6.7 Indefinite integrals

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is currently in development and is a work-in-progress.

Indefinite integrals

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What you’ll learn:

Difference between definite and indefinite integrals
Key integration rules

In earlier sections, you used antiderivatives to evaluate definite integrals. A definite integral gives the net area (accumulated change) under a curve over a specific interval.

If an integral has no bounds of integration (no upper and lower limits), it’s an indefinite integral.

A definite integral produces a single number.
An indefinite integral produces a whole family of antiderivatives.

Suppose there are 3 functions:

$F (x) = x^{2} - 1$

$G (x) = x^{2} + 100$

$H (x) = x^{2} + π$

All three are the same basic function $x^{2}$ , just shifted vertically by different constants. When you differentiate, those constants become $0$ , so all three derivatives are the same:

$F^{'} (x) = G^{'} (x) = H^{'} (x) = 2 x$

Now think about integrating $2 x$ . You’re looking for an antiderivative - but which one is it: $F (x)$ , $G (x)$ , or $H (x)$ ?

This is why we add $+ C$ . Here, $C$ is the constant of integration, and it represents any vertical shift. No matter what constant you add, the derivative of $x^{2} + C$ is still $2 x$ .

So,

$\int 2 x d x = x^{2} + C$

That answer represents the entire family of functions whose derivative is $2 x$ .

Sidenote

Constant of integration

Your answer for an indefinite integral must include $+ C$ at the end unless given an initial condition.

On the previous page, the reverse power rule was given as a way to find the antiderivative of a function involving powers.

Reverse power rule

For $n \neq = - 1$ ,

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$

The special case $n = - 1$

What happens when $n = - 1$ ? If we try to integrate $f (x) = \frac{1}{x}$ using the reverse power rule, we run into a division-by-zero problem:

$\int \frac{1}{x} d x = \int x^{- 1} d x = \frac{x ^{- 1 + 1}}{- 1 + 1} = \frac{x ^{0}}{0}$

This is a problem because nothing can be divided by $0$ . Instead, use a derivative you already know:

The derivative of $ln ∣ x ∣$ is $\frac{1}{x}$ .

So,

$\int \frac{1}{x} d x = ln ∣ x ∣ + C$

By working backward from derivative rules, we get a set of core integration rules:

$f (x)$	Indefinite integral
$x^{n}$	$\frac{x ^{n + 1}}{n + 1} + C$
$\frac{1}{x}$	$ln ∣ x ∣ + C$
$e^{x}$	$e^{x} + C$
$a^{x}$	$\frac{a ^{x}}{ln ( a )} + C$

Some trig integration rules:

$f (x)$	Indefinite integral
$sin (x)$	$- cos (x) + C$
$cos (x)$	$sin (x) + C$
$sec^{2} (x)$	$tan (x) + C$
$csc^{2} (x)$	$- cot (x) + C$
$sec (x) tan (x)$	$sec (x) + C$
$csc (x) cot (x)$	$- csc (x) + C$

Some antiderivatives derived from inverse trig functions:

$f (x)$	Indefinite integral
$\frac{1}{1 - x ^{2}}$	$sin^{- 1} (x) + C$
$\frac{- 1}{1 - x ^{2}}$	$cos^{- 1} (x) + C$
$\frac{1}{1 + x ^{2}}$	$tan^{- 1} (x) + C$
$\frac{- 1}{1 + x ^{2}}$	$cot^{- 1} (x) + C$
$\frac{1}{∣ x ∣ x ^{2} - 1}$	$sec^{- 1} (x) + C$
$\frac{- 1}{∣ x ∣ x ^{2} - 1}$	$csc^{- 1} (x) + C$

Although you’re expected to know these basic integration rules for the exam, most exam questions won’t stop at “find this antiderivative.” More often, you’ll be asked to set up a definite integral from a context (like a rate problem) and then evaluate it, with or without a calculator.

Examples

Water flows into a tank at a rate of $R (t) = e^{t} + e^{3}$ liters per hour. How much water enters the tank from $t = 0$ to $t = 10$ ?

Solution

(spoiler)

We want the total amount of water added, so we integrate the rate over the time interval:

$\int_{0}^{10} (e^{t} + e^{3}) d t$

Find an antiderivative term-by-term:

The antiderivative of $e^{t}$ is $e^{t}$ .
$e^{3}$ is a constant, so its antiderivative with respect to $t$ is $e^{3} \cdot t$ .

Now evaluate the definite integral:

$e^{t} + e^{3} (t)_{0}^{10}$

$= [e^{10} + e^{3} (10)] - [e^{0} + e^{3} (0)]$

$= e^{10} + 10 e^{3} - 1 - 0$

$\approx 22, 226.321 liters$

Sidenote

Definite integrals in Desmos

On calculator sections, you can use Desmos to compute definite integrals that might be tedious to integrate by hand. Type “int” and an integral symbol will appear.

Desmos requires you enter the bounds of integration - it won’t be able to show you the result of an indefinite integral.

Make sure to put parentheses around the integrand before the $d x$ .

A factory releases a pollutant into a river at a rate of $R (t) = \frac{500 t}{1 + e ^{- 0.2 t}}$ grams per hour, where time $t$ is the number of hours since the start of the day. To the nearest gram, how much pollutant was released between $5 \leq t \leq 12$ ?

Solution

(spoiler)

The rate function $R (t)$ is in grams per hour. To get the total amount released (in grams) over a time interval, integrate the rate over that interval.

Set up this definite integral in Desmos:

$\int_{5}^{12} \frac{500 t}{1 + e ^{- 0.2 t}} d t$

The value it gives is $8557$ grams.

The rate of change of the temperature in degrees Celsius per hour is given by

$T^{'} (t) = - t^{3} + 6 t^{2} - 9 t + 4$

where $t$ is the number of hours since midnight and $0 \leq t \leq 6$ .

Write an integral expression that gives the change in temperature during the time when the temperature is decreasing.

Solution

(spoiler)

The question gives a rate of change and asks for the change in temperature, so we integrate the rate.

But we only integrate over the times when the temperature is decreasing. That happens when $T^{'} (t) < 0$ .

If you graph $T^{'}$ in Desmos (using $x$ as the input variable), you’ll see the graph is below the $x$ -axis for $4 \leq x \leq \infty$ . In this problem, time is limited to $0 \leq t \leq 6$ , so the relevant interval is $4 \leq t \leq 6$ .

So the change in temperature (in degrees Celsius) during the time it’s decreasing is

$\int_{4}^{6} T^{'} (t) d t$

$= \int_{4}^{6} (- t^{3} + 6 t^{2} - 9 t + 4) d t$

Find a function that satisfies

$\frac{d y}{d x} = 2^{x} ln (2) - \frac{2}{x}$

and

$y (2) = 5$

Solution

(spoiler)

Because the derivative is given, integrate to recover $y$ :

$\int \frac{d y}{d x} d x = \int (2^{x} ln (2) - \frac{2}{x}) d x$

Integrate term-by-term:

$y = ln (2) \int 2^{x} d x - 2 \int \frac{1}{x} d x$

$= ln (2) \cdot \frac{2 ^{x}}{ln ( 2 )} - 2 ln ∣ x ∣$

$= 2^{x} - 2 ln ∣ x ∣ + C$

Now use the initial condition $y (2) = 5$ to find $C$ :

$5 = 2^{2} - 2 ln (2) + C$

$C = 2.386$

Therefore,

$y = 2^{x} - 2 ln ∣ x ∣ + 2.386$

Here is a problem combining both definite and indefinite integrals that’s similar to one from a past AP exam (1988 AB FRQ #6), with the numbers altered:

Let $f$ be a differentiable function, defined for all real numbers $x$ , with the following properties.

(i) $f^{'} (x) = a x^{2} + b x$
(ii) $f^{'} (2) = 10$ and $f^{''} (2) = 18$
(iii) $\int_{2}^{4} f (x) d x = 60$

Find $f (x)$ .

Solution

(spoiler)

From (i) and (ii), use $f^{'} (2) = 10$ :

$a (2)^{2} + b (2) = 10$

$4 a + 2 b = 10$

Differentiate $f^{'} (x)$ to get $f^{''} (x)$ :

$f^{''} (x) = 2 a x + b$

Now use $f^{''} (2) = 18$ :

$2 a (2) + b = 16$

$4 a + b = 18$

Now solve the system for $a$ and $b$ :

$4 a + 2 b = 10 4 a + b = 18$

Subtracting equation $(2)$ from $(1)$ gives $b = - 6$ . Substituting back gives $a = 6$ . Therefore,

$f^{'} (x) = 6 x^{2} - 6 x$

To find $f (x)$ , integrate $f^{'} (x)$ :

$f (x) = \int f^{'} (x) d x$

$= \int (6 x^{2} - 6 x) d x$

$= 2 x^{3} - 3 x^{2} + C$

The $+ C$ is required because this is an indefinite integral.

Now use (iii). Let $F (x)$ be an antiderivative of $f (x)$ :

$F (x) = \int (6 x^{3} - 3 x^{2} + C) d x$

$= \frac{6 x ^{4}}{4} - x^{3} + C x$

$= \frac{3}{2} x^{4} - x^{3} + C x$

We don’t add another constant here because we’re about to evaluate a definite integral using FTC Part 2:

$\int_{2}^{4} f (x) = F (4) - F (2)$

Using (iii),

$F (4) - F (2) = 60$

$[\frac{3}{2} (4)^{4} - (4)^{3} + C (4)] - [\frac{3}{2} (2)^{4} - (2)^{3} + C (2)] = 60$

$(384 - 64 + 4 C) - (24 - 8 + 2 C) = 60$

$304 + 2 C = 60$

$C = - 122$

Therefore,

$f (x) = 2 x^{3} - 3 x^{2} - 122$

Indefinite integrals

What you’ll learn:

Difference between definite and indefinite integrals
Key integration rules

In earlier sections, you used antiderivatives to evaluate definite integrals. A definite integral gives the net area (accumulated change) under a curve over a specific interval.

If an integral has no bounds of integration (no upper and lower limits), it’s an indefinite integral.

A definite integral produces a single number.
An indefinite integral produces a whole family of antiderivatives.

Suppose there are 3 functions:

$F (x) = x^{2} - 1$

$G (x) = x^{2} + 100$

$H (x) = x^{2} + π$

All three are the same basic function $x^{2}$ , just shifted vertically by different constants. When you differentiate, those constants become $0$ , so all three derivatives are the same:

$F^{'} (x) = G^{'} (x) = H^{'} (x) = 2 x$

Now think about integrating $2 x$ . You’re looking for an antiderivative - but which one is it: $F (x)$ , $G (x)$ , or $H (x)$ ?

This is why we add $+ C$ . Here, $C$ is the constant of integration, and it represents any vertical shift. No matter what constant you add, the derivative of $x^{2} + C$ is still $2 x$ .

So,

$\int 2 x d x = x^{2} + C$

That answer represents the entire family of functions whose derivative is $2 x$ .

Sidenote

Constant of integration

Your answer for an indefinite integral must include $+ C$ at the end unless given an initial condition.

On the previous page, the reverse power rule was given as a way to find the antiderivative of a function involving powers.

Reverse power rule

For $n \neq = - 1$ ,

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$

The special case $n = - 1$

What happens when $n = - 1$ ? If we try to integrate $f (x) = \frac{1}{x}$ using the reverse power rule, we run into a division-by-zero problem:

$\int \frac{1}{x} d x = \int x^{- 1} d x = \frac{x ^{- 1 + 1}}{- 1 + 1} = \frac{x ^{0}}{0}$

This is a problem because nothing can be divided by $0$ . Instead, use a derivative you already know:

The derivative of $ln ∣ x ∣$ is $\frac{1}{x}$ .

So,

$\int \frac{1}{x} d x = ln ∣ x ∣ + C$

By working backward from derivative rules, we get a set of core integration rules:

$f (x)$	Indefinite integral
$x^{n}$	$\frac{x ^{n + 1}}{n + 1} + C$
$\frac{1}{x}$	$ln ∣ x ∣ + C$
$e^{x}$	$e^{x} + C$
$a^{x}$	$\frac{a ^{x}}{ln ( a )} + C$

Some trig integration rules:

$f (x)$	Indefinite integral
$sin (x)$	$- cos (x) + C$
$cos (x)$	$sin (x) + C$
$sec^{2} (x)$	$tan (x) + C$
$csc^{2} (x)$	$- cot (x) + C$
$sec (x) tan (x)$	$sec (x) + C$
$csc (x) cot (x)$	$- csc (x) + C$

Some antiderivatives derived from inverse trig functions:

$f (x)$	Indefinite integral
$\frac{1}{1 - x ^{2}}$	$sin^{- 1} (x) + C$
$\frac{- 1}{1 - x ^{2}}$	$cos^{- 1} (x) + C$
$\frac{1}{1 + x ^{2}}$	$tan^{- 1} (x) + C$
$\frac{- 1}{1 + x ^{2}}$	$cot^{- 1} (x) + C$
$\frac{1}{∣ x ∣ x ^{2} - 1}$	$sec^{- 1} (x) + C$
$\frac{- 1}{∣ x ∣ x ^{2} - 1}$	$csc^{- 1} (x) + C$

Examples

Water flows into a tank at a rate of $R (t) = e^{t} + e^{3}$ liters per hour. How much water enters the tank from $t = 0$ to $t = 10$ ?

Solution

(spoiler)

We want the total amount of water added, so we integrate the rate over the time interval:

$\int_{0}^{10} (e^{t} + e^{3}) d t$

Find an antiderivative term-by-term:

The antiderivative of $e^{t}$ is $e^{t}$ .
$e^{3}$ is a constant, so its antiderivative with respect to $t$ is $e^{3} \cdot t$ .

Now evaluate the definite integral:

$e^{t} + e^{3} (t)_{0}^{10}$

$= [e^{10} + e^{3} (10)] - [e^{0} + e^{3} (0)]$

$= e^{10} + 10 e^{3} - 1 - 0$

$\approx 22, 226.321 liters$

Sidenote

Definite integrals in Desmos

On calculator sections, you can use Desmos to compute definite integrals that might be tedious to integrate by hand. Type “int” and an integral symbol will appear.

Desmos requires you enter the bounds of integration - it won’t be able to show you the result of an indefinite integral.

Make sure to put parentheses around the integrand before the $d x$ .

A factory releases a pollutant into a river at a rate of $R (t) = \frac{500 t}{1 + e ^{- 0.2 t}}$ grams per hour, where time $t$ is the number of hours since the start of the day. To the nearest gram, how much pollutant was released between $5 \leq t \leq 12$ ?

Solution

(spoiler)

The rate function $R (t)$ is in grams per hour. To get the total amount released (in grams) over a time interval, integrate the rate over that interval.

Set up this definite integral in Desmos:

$\int_{5}^{12} \frac{500 t}{1 + e ^{- 0.2 t}} d t$

The value it gives is $8557$ grams.

The rate of change of the temperature in degrees Celsius per hour is given by

$T^{'} (t) = - t^{3} + 6 t^{2} - 9 t + 4$

where $t$ is the number of hours since midnight and $0 \leq t \leq 6$ .

Write an integral expression that gives the change in temperature during the time when the temperature is decreasing.

Solution

(spoiler)

The question gives a rate of change and asks for the change in temperature, so we integrate the rate.

But we only integrate over the times when the temperature is decreasing. That happens when $T^{'} (t) < 0$ .

So the change in temperature (in degrees Celsius) during the time it’s decreasing is

$\int_{4}^{6} T^{'} (t) d t$

$= \int_{4}^{6} (- t^{3} + 6 t^{2} - 9 t + 4) d t$

Find a function that satisfies

$\frac{d y}{d x} = 2^{x} ln (2) - \frac{2}{x}$

and

$y (2) = 5$

Solution

(spoiler)

Because the derivative is given, integrate to recover $y$ :

$\int \frac{d y}{d x} d x = \int (2^{x} ln (2) - \frac{2}{x}) d x$

Integrate term-by-term:

$y = ln (2) \int 2^{x} d x - 2 \int \frac{1}{x} d x$

$= ln (2) \cdot \frac{2 ^{x}}{ln ( 2 )} - 2 ln ∣ x ∣$

$= 2^{x} - 2 ln ∣ x ∣ + C$

Now use the initial condition $y (2) = 5$ to find $C$ :

$5 = 2^{2} - 2 ln (2) + C$

$C = 2.386$

Therefore,

$y = 2^{x} - 2 ln ∣ x ∣ + 2.386$

Here is a problem combining both definite and indefinite integrals that’s similar to one from a past AP exam (1988 AB FRQ #6), with the numbers altered:

Let $f$ be a differentiable function, defined for all real numbers $x$ , with the following properties.

(i) $f^{'} (x) = a x^{2} + b x$
(ii) $f^{'} (2) = 10$ and $f^{''} (2) = 18$
(iii) $\int_{2}^{4} f (x) d x = 60$

Find $f (x)$ .

Solution

(spoiler)

From (i) and (ii), use $f^{'} (2) = 10$ :

$a (2)^{2} + b (2) = 10$

$4 a + 2 b = 10$

Differentiate $f^{'} (x)$ to get $f^{''} (x)$ :

$f^{''} (x) = 2 a x + b$

Now use $f^{''} (2) = 18$ :

$2 a (2) + b = 16$

$4 a + b = 18$

Now solve the system for $a$ and $b$ :

$4 a + 2 b = 10 4 a + b = 18$

Subtracting equation $(2)$ from $(1)$ gives $b = - 6$ . Substituting back gives $a = 6$ . Therefore,

$f^{'} (x) = 6 x^{2} - 6 x$

To find $f (x)$ , integrate $f^{'} (x)$ :

$f (x) = \int f^{'} (x) d x$

$= \int (6 x^{2} - 6 x) d x$

$= 2 x^{3} - 3 x^{2} + C$

The $+ C$ is required because this is an indefinite integral.

Now use (iii). Let $F (x)$ be an antiderivative of $f (x)$ :

$F (x) = \int (6 x^{3} - 3 x^{2} + C) d x$

$= \frac{6 x ^{4}}{4} - x^{3} + C x$

$= \frac{3}{2} x^{4} - x^{3} + C x$

We don’t add another constant here because we’re about to evaluate a definite integral using FTC Part 2:

$\int_{2}^{4} f (x) = F (4) - F (2)$

Using (iii),

$F (4) - F (2) = 60$

$[\frac{3}{2} (4)^{4} - (4)^{3} + C (4)] - [\frac{3}{2} (2)^{4} - (2)^{3} + C (2)] = 60$

$(384 - 64 + 4 C) - (24 - 8 + 2 C) = 60$

$304 + 2 C = 60$

$C = - 122$

Therefore,

$f (x) = 2 x^{3} - 3 x^{2} - 122$

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is currently in development and is a work-in-progress.

Indefinite integrals

8 min read

Font

Discuss

Feedback

What you’ll learn:

Difference between definite and indefinite integrals
Key integration rules

In earlier sections, you used antiderivatives to evaluate definite integrals. A definite integral gives the net area (accumulated change) under a curve over a specific interval.

If an integral has no bounds of integration (no upper and lower limits), it’s an indefinite integral.

A definite integral produces a single number.
An indefinite integral produces a whole family of antiderivatives.

Suppose there are 3 functions:

$F (x) = x^{2} - 1$

$G (x) = x^{2} + 100$

$H (x) = x^{2} + π$

All three are the same basic function $x^{2}$ , just shifted vertically by different constants. When you differentiate, those constants become $0$ , so all three derivatives are the same:

$F^{'} (x) = G^{'} (x) = H^{'} (x) = 2 x$

Now think about integrating $2 x$ . You’re looking for an antiderivative - but which one is it: $F (x)$ , $G (x)$ , or $H (x)$ ?

This is why we add $+ C$ . Here, $C$ is the constant of integration, and it represents any vertical shift. No matter what constant you add, the derivative of $x^{2} + C$ is still $2 x$ .

So,

$\int 2 x d x = x^{2} + C$

That answer represents the entire family of functions whose derivative is $2 x$ .

Sidenote

Constant of integration

Your answer for an indefinite integral must include $+ C$ at the end unless given an initial condition.

On the previous page, the reverse power rule was given as a way to find the antiderivative of a function involving powers.

Reverse power rule

For $n \neq = - 1$ ,

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$

The special case $n = - 1$

What happens when $n = - 1$ ? If we try to integrate $f (x) = \frac{1}{x}$ using the reverse power rule, we run into a division-by-zero problem:

$\int \frac{1}{x} d x = \int x^{- 1} d x = \frac{x ^{- 1 + 1}}{- 1 + 1} = \frac{x ^{0}}{0}$

This is a problem because nothing can be divided by $0$ . Instead, use a derivative you already know:

The derivative of $ln ∣ x ∣$ is $\frac{1}{x}$ .

So,

$\int \frac{1}{x} d x = ln ∣ x ∣ + C$

By working backward from derivative rules, we get a set of core integration rules:

$f (x)$	Indefinite integral
$x^{n}$	$\frac{x ^{n + 1}}{n + 1} + C$
$\frac{1}{x}$	$ln ∣ x ∣ + C$
$e^{x}$	$e^{x} + C$
$a^{x}$	$\frac{a ^{x}}{ln ( a )} + C$

Some trig integration rules:

$f (x)$	Indefinite integral
$sin (x)$	$- cos (x) + C$
$cos (x)$	$sin (x) + C$
$sec^{2} (x)$	$tan (x) + C$
$csc^{2} (x)$	$- cot (x) + C$
$sec (x) tan (x)$	$sec (x) + C$
$csc (x) cot (x)$	$- csc (x) + C$

Some antiderivatives derived from inverse trig functions:

$f (x)$	Indefinite integral
$\frac{1}{1 - x ^{2}}$	$sin^{- 1} (x) + C$
$\frac{- 1}{1 - x ^{2}}$	$cos^{- 1} (x) + C$
$\frac{1}{1 + x ^{2}}$	$tan^{- 1} (x) + C$
$\frac{- 1}{1 + x ^{2}}$	$cot^{- 1} (x) + C$
$\frac{1}{∣ x ∣ x ^{2} - 1}$	$sec^{- 1} (x) + C$
$\frac{- 1}{∣ x ∣ x ^{2} - 1}$	$csc^{- 1} (x) + C$

Examples

Water flows into a tank at a rate of $R (t) = e^{t} + e^{3}$ liters per hour. How much water enters the tank from $t = 0$ to $t = 10$ ?

Solution

(spoiler)

We want the total amount of water added, so we integrate the rate over the time interval:

$\int_{0}^{10} (e^{t} + e^{3}) d t$

Find an antiderivative term-by-term:

The antiderivative of $e^{t}$ is $e^{t}$ .
$e^{3}$ is a constant, so its antiderivative with respect to $t$ is $e^{3} \cdot t$ .

Now evaluate the definite integral:

$e^{t} + e^{3} (t)_{0}^{10}$

$= [e^{10} + e^{3} (10)] - [e^{0} + e^{3} (0)]$

$= e^{10} + 10 e^{3} - 1 - 0$

$\approx 22, 226.321 liters$

Sidenote

Definite integrals in Desmos

On calculator sections, you can use Desmos to compute definite integrals that might be tedious to integrate by hand. Type “int” and an integral symbol will appear.

Desmos requires you enter the bounds of integration - it won’t be able to show you the result of an indefinite integral.

Make sure to put parentheses around the integrand before the $d x$ .

A factory releases a pollutant into a river at a rate of $R (t) = \frac{500 t}{1 + e ^{- 0.2 t}}$ grams per hour, where time $t$ is the number of hours since the start of the day. To the nearest gram, how much pollutant was released between $5 \leq t \leq 12$ ?

Solution

(spoiler)

The rate function $R (t)$ is in grams per hour. To get the total amount released (in grams) over a time interval, integrate the rate over that interval.

Set up this definite integral in Desmos:

$\int_{5}^{12} \frac{500 t}{1 + e ^{- 0.2 t}} d t$

The value it gives is $8557$ grams.

The rate of change of the temperature in degrees Celsius per hour is given by

$T^{'} (t) = - t^{3} + 6 t^{2} - 9 t + 4$

where $t$ is the number of hours since midnight and $0 \leq t \leq 6$ .

Write an integral expression that gives the change in temperature during the time when the temperature is decreasing.

Solution

(spoiler)

The question gives a rate of change and asks for the change in temperature, so we integrate the rate.

But we only integrate over the times when the temperature is decreasing. That happens when $T^{'} (t) < 0$ .

So the change in temperature (in degrees Celsius) during the time it’s decreasing is

$\int_{4}^{6} T^{'} (t) d t$

$= \int_{4}^{6} (- t^{3} + 6 t^{2} - 9 t + 4) d t$

Find a function that satisfies

$\frac{d y}{d x} = 2^{x} ln (2) - \frac{2}{x}$

and

$y (2) = 5$

Solution

(spoiler)

Because the derivative is given, integrate to recover $y$ :

$\int \frac{d y}{d x} d x = \int (2^{x} ln (2) - \frac{2}{x}) d x$

Integrate term-by-term:

$y = ln (2) \int 2^{x} d x - 2 \int \frac{1}{x} d x$

$= ln (2) \cdot \frac{2 ^{x}}{ln ( 2 )} - 2 ln ∣ x ∣$

$= 2^{x} - 2 ln ∣ x ∣ + C$

Now use the initial condition $y (2) = 5$ to find $C$ :

$5 = 2^{2} - 2 ln (2) + C$

$C = 2.386$

Therefore,

$y = 2^{x} - 2 ln ∣ x ∣ + 2.386$

Here is a problem combining both definite and indefinite integrals that’s similar to one from a past AP exam (1988 AB FRQ #6), with the numbers altered:

Let $f$ be a differentiable function, defined for all real numbers $x$ , with the following properties.

(i) $f^{'} (x) = a x^{2} + b x$
(ii) $f^{'} (2) = 10$ and $f^{''} (2) = 18$
(iii) $\int_{2}^{4} f (x) d x = 60$

Find $f (x)$ .

Solution

(spoiler)

From (i) and (ii), use $f^{'} (2) = 10$ :

$a (2)^{2} + b (2) = 10$

$4 a + 2 b = 10$

Differentiate $f^{'} (x)$ to get $f^{''} (x)$ :

$f^{''} (x) = 2 a x + b$

Now use $f^{''} (2) = 18$ :

$2 a (2) + b = 16$

$4 a + b = 18$

Now solve the system for $a$ and $b$ :

$4 a + 2 b = 10 4 a + b = 18$

Subtracting equation $(2)$ from $(1)$ gives $b = - 6$ . Substituting back gives $a = 6$ . Therefore,

$f^{'} (x) = 6 x^{2} - 6 x$

To find $f (x)$ , integrate $f^{'} (x)$ :

$f (x) = \int f^{'} (x) d x$

$= \int (6 x^{2} - 6 x) d x$

$= 2 x^{3} - 3 x^{2} + C$

The $+ C$ is required because this is an indefinite integral.

Now use (iii). Let $F (x)$ be an antiderivative of $f (x)$ :

$F (x) = \int (6 x^{3} - 3 x^{2} + C) d x$

$= \frac{6 x ^{4}}{4} - x^{3} + C x$

$= \frac{3}{2} x^{4} - x^{3} + C x$

We don’t add another constant here because we’re about to evaluate a definite integral using FTC Part 2:

$\int_{2}^{4} f (x) = F (4) - F (2)$

Using (iii),

$F (4) - F (2) = 60$

$[\frac{3}{2} (4)^{4} - (4)^{3} + C (4)] - [\frac{3}{2} (2)^{4} - (2)^{3} + C (2)] = 60$

$(384 - 64 + 4 C) - (24 - 8 + 2 C) = 60$

$304 + 2 C = 60$

$C = - 122$

Therefore,

$f (x) = 2 x^{3} - 3 x^{2} - 122$

Indefinite integrals

What you’ll learn:

Difference between definite and indefinite integrals
Key integration rules

In earlier sections, you used antiderivatives to evaluate definite integrals. A definite integral gives the net area (accumulated change) under a curve over a specific interval.

If an integral has no bounds of integration (no upper and lower limits), it’s an indefinite integral.

A definite integral produces a single number.
An indefinite integral produces a whole family of antiderivatives.

Suppose there are 3 functions:

$F (x) = x^{2} - 1$

$G (x) = x^{2} + 100$

$H (x) = x^{2} + π$

All three are the same basic function $x^{2}$ , just shifted vertically by different constants. When you differentiate, those constants become $0$ , so all three derivatives are the same:

$F^{'} (x) = G^{'} (x) = H^{'} (x) = 2 x$

Now think about integrating $2 x$ . You’re looking for an antiderivative - but which one is it: $F (x)$ , $G (x)$ , or $H (x)$ ?

This is why we add $+ C$ . Here, $C$ is the constant of integration, and it represents any vertical shift. No matter what constant you add, the derivative of $x^{2} + C$ is still $2 x$ .

So,

$\int 2 x d x = x^{2} + C$

That answer represents the entire family of functions whose derivative is $2 x$ .

Sidenote

Constant of integration

Your answer for an indefinite integral must include $+ C$ at the end unless given an initial condition.

On the previous page, the reverse power rule was given as a way to find the antiderivative of a function involving powers.

Reverse power rule

For $n \neq = - 1$ ,

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$

The special case $n = - 1$

What happens when $n = - 1$ ? If we try to integrate $f (x) = \frac{1}{x}$ using the reverse power rule, we run into a division-by-zero problem:

$\int \frac{1}{x} d x = \int x^{- 1} d x = \frac{x ^{- 1 + 1}}{- 1 + 1} = \frac{x ^{0}}{0}$

This is a problem because nothing can be divided by $0$ . Instead, use a derivative you already know:

The derivative of $ln ∣ x ∣$ is $\frac{1}{x}$ .

So,

$\int \frac{1}{x} d x = ln ∣ x ∣ + C$

By working backward from derivative rules, we get a set of core integration rules:

$f (x)$	Indefinite integral
$x^{n}$	$\frac{x ^{n + 1}}{n + 1} + C$
$\frac{1}{x}$	$ln ∣ x ∣ + C$
$e^{x}$	$e^{x} + C$
$a^{x}$	$\frac{a ^{x}}{ln ( a )} + C$

Some trig integration rules:

$f (x)$	Indefinite integral
$sin (x)$	$- cos (x) + C$
$cos (x)$	$sin (x) + C$
$sec^{2} (x)$	$tan (x) + C$
$csc^{2} (x)$	$- cot (x) + C$
$sec (x) tan (x)$	$sec (x) + C$
$csc (x) cot (x)$	$- csc (x) + C$

Some antiderivatives derived from inverse trig functions:

$f (x)$	Indefinite integral
$\frac{1}{1 - x ^{2}}$	$sin^{- 1} (x) + C$
$\frac{- 1}{1 - x ^{2}}$	$cos^{- 1} (x) + C$
$\frac{1}{1 + x ^{2}}$	$tan^{- 1} (x) + C$
$\frac{- 1}{1 + x ^{2}}$	$cot^{- 1} (x) + C$
$\frac{1}{∣ x ∣ x ^{2} - 1}$	$sec^{- 1} (x) + C$
$\frac{- 1}{∣ x ∣ x ^{2} - 1}$	$csc^{- 1} (x) + C$

Examples

Water flows into a tank at a rate of $R (t) = e^{t} + e^{3}$ liters per hour. How much water enters the tank from $t = 0$ to $t = 10$ ?

Solution

(spoiler)

We want the total amount of water added, so we integrate the rate over the time interval:

$\int_{0}^{10} (e^{t} + e^{3}) d t$

Find an antiderivative term-by-term:

The antiderivative of $e^{t}$ is $e^{t}$ .
$e^{3}$ is a constant, so its antiderivative with respect to $t$ is $e^{3} \cdot t$ .

Now evaluate the definite integral:

$e^{t} + e^{3} (t)_{0}^{10}$

$= [e^{10} + e^{3} (10)] - [e^{0} + e^{3} (0)]$

$= e^{10} + 10 e^{3} - 1 - 0$

$\approx 22, 226.321 liters$

Sidenote

Definite integrals in Desmos

On calculator sections, you can use Desmos to compute definite integrals that might be tedious to integrate by hand. Type “int” and an integral symbol will appear.

Desmos requires you enter the bounds of integration - it won’t be able to show you the result of an indefinite integral.

Make sure to put parentheses around the integrand before the $d x$ .

A factory releases a pollutant into a river at a rate of $R (t) = \frac{500 t}{1 + e ^{- 0.2 t}}$ grams per hour, where time $t$ is the number of hours since the start of the day. To the nearest gram, how much pollutant was released between $5 \leq t \leq 12$ ?

Solution

(spoiler)

The rate function $R (t)$ is in grams per hour. To get the total amount released (in grams) over a time interval, integrate the rate over that interval.

Set up this definite integral in Desmos:

$\int_{5}^{12} \frac{500 t}{1 + e ^{- 0.2 t}} d t$

The value it gives is $8557$ grams.

The rate of change of the temperature in degrees Celsius per hour is given by

$T^{'} (t) = - t^{3} + 6 t^{2} - 9 t + 4$

where $t$ is the number of hours since midnight and $0 \leq t \leq 6$ .

Write an integral expression that gives the change in temperature during the time when the temperature is decreasing.

Solution

(spoiler)

The question gives a rate of change and asks for the change in temperature, so we integrate the rate.

But we only integrate over the times when the temperature is decreasing. That happens when $T^{'} (t) < 0$ .

So the change in temperature (in degrees Celsius) during the time it’s decreasing is

$\int_{4}^{6} T^{'} (t) d t$

$= \int_{4}^{6} (- t^{3} + 6 t^{2} - 9 t + 4) d t$

Find a function that satisfies

$\frac{d y}{d x} = 2^{x} ln (2) - \frac{2}{x}$

and

$y (2) = 5$

Solution

(spoiler)

Because the derivative is given, integrate to recover $y$ :

$\int \frac{d y}{d x} d x = \int (2^{x} ln (2) - \frac{2}{x}) d x$

Integrate term-by-term:

$y = ln (2) \int 2^{x} d x - 2 \int \frac{1}{x} d x$

$= ln (2) \cdot \frac{2 ^{x}}{ln ( 2 )} - 2 ln ∣ x ∣$

$= 2^{x} - 2 ln ∣ x ∣ + C$

Now use the initial condition $y (2) = 5$ to find $C$ :

$5 = 2^{2} - 2 ln (2) + C$

$C = 2.386$

Therefore,

$y = 2^{x} - 2 ln ∣ x ∣ + 2.386$

Here is a problem combining both definite and indefinite integrals that’s similar to one from a past AP exam (1988 AB FRQ #6), with the numbers altered:

Let $f$ be a differentiable function, defined for all real numbers $x$ , with the following properties.

(i) $f^{'} (x) = a x^{2} + b x$
(ii) $f^{'} (2) = 10$ and $f^{''} (2) = 18$
(iii) $\int_{2}^{4} f (x) d x = 60$

Find $f (x)$ .

Solution

(spoiler)

From (i) and (ii), use $f^{'} (2) = 10$ :

$a (2)^{2} + b (2) = 10$

$4 a + 2 b = 10$

Differentiate $f^{'} (x)$ to get $f^{''} (x)$ :

$f^{''} (x) = 2 a x + b$

Now use $f^{''} (2) = 18$ :

$2 a (2) + b = 16$

$4 a + b = 18$

Now solve the system for $a$ and $b$ :

$4 a + 2 b = 10 4 a + b = 18$

Subtracting equation $(2)$ from $(1)$ gives $b = - 6$ . Substituting back gives $a = 6$ . Therefore,

$f^{'} (x) = 6 x^{2} - 6 x$

To find $f (x)$ , integrate $f^{'} (x)$ :

$f (x) = \int f^{'} (x) d x$

$= \int (6 x^{2} - 6 x) d x$

$= 2 x^{3} - 3 x^{2} + C$

The $+ C$ is required because this is an indefinite integral.

Now use (iii). Let $F (x)$ be an antiderivative of $f (x)$ :

$F (x) = \int (6 x^{3} - 3 x^{2} + C) d x$

$= \frac{6 x ^{4}}{4} - x^{3} + C x$

$= \frac{3}{2} x^{4} - x^{3} + C x$

We don’t add another constant here because we’re about to evaluate a definite integral using FTC Part 2:

$\int_{2}^{4} f (x) = F (4) - F (2)$

Using (iii),

$F (4) - F (2) = 60$

$[\frac{3}{2} (4)^{4} - (4)^{3} + C (4)] - [\frac{3}{2} (2)^{4} - (2)^{3} + C (2)] = 60$

$(384 - 64 + 4 C) - (24 - 8 + 2 C) = 60$

$304 + 2 C = 60$

$C = - 122$

Therefore,

$f (x) = 2 x^{3} - 3 x^{2} - 122$

Indefinite integrals

What you’ll learn:

The special case n=−1

Examples

Solution

Solution

Solution

Solution

Solution

Indefinite integrals

What you’ll learn:

The special case n=−1

Examples

Solution

Solution

Solution

Solution

Solution

Indefinite integrals

What you’ll learn:

The special case n=−1

Examples

Solution

Solution

Solution

Solution

Solution

Indefinite integrals

What you’ll learn:

The special case n=−1

Examples

Solution

Solution

Solution

Solution

Solution

The special case $n = - 1$

The special case $n = - 1$

The special case $n = - 1$

The special case $n = - 1$