Indefinite integrals | Integration | Achievable AP Calculus AB

What you’ll learn:

Difference between definite and indefinite integrals
Key integration rules

In the previous sections, antiderivatives were used to evaluate definite integrals to find the net area under a curve over a specific interval.

On the other hand, if the bounds of integration are absent, and it’s just a single integral symbol followed by the integrand, then what you’re dealing with is an indefinite integral.

While definite integrals produce a number, the answer to an indefinite integrals is an entire family of antiderivatives.

Suppose there are 3 functions:

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

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

$H (x) = x^{2} + π$

All of these are the same $x^{2}$ with different vertical shifts. When differentiated, the constants become $0$ and the derivative of all three functions is $2 x$ .

Then to integrate $2 x$ , find its antiderivative. But which function - $F (x), G (x),$ or $H (x)$ - is the antiderivative?

This is where $+ C$ must be added. $C$ is the constant of integration that accounts for all vertical shifts of the antiderivative, because regardless of its value, the derivative of $x^{2} + C$ is $2 x$ . Then

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

The answer to this indefinite integral is $x^{2} + C$ , which represents the entire family of functions that have the derivative $2 x$ .

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 $n \neq = - 1$ ,

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

What about the case where $n = - 1$ ? If we wanted to integrate $f (x) = \frac{1}{x}$ , the reverse power rule can’t be applied because

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

This is problem because nothing can be divided by 0. Instead, recall that the derivative of $ln ∣ x ∣$ is $\frac{1}{x}$ .

Then

$\int \frac{1}{x} = ln ∣ x ∣ + C$

By working backwards from some of the derivative rules from the 1st part of calculus, we obtain some important 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, there aren’t too many straightforward questions on just these. Many problems will be worded for you to set up a definite integral and then evaluate it with or without a calculator. So we’ll continue working with such problems instead.

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’re looking for the accumulation of change in the water between $t = 0$ to $t = 10$ , or evaluating

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

The antiderivative of $e^{t}$ is itself - $e^{t}$ .

On the other hand, $e^{3}$ is a constant so its antiderivative with respect to $t$ is $e^{3} \cdot t$ (using the reverse power rule).

Therefore the definite integral becomes

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

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

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

$\approx 22, 226.321 liters$

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 problem gives a rate function that describes how fast, in grams per hour, the pollutant is being released. It’s looking for the amount of pollutant, in grams, over a specified time interval, which means integrate the rate for the accumulated change. Set up this 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)

Notice the question gives the rate of change and asks for the change in temperature. So we have to integrate the rate for the accumulated change.

This will be done over the time interval in which the temperature is decreasing, or when the rate of change is negative.

Graphing the rate function $T^{'}$ in Desmos (in terms of $x$ , since it won’t accept input variable $t$ when graphing), we see that the graph is below the $x$ -axis in $4 \leq x \leq \infty$ , but for this problem the time is capped at $t = 6$ hours. So the change in temperature when the temperature is decreasing, in degrees Celsius, 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)

Since the derivative is given, we can integrate to get the original function.

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

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

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

$= 2^{x} - 2 ln ∣ x ∣ + C$

The problem also gives $y (2) = 5$ , which we can plug in to find the specific $C$ such that the function passes through the point $(2, 5)$ .

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

$C = 2.386$

Therefore the function that satisfies those conditions is

$y = 2^{x} - 2 ln ∣ x ∣ + 2.386$

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)

Since $f^{'} (2) = 10$ , then

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

$4 a + 2 b = 10$

Next, $f^{''} (x) = 2 a x + b$ . Since $f^{''} (2) = 18$ ,

$2 a (2) + b = 16$

$4 a + b = 18$

Now we solve the system of equations for $a$ and $b$ :

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

Subtracting equation $(2)$ from $(1)$ , we find that $b = - 6$ and $a = 6$ . Therefore

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

To find $f (x)$ , integrate $f^{'} (x)$ with the reverse power rule.

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

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

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

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

Lastly, use property (iii) by further integrating $f (x)$ . Let $F (x)$ be an antiderivative of $f (x)$ .

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

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

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

No additional $+ C$ needs to be added because we can immediately use the FTC part 2 for definite integrals, which says that

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

Combined with property (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

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

Key points

A definite integral always produces a value while an indefinite integral produces a family of functions (don’t forget $+ C$ ).

What you’ll learn:

Difference between definite and indefinite integrals
Key integration rules

In the previous sections, antiderivatives were used to evaluate definite integrals to find the net area under a curve over a specific interval.

On the other hand, if the bounds of integration are absent, and it’s just a single integral symbol followed by the integrand, then what you’re dealing with is an indefinite integral.

While definite integrals produce a number, the answer to an indefinite integrals is an entire family of antiderivatives.

Suppose there are 3 functions:

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

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

$H (x) = x^{2} + π$

All of these are the same $x^{2}$ with different vertical shifts. When differentiated, the constants become $0$ and the derivative of all three functions is $2 x$ .

Then to integrate $2 x$ , find its antiderivative. But which function - $F (x), G (x),$ or $H (x)$ - is the antiderivative?

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

The answer to this indefinite integral is $x^{2} + C$ , which represents the entire family of functions that have the derivative $2 x$ .

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 $n \neq = - 1$ ,

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

What about the case where $n = - 1$ ? If we wanted to integrate $f (x) = \frac{1}{x}$ , the reverse power rule can’t be applied because

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

This is problem because nothing can be divided by 0. Instead, recall that the derivative of $ln ∣ x ∣$ is $\frac{1}{x}$ .

Then

$\int \frac{1}{x} = ln ∣ x ∣ + C$

By working backwards from some of the derivative rules from the 1st part of calculus, we obtain some important 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’re looking for the accumulation of change in the water between $t = 0$ to $t = 10$ , or evaluating

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

The antiderivative of $e^{t}$ is itself - $e^{t}$ .

On the other hand, $e^{3}$ is a constant so its antiderivative with respect to $t$ is $e^{3} \cdot t$ (using the reverse power rule).

Therefore the definite integral becomes

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

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

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

$\approx 22, 226.321 liters$

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)

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

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)

Notice the question gives the rate of change and asks for the change in temperature. So we have to integrate the rate for the accumulated change.

This will be done over the time interval in which the temperature is decreasing, or when the rate of change is negative.

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

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

Find a function that satisfies

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

and

$y (2) = 5$

Solution

(spoiler)

Since the derivative is given, we can integrate to get the original function.

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

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

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

$= 2^{x} - 2 ln ∣ x ∣ + C$

The problem also gives $y (2) = 5$ , which we can plug in to find the specific $C$ such that the function passes through the point $(2, 5)$ .

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

$C = 2.386$

Therefore the function that satisfies those conditions is

$y = 2^{x} - 2 ln ∣ x ∣ + 2.386$

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)

Since $f^{'} (2) = 10$ , then

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

$4 a + 2 b = 10$

Next, $f^{''} (x) = 2 a x + b$ . Since $f^{''} (2) = 18$ ,

$2 a (2) + b = 16$

$4 a + b = 18$

Now we solve the system of equations for $a$ and $b$ :

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

Subtracting equation $(2)$ from $(1)$ , we find that $b = - 6$ and $a = 6$ . Therefore

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

To find $f (x)$ , integrate $f^{'} (x)$ with the reverse power rule.

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

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

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

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

Lastly, use property (iii) by further integrating $f (x)$ . Let $F (x)$ be an antiderivative of $f (x)$ .

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

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

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

No additional $+ C$ needs to be added because we can immediately use the FTC part 2 for definite integrals, which says that

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

Combined with property (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

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

Key points

A definite integral always produces a value while an indefinite integral produces a family of functions (don’t forget $+ C$ ).