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Antiderivatives

An antiderivative of f is a function whose derivative is f. Finding one means reading the differentiation rules backwards, remembering that there is always a whole family of them, and using a given value to pick the one that fits. Velocity back to position is the standard application.

In this topic

Taught
Not in the course
Questions
11
Total marks
29
Suggested time
47 min
Difficulty

Introductory to Capstone

Show the 3 hardest

Key results

The general antiderivative
G′(x)=F′(x) for all x in an interval  ⟹  G(x)=F(x)+CG'(x) = F'(x) \text{ for all } x \text{ in an interval} \implies G(x) = F(x) + C

For some constant CC, on an interval only; on a domain made of separate intervals, such as x≠0x \neq 0, each interval gets its own constant.

Powers
f(x)=xn:F(x)=xn+1n+1+Cf(x) = x^{n}:\quad F(x) = \frac{x^{n+1}}{n + 1} + C

For any constant n≠−1n \neq -1, on an interval where xnx^{n} is defined.

The reciprocal
f(x)=1x:F(x)=log⁡∣x∣+Cf(x) = \frac{1}{x}:\quad F(x) = \log|x| + C

On x>0x > 0 or on x<0x < 0, one interval at a time.

Exponentials and trigonometric functions
f(x)F(x)eax1aeax+Ccos⁡(ax)1asin⁡(ax)+Csin⁡(ax)−1acos⁡(ax)+C11+x2arctan⁡x+C\begin{array}{c|c} f(x) & F(x) \\ \hline e^{ax} & \frac{1}{a}e^{ax} + C \\ \cos(ax) & \frac{1}{a}\sin(ax) + C \\ \sin(ax) & -\frac{1}{a}\cos(ax) + C \\ \frac{1}{1 + x^{2}} & \arctan x + C \end{array}

For a constant a≠0a \neq 0, with xx in radians.

Constant acceleration
v(t)=v0+at,s(t)=s0+v0t+12at2v(t) = v_{0} + a t, \qquad s(t) = s_{0} + v_{0} t + \tfrac{1}{2} a t^{2}

For a constant acceleration aa, with v0=v(0)v_{0} = v(0) and s0=s(0)s_{0} = s(0).

Read the theoryCLP-1 §4.1 Introduction to antiderivatives

Worked example

Standard3 marks · 4 min

Find the function ff with f′(x)=4cos⁡(2x)+exf'(x) = 4\cos(2x) + e^{x} and f(0)=1f(0) = 1, and evaluate f ⁣(π4)f\!\left(\dfrac{\pi}{4}\right).

Strategy

Antidifferentiate each term, compensating for the chain-rule factor in cos(2x), fix the constant, and evaluate.

Working

  1. f(x)=2sin⁡(2x)+ex+Cf(x) = 2\sin(2x) + e^{x} + C

    Check: derivative is 4cos(2x) + eˣ.

  2. f(0)=0+1+C=1⟹C=0f(0) = 0 + 1 + C = 1 \quad\Longrightarrow\quad C = 0

    Initial condition.

  3. f ⁣(π4)=2sin⁡π2+eπ/4=2+eπ/4≈4.19f\!\left(\tfrac{\pi}{4}\right) = 2\sin\tfrac{\pi}{2} + e^{\pi/4} = 2 + e^{\pi/4} \approx 4.19

    sin(π/2) = 1.

Answer

f(x)=2sin⁡(2x)+ex;f(π/4)=2+eπ/4f(x) = 2\sin(2x) + e^{x};\quad f(\pi/4) = 2 + e^{\pi/4}

The answer is 2 plus e to the pi over 4, about 4.19.

Check

Differentiating 2 sin(2x) + eˣ gives back 4 cos(2x) + eˣ, and its value at x = 0 is 0 + 1 = 1, as required.

Where this goes wrong

Writing the antiderivative of 4cos(2x) as 4 sin(2x), which differentiates to 8cos(2x). The inner factor 2 must be divided out.

On the paper

Write the general antiderivative with its + C+\,C, find CC from the value given, then evaluate. Differentiate your answer to check it.

Leans on

The chain rule, from The chain rule and logarithms

Your turn

Standard3 marks · 4 min

A function FF satisfies F′(x)=31+9x2F'(x) = \dfrac{3}{1 + 9x^{2}} and F(0)=0F(0) = 0. Find F ⁣(13)F\!\left(\dfrac{1}{3}\right).

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Practice questions

5 of the 11, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

  • Introductory2 marks · 3 min

    A function FF satisfies F′(x)=6x2−4x+1F'(x) = 6x^{2} - 4x + 1 and F(0)=2F(0) = 2. Find F(1)F(1).

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  • Introductory1 mark · 2 min

    Which of the following is the most general antiderivative of f(x)=3sin⁡x−2f(x) = 3\sin x - 2?

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  • Standard3 marks · 5 min

    A ball is thrown straight up from a height of 1 m with an initial velocity of 14.7 m/s. Its acceleration is a constant −9.8-9.8 m/s². Find the maximum height it reaches.

    Give the answer in metres.

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  • Challenging3 marks · 5 min

    A function FF satisfies F′(x)=6x2ex3F'(x) = 6x^{2}e^{x^{3}} and F(0)=0F(0) = 0. Find F(1)F(1).

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  • Capstone5 marks · 8 min

    A function ff is defined for all x≠0x \neq 0 and satisfies f′(x)=1xf'(x) = \dfrac{1}{x} for every x≠0x \neq 0, with f(−1)=0f(-1) = 0 and f(1)=3f(1) = 3. Find f(−e)+f(e)f(-e) + f(e).

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