maths.free › Calculus › 1. Functions and Graphs › Inverse Functions

Inverse Functions

Determine the conditions for when a function has an inverse.

Existence of an Inverse Function

We begin with an example. Given a function $f$ and an output $y=f(x),$ we are often interested in finding what value or values $x$ were mapped to $y$ by $f.$ For example, consider the function $f(x)={x}^{3}+4.$ Since any output $y={x}^{3}+4,$ we can solve this equation for $x$ to find that the input is $x=\sqrt[3]{y-4}.$ This equation defines $x$ as a function of $y.$ Denoting this function as ${f}^{-1},$ and writing $x={f}^{-1}(y)=\sqrt[3]{y-4},$ we see that for any $x$ in the domain of $f,{f}^{-1}(f(x))={f}^{-1}({x}^{3}+4)=x.$ Thus, this new function, ${f}^{-1},$ “undid” what the original function $f$ did. A function with this property is called the inverse function of the original function.

Note that ${f}^{-1}$ is read as “f inverse.” Here, the $-1$ is not used as an exponent and ${f}^{-1}(x)\ne 1\text{/}f(x).$ shows the relationship between the domain and range of f and the domain and range of ${f}^{-1}.$

Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each input to exactly one output. For example, let’s try to find the inverse function for $f(x)={x}^{2}.$ Solving the equation $y={x}^{2}$ for $x,$ we arrive at the equation $x=\pm \sqrt{y}.$ This equation does not describe $x$ as a function of $y$ because there are two solutions to this equation for every $y>0.$ The problem with trying to find an inverse function for $f(x)={x}^{2}$ is that two inputs are sent to the same output for each output $y>0.$ The function $f(x)={x}^{3}+4$ discussed earlier did not have this problem. For that function, each input was sent to a different output. A function that sends each input to a different output is called a one-to-one function.

Example

Try it.

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

Solution

Since the horizontal line $y=n$ for any integer $n\ge 0$ intersects the graph more than once, this function is not one-to-one.
Since every horizontal line intersects the graph once (at most), this function is one-to-one.

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

Finding a Function’s Inverse

We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of $f$ to elements in the range of $f.$ The inverse function maps each element from the range of $f$ back to its corresponding element from the domain of $f.$ Therefore, to find the inverse function of a one-to-one function $f,$ given any $y$ in the range of $f,$ we need to determine which $x$ in the domain of $f$ satisfies $f(x)=y.$ Since $f$ is one-to-one, there is exactly one such value $x.$ We can find that value $x$ by solving the equation $f(x)=y$ for $x.$ Doing so, we are able to write $x$ as a function of $y$ where the domain of this function is the range of $f$ and the range of this new function is the domain of $f.$ Consequently, this function is the inverse of $f,$ and we write $x={f}^{-1}(y).$ Since we typically use the variable $x$ to denote the independent variable and $y$ to denote the dependent variable, we often interchange the roles of $x$ and $y,$ and write $y={f}^{-1}(x).$ Representing the inverse function in this way is also helpful later when we graph a function $f$ and its inverse ${f}^{-1}$ on the same axes.

Example

Try it.

Find the inverse for the function $f(x)=3x-4.$ State the domain and range of the inverse function. Verify that ${f}^{-1}(f(x))=x.$

Solution

Follow the steps outlined in the strategy.

Step 1. If $y=3x-4,$ then $3x=y+4$ and $x=\frac{1}{3}y+\frac{4}{3}.$

Step 2. Rewrite as $y=\frac{1}{3}x+\frac{4}{3}$ and let $y={f}^{-1}(x).$

Therefore, ${f}^{-1}(x)=\frac{1}{3}x+\frac{4}{3}.$

Since the domain of $f$ is $(\text{-}\infty ,\infty ),$ the range of ${f}^{-1}$ is $(\text{-}\infty ,\infty ).$ Since the range of $f$ is $(\text{-}\infty ,\infty ),$ the domain of ${f}^{-1}$ is $(\text{-}\infty ,\infty ).$

You can verify that ${f}^{-1}(f(x))=x$ by writing

\[{f}^{-1}(f(x))={f}^{-1}(3x-4)=\frac{1}{3}(3x-4)+\frac{4}{3}=x-\frac{4}{3}+\frac{4}{3}=x.\]

Note that for ${f}^{-1}(x)$ to be the inverse of $f(x),$ both ${f}^{-1}(f(x))=x$ and $f({f}^{-1}(x))=x$ for all x in the domain of the inside function.

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

Inverse Trigonometric Functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function (). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval $[-\frac{\pi }{2},\frac{\pi }{2}].$ By doing so, we define the inverse sine function on the domain $[-1,1]$ such that for any $x$ in the interval $[-1,1],$ the inverse sine function tells us which angle $\theta$ in the interval $[-\frac{\pi }{2},\frac{\pi }{2}]$ satisfies $\text{sin}\ \theta =x.$ Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions, which are functions that tell us which angle in a certain interval has a specified trigonometric value.

Definition

The inverse sine function, denoted ${\text{sin}}^{-1}$ or arcsin, and the inverse cosine function, denoted ${\text{cos}}^{-1}$ or arccos, are defined on the domain $D=\text{\{}x|-1\le x\le 1\}$ as follows:

\[\begin{array}{l}{\text{sin}}^{-1}(x)=y\ \text{if and only if}\ \text{sin}(y)=x\ \text{and}\ -\frac{\pi }{2}\le y\le \frac{\pi }{2}; \\ {\text{cos}}^{-1}(x)=y\ \text{if and only if}\ \text{cos}(y)=x\ \text{and}\ 0\le y\le \pi .\end{array}\]

The inverse tangent function, denoted ${\text{tan}}^{-1}$ or arctan, and inverse cotangent function, denoted ${\text{cot}}^{-1}$ or arccot, are defined on the domain $D=\{x|-\infty \[\begin{array}{l}{\text{tan}}^{-1}(x)=y\ \text{if and only if}\ \text{tan}(y)=x\ \text{and}\ -\frac{\pi }{2}The inverse cosecant function, denoted \({\text{csc}}^{-1}$ or arccsc, and inverse secant function, denoted ${\text{sec}}^{-1}$ or arcsec, are defined on the domain $D=\{x||x|\ge 1\}$ as follows:

\[\begin{array}{l}{\text{csc}}^{-1}(x)=y\ \text{if and only if}\ \text{csc}(y)=x\ \text{and}\ -\frac{\pi }{2}\le y\le \frac{\pi }{2},y\ne 0; \\ {\text{sec}}^{-1}(x)=y\ \text{if and only if}\ \text{sec}(y)=x\ \text{and}\ 0\le y\le \pi ,y\ne \pi \text{/}2.\end{array}\]

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line $y=x$ ().

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions $\text{sin}({\text{sin}}^{-1}(\frac{\sqrt{2}}{\ \ 2}))$ and ${\text{sin}}^{-1}(\text{sin}(\pi )).$ For the first one, we simplify as follows:

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

Key Concepts

For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine whether the function is one-to-one by using the horizontal line test.
If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and then define the inverse of the function on the smaller domain.
For a function $f$ and its inverse ${f}^{-1},f({f}^{-1}(x))=x$ for all $x$ in the domain of ${f}^{-1}$ and ${f}^{-1}(f(x))=x$ for all $x$ in the domain of $f.$
Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric functions.
The graph of a function $f$ and its inverse ${f}^{-1}$ are symmetric about the line $y=x.$

Key Equations

Inverse functions

${f}^{-1}(f(x))=x\ \text{for all}\ x\ \text{in}\ D,\text{and}\ f({f}^{-1}(y))=y\ \text{for all}\ y\ \text{in}\ R.$

Inverse Functions

For the following exercises, use the horizontal line test to determine whether each of the given graphs is one-to-one.

For the following exercises, a. find the inverse function, and b. find the domain and range of the inverse function.

For the following exercises, use the graph of $f$ to sketch the graph of its inverse function.

For the following exercises, use composition to determine which pairs of functions are inverses.

For the following exercises, evaluate the functions. Give the exact value.

Practice (40)

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

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.
જવાબ બતાવો
1. Since the horizontal line $y=n$ for any integer $n\ge 0$ intersects the graph more than once, this function is not one-to-one.
2. Since every horizontal line intersects the graph once (at most), this function is one-to-one.
Is the function $f$ graphed in the following image one-to-one?

જવાબ બતાવો

No.
Find the inverse for the function $f(x)=3x-4.$ State the domain and range of the inverse function. Verify that ${f}^{-1}(f(x))=x.$

જવાબ બતાવો

Follow the steps outlined in the strategy.
Step 1. If $y=3x-4,$ then $3x=y+4$ and $x=\frac{1}{3}y+\frac{4}{3}.$
Step 2. Rewrite as $y=\frac{1}{3}x+\frac{4}{3}$ and let $y={f}^{-1}(x).$
Therefore, ${f}^{-1}(x)=\frac{1}{3}x+\frac{4}{3}.$
Since the domain of $f$ is $(\text{-}\infty ,\infty ),$ the range of ${f}^{-1}$ is $(\text{-}\infty ,\infty ).$ Since the range of $f$ is $(\text{-}\infty ,\infty ),$ the domain of ${f}^{-1}$ is $(\text{-}\infty ,\infty ).$
You can verify that ${f}^{-1}(f(x))=x$ by writing
\[{f}^{-1}(f(x))={f}^{-1}(3x-4)=\frac{1}{3}(3x-4)+\frac{4}{3}=x-\frac{4}{3}+\frac{4}{3}=x.\]
Note that for ${f}^{-1}(x)$ to be the inverse of $f(x),$ both ${f}^{-1}(f(x))=x$ and $f({f}^{-1}(x))=x$ for all x in the domain of the inside function.
Find the inverse of the function $f(x)=3x\text{/}(x-2).$ State the domain and range of the inverse function.

જવાબ બતાવો

${f}^{-1}(x)=\frac{2x}{x-3}.$ The domain of ${f}^{-1}$ is $\{x|x\ne 3\}.$ The range of ${f}^{-1}$ is $\{y|y\ne 2\}.$
For the graph of $f$ in the following image, sketch a graph of ${f}^{-1}$ by sketching the line $y=x$ and using symmetry. Identify the domain and range of ${f}^{-1}.$

જવાબ બતાવો

Reflect the graph about the line $y=x.$ The domain of ${f}^{-1}$ is $[0,\infty ).$ The range of ${f}^{-1}$ is $[-2,\infty ).$ By using the preceding strategy for finding inverse functions, we can verify that the inverse function is ${f}^{-1}(x)={x}^{2}-2,$ as shown in the graph.
Sketch the graph of $f(x)=2x+3$ and the graph of its inverse using the symmetry property of inverse functions.

જવાબ બતાવો
Consider the function $f(x)={(x+1)}^{2}.$
1. Sketch the graph of $f$ and use the horizontal line test to show that $f$ is not one-to-one.
2. Show that $f$ is one-to-one on the restricted domain $[-1,\infty ).$ Determine the domain and range for the inverse of $f$ on this restricted domain and find a formula for ${f}^{-1}.$
જવાબ બતાવો
1. The graph of $f$ is the graph of $y={x}^{2}$ shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, $f$ is not one-to-one.
2. On the interval $[-1,\infty ),\ f$ is one-to-one.
  
  The domain and range of ${f}^{-1}$ are given by the range and domain of $f,$ respectively. Therefore, the domain of ${f}^{-1}$ is $[0,\infty )$ and the range of ${f}^{-1}$ is $[-1,\infty ).$ To find a formula for ${f}^{-1},$ solve the equation $y={(x+1)}^{2}$ for $x.$ If $y={(x+1)}^{2},$ then $x=-1\pm \sqrt{y}.$ Since we are restricting the domain to the interval where $x\ge -1,$ we need $\pm \sqrt{y}\ge 0.$ Therefore, $x=-1+\sqrt{y}.$ Interchanging $x$ and $y,$ we write $y=-1+\sqrt{x}$ and conclude that ${f}^{-1}(x)=-1+\sqrt{x}.$
Consider $f(x)=1\text{/}{x}^{2}$ restricted to the domain $(\text{-}\infty ,0).$ Verify that $f$ is one-to-one on this domain. Determine the domain and range of the inverse of $f$ and find a formula for ${f}^{-1}.$

જવાબ બતાવો

The domain of ${f}^{-1}$ is $(0,\infty ).$ The range of ${f}^{-1}$ is $(\text{-}\infty ,0).$ The inverse function is given by the formula ${f}^{-1}(x)=-1\text{/}\sqrt{x}.$
Evaluate each of the following expressions.
1. ${\text{sin}}^{-1}(-\frac{\sqrt{3}}{2})$
2. $\text{tan}({\text{tan}}^{-1}(-\frac{1}{\sqrt{3}}))$
3. ${\text{cos}}^{-1}(\text{cos}(\frac{5\pi }{4}))$
4. ${\text{sin}}^{-1}(\text{cos}(\frac{2\pi }{3}))$
જવાબ બતાવો
1. Evaluating ${\text{sin}}^{-1}(\text{-}\sqrt{3}\text{/}2)$ is equivalent to finding the angle $\theta$ such that $\text{sin}\ \theta =\text{-}\sqrt{3}\text{/}2$ and $\text{-}\pi \text{/}2\le \theta \le \pi \text{/}2.$ The angle $\theta =\text{-}\pi \text{/}3$ satisfies these two conditions. Therefore, ${\text{sin}}^{-1}(\text{-}\sqrt{3}\text{/}2)=\text{-}\pi \text{/}3.$
2. First we use the fact that ${\text{tan}}^{-1}(-1\text{/}\sqrt{3})=\text{-}\pi \text{/}6.$ Then $\text{tan}(-\pi \text{/}6)=-1\text{/}\sqrt{3}.$ Therefore, $\text{tan}({\text{tan}}^{-1}(-1\text{/}\sqrt{3}))=-1\text{/}\sqrt{3}.$
3. To evaluate ${\text{cos}}^{-1}(\text{cos}(5\pi \text{/}4)),$ first use the fact that $\text{cos}(5\pi \text{/}4)=\text{-}\sqrt{2}\text{/}2.$ Then we need to find the angle $\theta$ such that $\text{cos}(\theta )=\text{-}\sqrt{2}\text{/}2$ and $0\le \theta \le \pi .$ Since $3\pi \text{/}4$ satisfies both these conditions, we have ${\text{cos}}^{-1}(\text{cos}(5\pi \text{/}4))={\text{cos}}^{-1}(\text{-}\sqrt{2}\text{/}2)=3\pi \text{/}4.$
4. Since $\text{cos}(2\pi \text{/}3)=-1\text{/}2,$ we need to evaluate ${\text{sin}}^{-1}(-1\text{/}2).$ That is, we need to find the angle $\theta$ such that $\text{sin}(\theta )=-1\text{/}2$ and $\text{-}\pi \text{/}2\le \theta \le \pi \text{/}2.$ Since $\text{-}\pi \text{/}6$ satisfies both these conditions, we can conclude that ${\text{sin}}^{-1}(\text{cos}(2\pi \text{/}3))={\text{sin}}^{-1}(-1\text{/}2)=\text{-}\pi \text{/}6.$
$f(x)={x}^{2}-4,x\ge 0$

જવાબ બતાવો

a. ${f}^{-1}(x)=\sqrt{x+4}$ b. Domain $\text{:}\ x\ge -4,\text{range}\text{:}\ y\ge 0$
$f(x)=\sqrt[3]{x-4}$
$f(x)={x}^{3}+1$

જવાબ બતાવો

a. ${f}^{-1}(x)=\sqrt[3]{x-1}$ b. Domain: all real numbers, range: all real numbers
$f(x)={(x-1)}^{2},x\le 1$
$f(x)=\sqrt{x-1}$

જવાબ બતાવો

a. ${f}^{-1}(x)={x}^{2}+1,$ b. Domain: $x\ge 0,$ range: $y\ge 1$
$f(x)=\frac{1}{x+2}$
$f(x)=8x,g(x)=\frac{x}{8}$

જવાબ બતાવો

These are inverses.
$f(x)=8x+3,g(x)=\frac{x-3}{8}$
$f(x)=5x-7,g(x)=\frac{x+5}{7}$

જવાબ બતાવો

These are not inverses.
$f(x)=\frac{2}{3}x+2,g(x)=\frac{3}{2}x+3$
$f(x)=\frac{1}{x-1},x\ne 1,g(x)=\frac{1}{x}+1,x\ne 0$

જવાબ બતાવો

These are inverses.
$f(x)={x}^{3}+1,g(x)={(x-1)}^{1\text{/}3}$
$f(x)={x}^{2}+2x+1,x\ge -1,\ g(x)=-1+\sqrt{x},x\ge 0$

જવાબ બતાવો

These are inverses.
$f(x)=\sqrt{4-{x}^{2}},0\le x\le 2,g(x)=\sqrt{4-{x}^{2}},0\le x\le 2$
${\text{tan}}^{-1}(\frac{\sqrt{3}}{3})$

જવાબ બતાવો

$\frac{\pi }{6}$
${\text{cos}}^{-1}(-\frac{\sqrt{2}}{2})$
${\text{cot}}^{-1}(1)$

જવાબ બતાવો

$\frac{\pi }{4}$
${\text{sin}}^{-1}(-1)$
${\text{cos}}^{-1}(\frac{\sqrt{3}}{2})$

જવાબ બતાવો

$\frac{\pi }{6}$
$\text{cos}({\text{tan}}^{-1}(\sqrt{3}))$
$\text{sin}({\text{cos}}^{-1}(\frac{\sqrt{2}}{2}))$

જવાબ બતાવો

$\frac{\sqrt{2}}{2}$
${\text{sin}}^{-1}(\text{sin}(\frac{\pi }{3}))$
${\text{tan}}^{-1}(\text{tan}(-\frac{\pi }{6}))$

જવાબ બતાવો

$-\frac{\pi }{6}$
The function $C=T(F)=(5\text{/}9)(F-32)$ converts degrees Fahrenheit to degrees Celsius.
1. Find the inverse function $F={T}^{-1}(C)$
2. What is the inverse function used for?
[T] The velocity V (in centimeters per second) of blood in an artery at a distance x cm from the center of the artery can be modeled by the function $V=f(x)=500(0.04-{x}^{2})$ for $0\le x\le 0.2.$
1. Find $x={f}^{-1}(V).$
2. Interpret what the inverse function is used for.
3. Find the distance from the center of an artery with a velocity of 15 cm/sec, 10 cm/sec, and 5 cm/sec.
જવાબ બતાવો

a. $x={f}^{-1}(V)=\sqrt{0.04-\frac{V}{500}}$ b. The inverse function determines the distance from the center of the artery at which blood is flowing with velocity V. c. 0.1 cm; 0.14 cm; 0.17 cm
A function that converts dress sizes in the United States to those in Europe is given by $D(x)=2x+24.$
1. Find the European dress sizes that correspond to sizes 6, 8, 10, and 12 in the United States.
2. Find the function that converts European dress sizes to U.S. dress sizes.
3. Use part b. to find the dress sizes in the United States that correspond to 46, 52, 62, and 70.
[T] The cost to remove a toxin from a lake is modeled by the function
$C(p)=75p\text{/}(85-p),$ where $C$ is the cost (in thousands of dollars) and $p$ is the amount of toxin in a small lake (measured in parts per billion [ppb]). This model is valid only when the amount of toxin is less than 85 ppb.
1. Find the cost to remove 25 ppb, 40 ppb, and 50 ppb of the toxin from the lake.
2. Find the inverse function. c. Use part b. to determine how much of the toxin is removed for $50,000.
જવાબ બતાવો

a. $31,250, $66,667, $107,143 b. $(p=\frac{85C}{C+75})$ c. 34 ppb
[T] A race car is accelerating at a velocity given by
$v(t)=\frac{25}{4}t+54,$ where v is the velocity (in feet per second) at time t.
1. Find the velocity of the car at 10 sec.
2. Find the inverse function.
3. Use part b. to determine how long it takes for the car to reach a speed of 150 ft/sec.
[T] An airplane’s Mach number M is the ratio of its speed to the speed of sound. When a plane is flying at a constant altitude, then its Mach angle is given by $\mu =2{\text{sin}}^{-1}(\frac{1}{M}).$
Find the Mach angle (to the nearest degree) for the following Mach numbers.
1. $M=1.4$
2. $M=2.8$
3. $M=4.3$
જવાબ બતાવો

a. $\sim 92\text{^{\circ}}$ b. $\sim 42\text{^{\circ}}$ c. $\sim 27\text{^{\circ}}$
[T] Using $\mu =2{\text{sin}}^{-1}(\frac{1}{M}),$ find the Mach number M for the following angles.
1. $\mu =\frac{\pi }{6}$
2. $\mu =\frac{2\pi }{7}$
3. $\mu =\frac{3\pi }{8}$
[T] The average temperature (in degrees Celsius) of a city in the northern United States can be modeled by the function
$T(x)=5+18\ \text{sin}[\frac{\pi }{6}(x-4.6)],$ where $x$ is time in months and $x=1.00$ corresponds to January 1. Determine the day(s) (month and day) when the average temperature is $21\text{^{\circ}}\text{C}.$ Use the integer portion of your answer(s) as the month and calculate the day of the month from the decimal portion.

જવાબ બતાવો

$x\approx 6.69,8.51;$ so, the temperature occurs on June 21 and August 15

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.

\pi

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

\theta

theta
The usual name for an angle.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

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).

How to: Inverse Functions

Determine the conditions for when a function has an inverse.
Use the horizontal line test to recognize when a function is one-to-one.
Find the inverse of a given function.
Draw the graph of an inverse function.
Evaluate inverse trigonometric functions.
Since the horizontal line
Since every horizontal line intersects the graph once (at most), this function is one-to-one.
Solve the equation

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.

તમારા પોતાના પ્રયત્ન કરો

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.

આમાં વધુ 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