through (0,0) but the area under the sine function must be zero at $x=0$. The integral of sine is negative cosine minus $-cos(0)$. Negative cosine at $x=0$ is -1. We add the negative of that value to the antiderivative to get the integral. Thus the constant of integration is simply 1.

## Sine and cosine

Sine and cosine are a pair of trigonometric functions we shall call the "trig pair". In calculus, the trig pair have a relationship with each other. It helps to visualize both functions on one graph. At (0,0), the slope of the $sin(x)$ function equals 1. As $x$ increases to $\frac{\pi}{2}$, the reader can guess that the slope of $sin(x)$ is $cos(x)$. One can follow the same approach and see that the slope of $cos(x)$ is $-sin(x)$. The integral of $cos(x)$ is $sin(x)$, but the integral of $sin(x)$ is trickier. It is $1-cos(x)$. Think of that as $-cos(x)$ with an constant of integration equal to 1. Visualize that curve: start with $cos(x)$, flip it around the x-axis to get $-cos(x)$ and then shift the curve up by 1. Now you have a curve that starts at (0,0) and then forms a series of hills that rise up from the x-axis and then fall back down to zero as x increases. The trig pair have a four-iteration cycle period for higher orders of dirivatives:

 Slope Value 4th derivative 3rd derivative 2nd derivative $sin(x)$ $-cos(x)$ $-sin(x)$ $cos(x)$ $sin(x)$

The fourth derivative of $sin(x)$ is just $sin(x)$ again and the cycle repeats. Note that the physical significance of the fact that the second derivative of $sin(x)$ is $-sin(x)$ is that of simple harmonic motion. For a heavy weight hanging from a spring, the displacement from the resting state is proportional to the restoring force (which results in the acceleration) but in the negative (opposite) direction. For an application of such in analog electronics, see Electronic oscillator#Harmonic oscillator.

Repeated integration is not as simple:

 3rd integral 4th integral Value Area 2nd integral $sin(x)$ $1-cos(x)$ $x-sin(x)$ $1+\frac{1}{2}x^2+cos(x)$ $x+\frac{1}{6}x^3+sin(x)$

While the trig pair component of these formulas does cycle around, the terms that are powers of x accumulate.

## Definite integral

We initially defined the integral as the area under the cutve from (0,0) to $(x,f(x))$. We expand the definition of the area under the curve from $x_1$ to $x_2$ as the area bounded by:

• the curve from $(x_1,f(x_1))$ to $(x_2,f(x_2))$
• the vertical line segment from $(x_2,f(x_2))$ to $(x_2,0)$
• the horizontal line segment $(x_2,0)$ to $(x_1,0)$
• the vertical line from $(x_1,0)$ to $(x_1,f(x_1))$.

The definite integral is the area under the curve from $(x_1,f(x_1))$ to $(x_2,f(x_2))$. To evaluate it, compute the area under (0,0) to $(x_2,f(x_2))$ and subtract the area under (0,0) to $(x_1,f(x_1))$. In standard calculus notation, the definite integral is:

$\int_{x_1}^{x_2} f(x)\,dx = \int_0^{x_2} f(x)\,dx - \int_0^{x_1} f(x)\,dx$

If you sum up the area in the direction of negative $x$ then what looks like a positive area counts as negative. In other words, if $x_1>x_2$ then even if the function value is always positive, then the integral evaluates to a negative value. In short

$\int_{x_1}^{x_2} f(x)\,dx = - \int_{x_2}^{x_1} f(x)\,dx$

Pleaae review Fundamental theorem of calculus again.Template:Sfn

## Exponentiation

The natural logarithm is the area under the curve $y=\frac{1}{x}$ starting at $x=1$.Template:Sfn This is different from the other areas we have discussed, where we started at $x=0$:

$\ln(x_1)=\int_1^{x_1} \frac{1}{x}\,dx.$

This is that special case for the integral of $y=x^n$ where $n=(-1)$. Since integral and the derivative are inverses of each other, then at the point $(x_1,\ln(x_1))$, the slope is:

$\frac{d}{dt}\ln(x)=\frac{1}{x_1}.$

In our triple table:

Slope Value Area
$\frac{1}{x}$ $\ln x$ $x\ln x-x$

The base of the natural logarithm is the mathematical constant $e$. It is used extensively in studies of exponential growth and exponential decay. The slope and the area (except for the non-zero constant of integration) of the canonical exponential curve $y=e^x$ are remarkably, both identities:

Slope Value Area
$e^x$ $e^x$ $e^x-1$

On that curve, at the point $(a,e^a)$ in standard calculus notation:

$\frac{d}{dx}e^x=e^a$
$\int_0^a e^x\,dx=e^a-1$

## Other rules of differentiatoin and integration

There are several other rules beyond the power rule that Calculus I covers:

Differentiation rules
• Product rule - Consider the area of a rectangle with sides $x(t)$ and $y(t)$ that is growing on the two sides opposite the x-axis and y-axis. The area is $A=xy$. The growth of the area is the growth along the side opposite the axes of the x axis and that opposite the y axis and the little rectangle in the corner opposite (0,0):
$dA=x(\frac{d}{dt}y)+(\frac{d}{dt}x)y+(\frac{d}{dt}x)(\frac{d}{dt}y)$
$\frac{d}{dt}A=x(dy)+(dx)y+(dx)(dy)$

but as $dt, dx, dx$ are infinitesimals and the product of them vanishes, thus the growth is:

$\frac{d}{dt}A=x(dy)+(dx)y$th
$\frac{d}{dt}(xy)=x(dy)+(dx)y$
• Chain rule - Not as easy to visualize, but the math is easy to remember. A simple way to partially verify the chain rule is to apply the chain rule to calculate the derivative of $(x^2)^3$ and verify that it is the same results as with calculating the derivative by directly applying the power rule to $x^6$.
• Quotient rule - can be easily proven using either the product rule or the chain rule.
• Implicit function#Implicit differentiation - using chain rule
Integration rules

## Other Calculus I topics

Calculus I incudes:

• Maxima and minima - finding where the derivative equals zero, sometimes used in optimization, such as minimizing engineering or financial cost
• Linear approximation - simple application of the derivative
• Related rates - often uses the chain rule
• Critical point (mathematics) - a place where the derivative is zero or undefined (or infinite?)
• Derivative test
• Curvature of curves - follows the sign of the second derivative
• Functions that have x in its denominator may have asymptotes.
• Sometimes it is easier to calculate the integral with respect to y rather than x, i.e. using horizontal rectangles rather than vertical rectangles or $\int f(y)\,dy$.

## Later topics

Calculus can be applied in many ways beyond what can be easily visualized and beyond what is covered in most Calculus I courses: