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Exponential and Logarithmic Functions

Identify the form of an exponential function.

Exponential Functions

Exponential functions arise in many applications. One common example is population growth.

For example, if a population starts with ${P}_{0}$ individuals and then grows at an annual rate of $2\%,$ its population after 1 year is

\[P(1)={P}_{0}+0.02{P}_{0}={P}_{0}(1+0.02)={P}_{0}(1.02).\]

Its population after 2 years is

\[P(2)=P(1)+0.02P(1)=P(1)(1.02)={P}_{0}{(1.02)}^{2}.\]

In general, its population after $t$ years is

\[P(t)={P}_{0}{(1.02)}^{t},\]

which is an exponential function. More generally, any function of the form $f(x)={b}^{x},$ where $b>0,b\ne 1,$ is an exponential function with base $b$ and exponent x. Exponential functions have constant bases and variable exponents. Note that a function of the form $f(x)={x}^{b}$ for some constant $b$ is not an exponential function but a power function.

To see the difference between an exponential function and a power function, we compare the functions $y={x}^{2}$ and $y={2}^{x}.$ In , we see that both ${2}^{x}$ and ${x}^{2}$ approach infinity as $x\to \infty .$ Eventually, however, ${2}^{x}$ becomes larger than ${x}^{2}$ and grows more rapidly as $x\to \infty .$ In the opposite direction, as $x\to \text{-}\infty ,{x}^{2}\to \infty ,$ whereas ${2}^{x}\to 0.$ The line $y=0$ is a horizontal asymptote for $y={2}^{x}.$

$\text{x}$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
${\text{x}}^{2}$	$9$	$4$	$1$	$0$	$1$	$4$	$9$	$16$	$25$	$36$
${2}^{\text{x}}$	$1\text{/}8$	$1\text{/}4$	$1\text{/}2$	$1$	$2$	$4$	$8$	$16$	$32$	$64$

In , we graph both $y={x}^{2}$ and $y={2}^{x}$ to show how the graphs differ.

Condensed — the full section is in OpenStax Calculus Volume 1.

The Number

A special type of exponential function appears frequently in real-world applications. To describe it, consider the following example of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests $P$ dollars in a savings account with an annual interest rate $r,$ compounded annually. The amount of money after 1 year is

\[A(1)=P+rP=P(1+r).\]

The amount of money after $2$ years is

\[A(2)=A(1)+rA(1)=P(1+r)+rP(1+r)=P{(1+r)}^{2}.\]

More generally, the amount after $t$ years is

\[A(t)=P{(1+r)}^{t}.\]

If the money is compounded 2 times per year, the amount of money after half a year is

\[A(\frac{1}{2})=P+(\frac{r}{2})P=P(1+(\frac{r}{2})).\]

The amount of money after $1$ year is

\[A(1)=A(\frac{1}{2})+(\frac{r}{2})A(\frac{1}{2})=P(1+\frac{r}{2})+\frac{r}{2}(P(1+\frac{r}{2}))=P{(1+\frac{r}{2})}^{2}.\]

After $t$ years, the amount of money in the account is

\[A(t)=P{(1+\frac{r}{2})}^{2t}.\]

More generally, if the money is compounded $n$ times per year, the amount of money in the account after $t$ years is given by the function

\[A(t)=P{(1+\frac{r}{n})}^{nt}.\]\[{(1+\frac{r}{n})}^{nt}={(1+\frac{1}{m})}^{mrt},\]

$\text{m}$	$10$	$100$	$1000$	$10,000$	$100,000$	$1,000,000$
${(1+\frac{1}{\text{m}})}^{\text{m}}$	$2.5937$	$2.7048$	$2.71692$	$2.71815$	$2.718268$	$2.718280$

\[e\approx 2.718282.\]

Example

Try it.

Suppose $\text{\$}500$ is invested in an account at an annual interest rate of $r=5.5\%,$ compounded continuously.

Let $t$ denote the number of years after the initial investment and $A(t)$ denote the amount of money in the account at time $t.$ Find a formula for $A(t).$
Find the amount of money in the account after $10$ years and after $20$ years.

Solution

If $P$ dollars are invested in an account at an annual interest rate $r,$ compounded continuously, then $A(t)=P{e}^{rt}.$ Here $P=\text{\$}500$ and $r=0.055.$ Therefore, $A(t)=500{e}^{0.055t}.$
After $10$ years, the amount of money in the account is \[A(10)=500{e}^{0.055\cdot 10}=500{e}^{0.55}\approx \text{\$}866.63.\]
After $20$ years, the amount of money in the account is \[A(20)=500{e}^{0.055\cdot 20}=500{e}^{1.1}\approx \text{\$}1,502.08.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Logarithmic Functions

Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. These come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.

The exponential function $f(x)={b}^{x}$ is one-to-one, with domain $(\text{-}\infty ,\infty )$ and range $(0,\infty ).$ Therefore, it has an inverse function, called the logarithmic function with base $b.$ For any $b>0,b\ne 1,$ the logarithmic function with base b, denoted ${\text{log}}_{b},$ has domain $(0,\infty )$ and range $(\text{-}\infty ,\infty ),$ and satisfies

\[{\text{log}}_{b}(x)=y\ \text{if and only if}\ {b}^{y}=x.\]

For example,

\[\begin{array}{llll}{\text{log}}_{2}(8)=3 & & & \text{since}\ {2}^{3}=8, \\ {\text{log}}_{10}(\frac{1}{100})=-2 & & & \text{since}\ {10}^{-2}=\frac{1}{{10}^{2}}=\frac{1}{100}, \\ {\text{log}}_{b}(1)=0 & & & \text{since}\ {b}^{0}=1\ \text{for any base}\ b>0.\end{array}\]

Furthermore, since $y={\text{log}}_{b}(x)$ and $y={b}^{x}$ are inverse functions,

\[{\text{log}}_{b}({b}^{x})=x\ \text{and}\ {b}^{{\text{log}}_{b}(x)}=x.\]

The most commonly used logarithmic function is the function ${\text{log}}_{e}.$ Since this function uses natural $e$ as its base, it is called the natural logarithm. Here we use the notation $\text{ln}(x)$ or $\text{ln}\ x$ to mean ${\text{log}}_{e}(x).$ For example,

\[\text{ln}(e)={\text{log}}_{e}(e)=1,\text{ln}({e}^{3})={\text{log}}_{e}({e}^{3})=3,\text{ln}(1)={\text{log}}_{e}(1)=0.\]

Since the functions $f(x)={e}^{x}$ and $g(x)=\text{ln}(x)$ are inverses of each other,

\[\text{ln}({e}^{x})=x\ \text{and}\ {e}^{\text{ln}\ x}=x,\]

and their graphs are symmetric about the line $y=x$ ().

Condensed — the full section is in OpenStax Calculus Volume 1.

Hyperbolic Functions

The hyperbolic functions are defined in terms of certain combinations of ${e}^{x}$ and ${e}^{\text{-}x}.$ These functions arise naturally in various engineering and physics applications, including the study of water waves and vibrations of elastic membranes. Another common use for a hyperbolic function is the representation of a hanging chain or cable, also known as a catenary (). If we introduce a coordinate system so that the low point of the chain lies along the $y$-axis, we can describe the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.

The name cosh rhymes with “gosh,” whereas the name sinh is pronounced “cinch.” Tanh, sech, csch, and coth are pronounced “tanch,” “seech,” “coseech,” and “cotanch,” respectively.

Using the definition of $\text{cosh}(x)$ and principles of physics, it can be shown that the height of a hanging chain, such as the one in , can be described by the function $h(x)=a\ \text{cosh}(x\text{/}a)+c$ for certain constants $a$ and $c.$

But why are these functions called hyperbolic functions? To answer this question, consider the quantity ${\text{cosh}}^{2}t-{\text{sinh}}^{2}t.$ Using the definition of $\text{cosh}$ and $\text{sinh},$ we see that

\[{\text{cosh}}^{2}t-{\text{sinh}}^{2}t=\frac{{e}^{2t}+2+{e}^{-2t}}{4}-\frac{{e}^{2t}-2+{e}^{-2t}}{4}=1.\]

This identity is the analog of the trigonometric identity ${\text{cos}}^{2}t+{\text{sin}}^{2}t=1.$ Here, given a value $t,$ the point $(x,y)=(\text{cosh}\ t,\text{sinh}\ t)$ lies on the unit hyperbola ${x}^{2}-{y}^{2}=1$ ().

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

The exponential function $y={b}^{x}$ is increasing if $b>1$ and decreasing if \(0
The logarithmic function $y={\text{log}}_{b}(x)$ is the inverse of $y={b}^{x}.$ Its domain is $(0,\infty )$ and its range is $(\text{-}\infty ,\infty ).$
The natural exponential function is $y={e}^{x}$ and the natural logarithmic function is $y=\text{ln}\ x={\text{log}}_{e}x.$
Given an exponential function or logarithmic function in base $a,$ we can make a change of base to convert this function to any base $b>0,b\ne 1.$ We typically convert to base $e.$
The hyperbolic functions involve combinations of the exponential functions ${e}^{x}$ and ${e}^{\text{-}x}.$ As a result, the inverse hyperbolic functions involve the natural logarithm.

Exponential and Logarithmic Functions

For the following exercises, evaluate the given exponential functions as indicated, accurate to two significant digits after the decimal.

For the following exercises, match the exponential equation to the correct graph.

$y={4}^{\text{-}x}$
$y={3}^{x-1}$
$y={2}^{x+1}$
$y={(\frac{1}{2})}^{x}+2$
$y=\text{-}{3}^{\text{-}x}$
$y=1-{5}^{x}$

For the following exercises, sketch the graph of the exponential function. Determine the domain, range, and horizontal asymptote.

For the following exercises, write the equation in equivalent exponential form.

For the following exercises, write the equation in equivalent logarithmic form.

For the following exercises, sketch the graph of the logarithmic function. Determine the domain, range, and vertical asymptote.

For the following exercises, use properties of logarithms to write the expressions as a sum, difference, and/or product of logarithms.

Condensed — the full section is in OpenStax Calculus Volume 1.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Suppose a particular population of bacteria is known to double in size every $4$ hours. If a culture starts with $1000$ bacteria, the number of bacteria after $4$ hours is $n(4)=1000\cdot 2.$ The number of bacteria after $8$ hours is $n(8)=n(4)\cdot 2=1000\cdot {2}^{2}.$ In general, the number of bacteria after $4m$ hours is $n(4m)=1000\cdot {2}^{m}.$ Letting $t=4m,$ we see that the number of bacteria after $t$ hours is $n(t)=1000\cdot {2}^{t\text{/}4}.$ Find the number of bacteria after $6$ hours, $10$ hours, and $24$ hours.

Asehoy ny valinteny

The number of bacteria after 6 hours is given by $n(6)=1000\cdot {2}^{6\text{/}4}\approx 2828$ bacteria. The number of bacteria after $10$ hours is given by $n(10)=1000\cdot {2}^{10\text{/}4}\approx 5657$ bacteria. The number of bacteria after $24$ hours is given by $n(24)=1000\cdot {2}^{6}=64,000$ bacteria.
Given the exponential function $f(x)=100\cdot {3}^{x\text{/}2},$ evaluate $f(4)$ and $f(10).$

Asehoy ny valinteny

$f(4)=900;f(10)=24,300.$
Use the laws of exponents to simplify each of the following expressions.
1. $\frac{{(2{x}^{2\text{/}3})}^{3}}{{(4{x}^{-1\text{/}3})}^{2}}$
2. $\frac{{({x}^{3}{y}^{-1})}^{2}}{{(x{y}^{2})}^{-2}}$
Asehoy ny valinteny
1. We can simplify as follows: \[\frac{{(2{x}^{2\text{/}3})}^{3}}{{(4{x}^{-1\text{/}3})}^{2}}=\frac{{2}^{3}{({x}^{2\text{/}3})}^{3}}{{4}^{2}{({x}^{-1\text{/}3})}^{2}}=\frac{8{x}^{2}}{16{x}^{-2\text{/}3}}=\frac{{x}^{2}{x}^{2\text{/}3}}{2}=\frac{{x}^{8\text{/}3}}{2}.\]
2. We can simplify as follows: \[\frac{{({x}^{3}{y}^{-1})}^{2}}{{(x{y}^{2})}^{-2}}=\frac{{({x}^{3})}^{2}{({y}^{-1})}^{2}}{{x}^{-2}{({y}^{2})}^{-2}}=\frac{{x}^{6}{y}^{-2}}{{x}^{-2}{y}^{-4}}={x}^{6}{x}^{2}{y}^{-2}{y}^{4}={x}^{8}{y}^{2}.\]
Use the laws of exponents to simplify $(6{x}^{-3}{y}^{2})\text{/}(12{x}^{-4}{y}^{5}).$

Asehoy ny valinteny

$x\text{/}(2{y}^{3})$
Suppose $\text{\$}500$ is invested in an account at an annual interest rate of $r=5.5\%,$ compounded continuously.
1. Let $t$ denote the number of years after the initial investment and $A(t)$ denote the amount of money in the account at time $t.$ Find a formula for $A(t).$
2. Find the amount of money in the account after $10$ years and after $20$ years.
Asehoy ny valinteny
1. If $P$ dollars are invested in an account at an annual interest rate $r,$ compounded continuously, then $A(t)=P{e}^{rt}.$ Here $P=\text{\$}500$ and $r=0.055.$ Therefore, $A(t)=500{e}^{0.055t}.$
2. After $10$ years, the amount of money in the account is \[A(10)=500{e}^{0.055\cdot 10}=500{e}^{0.55}\approx \text{\$}866.63.\]
  After $20$ years, the amount of money in the account is \[A(20)=500{e}^{0.055\cdot 20}=500{e}^{1.1}\approx \text{\$}1,502.08.\]
If $\text{\$}750$ is invested in an account at an annual interest rate of $4\%,$ compounded continuously, find a formula for the amount of money in the account after $t$ years. Find the amount of money after $30$ years.

Asehoy ny valinteny

$A(t)=750{e}^{0.04t}.$ After $30$ years, there will be approximately $\text{\$}2,490.09.$
Solve each of the following equations for $x.$
1. ${5}^{x}=2$
2. ${e}^{x}+6{e}^{\text{-}x}=5$
Asehoy ny valinteny
1. Applying the natural logarithm function to both sides of the equation, we have \[\text{ln}{5}^{x}=\text{ln}\ 2.\]
  Using the power property of logarithms, \[x\ \text{ln}\ 5=\text{ln}\ 2.\]
  Therefore, $x=\text{ln}\ 2\text{/}\text{ln}\ 5.$
2. Multiplying both sides of the equation by ${e}^{x},$ we arrive at the equation \[{e}^{2x}+6=5{e}^{x}.\]
  Rewriting this equation as \[{e}^{2x}-5{e}^{x}+6=0,\]
  we can then rewrite it as a quadratic equation in ${e}^{x}\text{:}$
  \[{({e}^{x})}^{2}-5({e}^{x})+6=0.\]
  Now we can solve the quadratic equation. Factoring this equation, we obtain \[({e}^{x}-3)({e}^{x}-2)=0.\]
  Therefore, the solutions satisfy ${e}^{x}=3$ and ${e}^{x}=2.$ Taking the natural logarithm of both sides gives us the solutions $x=\text{ln}\ 3,\text{ln}\ 2.$
Solve ${e}^{2x}\text{/}(3+{e}^{2x})=1\text{/}2.$

Asehoy ny valinteny

$x=\frac{\text{ln}\ 3}{2}$
Solve each of the following equations for $x.$
1. $\text{ln}(\frac{1}{x})=4$
2. ${\text{log}}_{10}\sqrt{x}+{\text{log}}_{10}x=2$
3. $\text{ln}(2x)-3\ \text{ln}({x}^{2})=0$
Asehoy ny valinteny
1. By the definition of the natural logarithm function, \[\text{ln}(\frac{1}{x})=4\ \text{if and only if}\ {e}^{4}=\frac{1}{x}.\]
  Therefore, the solution is $x=1\text{/}{e}^{4}.$
2. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as \[{\text{log}}_{10}\sqrt{x}+{\text{log}}_{10}x={\text{log}}_{10}x\sqrt{x}={\text{log}}_{10}{x}^{3\text{/}2}=\frac{3}{2}{\text{log}}_{10}x.\]
  Therefore, the equation can be rewritten as \[\frac{3}{2}{\text{log}}_{10}x=2\ \text{or}\ {\text{log}}_{10}x=\frac{4}{3}.\]
  The solution is $x={10}^{4\text{/}3}=10\sqrt[3]{10}.$
3. Using the power property of logarithmic functions, we can rewrite the equation as $\text{ln}(2x)-\text{ln}({x}^{6})=0.$
  Using the quotient property, this becomes \[\text{ln}(\frac{2}{{x}^{5}})=0.\]
  Therefore, $2\text{/}{x}^{5}=1,$ which implies $x=\sqrt[5]{2}.$ We should then check for any extraneous solutions.
Solve $\text{ln}({x}^{3})-4\ \text{ln}(x)=1.$

Asehoy ny valinteny

$x=\frac{1}{e}$
Use a calculating utility to evaluate ${\text{log}}_{3}7$ with the change-of-base formula presented earlier.

Asehoy ny valinteny

Use the second equation with $a=3$ and $e=3\text{:}$
${\text{log}}_{3}7=\frac{\text{ln}\ 7}{\text{ln}\ 3}\approx 1.77124.$
Use the change-of-base formula and a calculating utility to evaluate ${\text{log}}_{4}6.$

Asehoy ny valinteny

$1.29248$
Compare the relative severity of a magnitude $8.4$ earthquake with a magnitude $7.4$ earthquake.

Asehoy ny valinteny

The magnitude $8.4$ earthquake is roughly $10$ times as severe as the magnitude $7.4$ earthquake.
1. Simplify $\text{sinh}(5\ \text{ln}\ x).$
2. If $\text{sinh}\ x=3\text{/}4,$ find the values of the remaining five hyperbolic functions.
Asehoy ny valinteny
1. Using the definition of the $\text{sinh}$ function, we write \[\text{sinh}(5\ \text{ln}\ x)=\frac{{e}^{5\ \text{ln}\ x}-{e}^{-5\ \text{ln}\ x}}{2}=\frac{{e}^{\text{ln}({x}^{5})}-{e}^{\text{ln}({x}^{-5})}}{2}=\frac{{x}^{5}-{x}^{-5}}{2}.\]
2. Using the identity ${\text{cosh}}^{2}x-{\text{sinh}}^{2}x=1,$ we see that \[{\text{cosh}}^{2}x=1+{(\frac{3}{4})}^{2}=\frac{25}{16}.\]
  Since $\text{cosh}\ x\ge 1$ for all $x,$ we must have $\text{cosh}\ x=5\text{/}4.$ Then, using the definitions for the other hyperbolic functions, we conclude that $\text{tanh}\ x=3\text{/}5,\text{csch}\ x=4\text{/}3,\text{sech}\ x=4\text{/}5,$ and $\text{coth}\ x=5\text{/}3.$
Simplify $\text{cosh}(2\ \text{ln}\ x).$

Asehoy ny valinteny

$({x}^{2}+{x}^{-2})\text{/}2$
Evaluate each of the following expressions.
\[{\text{sinh}}^{-1}(2)\]\[{\text{tanh}}^{-1}(1\text{/}4)\]

Asehoy ny valinteny

${\text{sinh}}^{-1}(2)=\text{ln}(2+\sqrt{{2}^{2}+1})=\text{ln}(2+\sqrt{5})\approx 1.4436$
${\text{tanh}}^{-1}(1\text{/}4)=\frac{1}{2}\text{ln}(\frac{1+1\text{/}4}{1-1\text{/}4})=\frac{1}{2}\text{ln}(\frac{5\text{/}4}{3\text{/}4})=\frac{1}{2}\text{ln}(\frac{5}{3})\approx 0.2554$
Evaluate ${\text{tanh}}^{-1}(1\text{/}2).$

Asehoy ny valinteny

$\frac{1}{2}\text{ln}(3)\approx 0.5493.$
$f(x)={5}^{x}$ a. $x=3$ b. $x=\frac{1}{2}$ c. $x=\sqrt{2}$

Asehoy ny valinteny

a. 125 b. 2.24 c. 9.74
$f(x)={(0.3)}^{x}$ a. $x=-1$ b. $x=4$ c. $x=-1.5$
$f(x)={10}^{x}$ a. $x=-2$ b. $x=4$ c. $x=\frac{5}{3}$

Asehoy ny valinteny

a. 0.01 b. 10,000 c. 46.42
$f(x)={e}^{x}$ a. $x=2$ b. $x=-3.2$ c. $x=\pi$
$f(x)={e}^{x}+2$

Asehoy ny valinteny

Domain: all real numbers, range: $(2,\infty ),y=2$
$f(x)=\text{-}{2}^{x}$
$f(x)={3}^{x+1}$

Asehoy ny valinteny

Domain: all real numbers, range: $(0,\infty ),y=0$
$f(x)={4}^{x}-1$
$f(x)=1-{2}^{\text{-}x}$

Asehoy ny valinteny

Domain: all real numbers, range: $(\text{-}\infty ,1),y=1$
$f(x)={5}^{x+1}+2$
$f(x)={e}^{\text{-}x}-1$

Asehoy ny valinteny

Domain: all real numbers, range: $(-1,\infty ),y=-1$
${\text{log}}_{3}81=4$
${\text{log}}_{8}2=\frac{1}{3}$

Asehoy ny valinteny

${8}^{1\text{/}3}=2$
${\text{log}}_{5}1=0$
${\text{log}}_{5}25=2$

Asehoy ny valinteny

${5}^{2}=25$
$\text{log}\ 0.1=-1$
$\text{ln}(\frac{1}{{e}^{3}})=-3$

Asehoy ny valinteny

${e}^{-3}=\frac{1}{{e}^{3}}$
${\text{log}}_{9}3=0.5$
$\text{ln}\ 1=0$

Asehoy ny valinteny

${e}^{0}=1$
${2}^{3}=8$
${4}^{-2}=\frac{1}{16}$

Asehoy ny valinteny

${\text{log}}_{4}(\frac{1}{16})=-2$
${10}^{2}=100$
${9}^{0}=1$

Asehoy ny valinteny

${\text{log}}_{9}1=0$

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\approx

approximately equal
Equal to the precision shown, not exactly.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Exponential and Logarithmic Functions

Identify the form of an exponential function.
Explain the difference between the graphs of
Recognize the significance of the number
Identify the form of a logarithmic function.
Explain the relationship between exponential and logarithmic functions.
Describe how to calculate a logarithm to a different base.
Identify the hyperbolic functions, their graphs, and basic identities.
We can simplify as follows:

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Andramo ny anao manokana

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Mbola maro ao Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests

\(\text{x}\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)
\({\text{x}}^{2}\)	\(9\)	\(4\)	\(1\)	\(0\)	\(1\)	\(4\)	\(9\)	\(16\)	\(25\)	\(36\)
\({2}^{\text{x}}\)	\(1\text{/}8\)	\(1\text{/}4\)	\(1\text{/}2\)	\(1\)	\(2\)	\(4\)	\(8\)	\(16\)	\(32\)	\(64\)

\(\text{m}\)	\(10\)	\(100\)	\(1000\)	\(10,000\)	\(100,000\)	\(1,000,000\)
\({(1+\frac{1}{\text{m}})}^{\text{m}}\)	\(2.5937\)	\(2.7048\)	\(2.71692\)	\(2.71815\)	\(2.718268\)	\(2.718280\)