SL 5.8 Find antiderivatives and apply initial conditions | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Calculus

Finding a function from its rate of change

Differentiation describes how a function changes. Antidifferentiation reverses that process: given a rate of change, we find functions that could have produced it. Since several functions can have the same derivative, an extra piece of information—an initial condition—can identify one particular function. This lesson reviews the needed derivative facts, develops the antiderivative rules, and connects the algebra to graphs and a simple motion context.

What you will learn

Explain what an antiderivative is and why a family of antiderivatives includes a constant.
Find antiderivatives of common power, exponential, and trigonometric functions.
Use an initial condition to select one function from a family.
Check an answer by differentiating it and interpret the result in context.

1. Prior knowledge: derivatives and function values

A derivative gives the gradient of a function at each input. For example, if

F(x)=x^3

, then

F'(x)=3x^2

. So, when we are given

3x^2

and asked to find an antiderivative, we are looking for a function whose derivative is

3x^2

.

You will need to rearrange simple equations and substitute a known input into a function. The notation

F'(x)

means the derivative of

F

with respect to

x

. An initial condition such as

F(2)=7

tells us the value of the unknown function at a particular input.

\frac{d}{dx}(x^n)=nx^{n-1}

An antiderivative reverses differentiation.
Check algebra by substituting the stated input into the function.

2. Antiderivative rules and the constant

If

F'(x)=f(x)

, then

F

is an antiderivative of

f

. We write the set of antiderivatives as

\int f(x)\,dx=F(x)+C

. The symbol

\int

indicates antidifferentiation,

dx

names the variable, and

C

is an arbitrary constant.

The constant is necessary because adding any fixed number to a function does not change its derivative. For example, the functions

x^3

,

x^3+4

, and

x^3-10

all have derivative

3x^2

. An initial condition can determine which member of this family is required.

For a power

x^n

with

n\ne -1

, increase the exponent by one and divide by the new exponent. A constant multiplier stays as a multiplier, and sums can be integrated term by term. The special case

1/x

has antiderivative

\ln|x|+C

on intervals that do not include zero. Other useful basic pairs are that

e^x

is its own antiderivative, the antiderivative of

\cos x

is

\sin x

, and the antiderivative of

\sin x

is

-\cos x

. Trigonometric angles here are in radians.

These rules connect symbolic work to graphs. If

F'(x)=f(x)

, the gradient of the graph of

F

at each

x

equals the height of the graph of

f

there. Antiderivatives that differ only by

C

have the same shape, shifted vertically. An initial condition pins one of these shifted graphs to a specified point.

\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\quad n\ne -1

Include $+C$ when writing a general antiderivative.
Use the power rule only when the exponent is not $-1$ .
Differentiate the result to verify the antiderivative.

3. Applying an initial condition and checking representations

First find the general antiderivative, including

C

. Then substitute the input and output from the initial condition, solve for

C

, and write the resulting particular function. Do not discard

C

before using the condition: doing so may force the wrong graph.

A graphing calculator can provide a purposeful check. Plot the given rate function and a candidate antiderivative, then use a numerical derivative feature or compare estimated slopes on the candidate graph with the rate-function values. The candidate should pass through the point given by the initial condition. Calculator evidence supports the reasoning, but it does not replace finding the antiderivative and solving for

C

.

In a context, units help interpret the result. If a rate is measured in metres per second and the input is time in seconds, an antiderivative represents position in metres, up to a starting position. The initial condition supplies that starting value. This interpretation assumes the rate function and initial value apply over the interval being considered.

F'(x)=f(x),\quad F(a)=b

The initial condition determines the constant of integration.
A derivative check and a point check test different parts of the answer.
Keep units consistent when interpreting a function and its rate.

Worked example

A polynomial antiderivative

Find the function

F

satisfying

F'(x)=6x^2-4x+3

and

F(1)=5

.

Integrate each term
Apply the power rule to each term, keeping the constant of integration because the derivative alone does not specify a unique function.
$F(x)=2x^3-2x^2+3x+C$
Use the condition
Substitute $x=1$ and $F(1)=5$ . This gives an equation for the one unknown constant.
$5=2-2+3+C$
Solve and state the function
The equation gives $C=2$ . Substituting this value selects the required member of the family.
$F(x)=2x^3-2x^2+3x+2$

Answer:

F(x)=2x^3-2x^2+3x+2

Check: Differentiating gives

6x^2-4x+3

, and substituting

x=1

gives

F(1)=5

.

Worked example

An exponential antiderivative

Find

G(x)

if

G'(x)=4e^x-2

and

G(0)=1

.

Find the general antiderivative
The antiderivative of $e^x$ is $e^x$ , while the antiderivative of the constant $-2$ is $-2x$ . Include $C$ .
$G(x)=4e^x-2x+C$
Determine the constant
Use $G(0)=1$ and $e^0=1$ to form and solve the condition.
$1=4-0+C,\quad C=-3$
Write the particular function
Replace $C$ with its value. The result is the function that has the required derivative and passes through the specified point.
$G(x)=4e^x-2x-3$

Answer:

G(x)=4e^x-2x-3

Check: Its derivative is

4e^x-2

, and

G(0)=4-3=1

.

Worked example

Interpreting an initial value

A particle moves along a straight path. Its velocity is

v(t)=6t-2

metres per second, where

t

is in seconds. Its position at

t=0

is

5

metres. Find its position

s(t)

for

t\ge 0

.

Antidifferentiate velocity
Position has velocity as its derivative, so integrate the given expression with respect to time. The resulting position includes an unknown constant.
$s(t)=3t^2-2t+C$
Apply the starting position
The condition says the position is $5$ metres when time is zero. Substitution determines the constant, with the position measured in metres.
$5=3(0)^2-2(0)+C,\quad C=5$
State and interpret the result
The constant represents the starting position. The domain $t\ge 0$ reflects the stated time interval.
$s(t)=3t^2-2t+5$

Answer: The position is

s(t)=3t^2-2t+5

metres for

t\ge 0

.

Check: Differentiating position gives

6t-2

metres per second, and substituting

t=0

gives

5

metres.

Common mistakes and how to avoid them

Leaving out the constant of integration in a general antiderivative.

Correction: Write

+C

until an initial condition determines its value.

Dividing by the original exponent instead of the new exponent.

Correction: Increase the power by one first, then divide by that new power.

Using the power rule for

1/x

as though its exponent were an ordinary power.

Correction: Use the separate antiderivative

\ln|x|+C

on an interval that excludes zero.

Checking only the derivative and ignoring the initial condition.

Correction: Differentiate the answer and substitute the initial input; both checks are needed.

Lesson summary

An antiderivative of $f$ is a function whose derivative is $f$ .
The power rule, basic exponential and trigonometric pairs, and the separate $1/x$ rule provide common antiderivatives.
Include $C$ in the general answer, then use the initial condition to find its value.
Verify a particular answer by differentiating and checking the stated function value.

Check your understanding

Question 1

If

H'(x)=8x^3

and

H(1)=6

, which function is

H(x)

?

$2x^4+4$
$2x^4+6$
$8x^4+4$
$2x^4+5$

Show answer and explanation

2x^4+4

The general antiderivative is

2x^4+C

. Since

H(1)=6

, we have

2+C=6

, so

C=4

.

Question 2

Which expression is an antiderivative of

\cos x

?

$-\sin x+C$
$\sin x+C$
$\cos x+C$
$-\cos x+C$

Show answer and explanation

\sin x+C

The derivative of

\sin x

is

\cos x

, so all antiderivatives are

\sin x+C

.

Question 3

A function

P

has derivative

P'(x)=5

and satisfies

P(2)=9

. What is

P(x)

?

$5x-1$
$5x+9$
$x+7$
$5x+1$

Show answer and explanation

5x-1

Integrating the constant gives

P(x)=5x+C

. Substitution gives

9=10+C

, so

C=-1

.

Key terms

Antiderivative: A function whose derivative is the given function.
Constant of integration: The arbitrary constant added to a general antiderivative because constants have derivative zero.
Initial condition: A stated value of the unknown function at a particular input, used to determine the constant.

Continue through IB AA SL

View the complete IB AA SL International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL, study topic SL 5.8. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question

SL 5.8 · Find antiderivatives and apply initial conditions

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Finding a function from its rate of change

What you will learn

1. Prior knowledge: derivatives and function values

2. Antiderivative rules and the constant

3. Applying an initial condition and checking representations

A polynomial antiderivative

An exponential antiderivative

Interpreting an initial value

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review