Chapter 7 · Integrals

Example 4: The anti derivative of $4x^3 - 6$ with $F(0) = 3$

ExamplesIntegration as anti-differentiation · Section 7.2

The question

Find the anti derivative $F$ of $f$ defined by $f(x) = 4x^3 - 6$ ⁠, where $F(0) = 3$ ⁠.

Exam answer

Since $\frac{d}{dx}(x^4 - 6x) = 4x^3 - 6$ ⁠,

F(x) = x^4 - 6x + C

Putting $x = 0$ and using $F(0) = 3$ ⁠:

3 = 0^4 - 6\times 0 + C \Rightarrow C = 3

$\therefore\ F(x) = x^4 - 6x + 3$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A function has infinitely many anti derivatives, and they differ only by a constant: $F(x) = (\text{one anti derivative}) + C$ ⁠. Geometrically they are copies of one curve shifted up and down. An extra condition such as $F(0) = 3$ says which point the curve must pass through, and that picks out exactly one value of $C$ ⁠.

Given, to find, and the plan

Given: $f(x) = 4x^3 - 6$ for all real $x$ ⁠, and the condition $F(0) = 3$ ⁠.
To find: The function $F$ with $F'(x) = f(x)$ for every $x$ and $F(0) = 3$ ⁠.
Plan: (1) Find one anti derivative by inspection; (2) write the general one with $+\,C$ ⁠; (3) substitute $x = 0$ to find $C$ ⁠; (4) check both conditions.

Basics you need

1Power rule

$\frac{d}{dx}x^4 = 4x^3$ and $\frac{d}{dx}(6x) = 6$ ⁠; in integral form $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ ⁠, $n \ne -1$ ⁠. Basics: powers of x and the power rule

2All anti derivatives differ by a constant

If $G' = H'$ on an interval, then $G - H$ has derivative 0 there, so it is a constant (Remark in Section 7.2). Hence $F + C$ ⁠, $C \in \R$ ⁠, lists every anti derivative. Basics: integration undoes differentiation

3Evaluating a function

$F(0)$ means: replace every $x$ in the formula for $F$ by 0.

Solution

Find one anti derivative

By the power rule,
$\frac{d}{dx}\left(x^4 - 6x\right) = 4x^3 - 6.$
So $x^4 - 6x$ is one anti derivative of $f$ ⁠.

The $4x^3$ is exactly the derivative of $x^4$ ⁠, and the constant $-6$ is the derivative of $-6x$ (a straight line of slope $-6$ ⁠). Equivalently, with formula (i): $\int 4x^3dx = 4\cdot\frac{x^4}{4} = x^4$ and $\int 6\,dx = 6x$ ⁠.

Standard formula (i), Section 7.2. Basics: the power rule
Write down every anti derivative
$F(x) = x^4 - 6x + C, \qquad C \text{ a constant}.$
Adding a constant does not change the derivative, so each $x^4 - 6x + C$ is an anti derivative. Conversely, because $f$ is defined on the whole real line (one interval), any anti derivative differs from $x^4 - 6x$ by a constant, so this formula misses none of them. That is why the answer found in the next step is the unique one.

Remark on functions with the same derivative, Section 7.2. Lesson: integration as anti-differentiation
Use $F(0) = 3$ to find $C$

Put $x = 0$ ⁠:
$F(0) = 0^4 - 6\times 0 + C = C.$
The condition $F(0) = 3$ therefore gives $C = 3$ ⁠.

Every term containing $x$ vanishes at $x = 0$ ⁠, so $F(0)$ is just the constant. That is why the condition at 0 reads off $C$ directly. With a condition at another point, say $F(1) = 3$ ⁠, we would get $1 - 6 + C = 3$ ⁠, so $C = 8$ ⁠: the same method, one more line of arithmetic.
Write the answer
$F(x) = x^4 - 6x + 3.$
This is the one member of the family of Step 2 that passes through the point $(0, 3)$ ⁠. The figure shows the family and the chosen curve.

The curves $y = x^4 - 6x + C$ for $C = -3, 0, 3, 6$ ⁠, drawn for $0 \le x \le 2.1$ ⁠. They are vertical shifts of one another, with the same slope at every $x$ ⁠. The condition $F(0) = 3$ picks out the blue one, through $(0, 3)$ ⁠.
Check both conditions
$\frac{d}{dx}\left(x^4 - 6x + 3\right) = 4x^3 - 6 = f(x) \ \checkmark$ $F(0) = 0 - 0 + 3 = 3 \ \checkmark$
The question asks for two things, an anti derivative and a value at 0, so both must be checked.

Result

The required anti derivative is $F(x) = x^4 - 6x + 3$ ⁠. This agrees with the textbook's solution.

Check it with numbers

$F(1) = 1 - 6 + 3 = -2$ ⁠, so $F(1) - F(0) = -2 - 3 = -5$ ⁠. Independently, a computer adding very many thin strips under $y = 4x^3 - 6$ from 0 to 1 gets $-5$ as well (the strips lie below the axis, since $4x^3 - 6 < 0$ there). Section 7.8 explains why these must agree.

Slope check at $x = 1$ ⁠: $f(1) = 4 - 6 = -2$ ⁠, and $\frac{F(1.001) - F(0.999)}{0.002} = -1.999996$ ⁠, which is $-2$ to within the size of the step.

Common mistakes

Forgetting $C$ and answering $x^4 - 6x$ ⁠. That function has $F(0) = 0$ ⁠, not 3.
Setting the answer equal to 3 everywhere. $F(0) = 3$ is a condition at one point; it fixes $C$ ⁠, not $F(x)$ for all $x$ ⁠.
Integrating $-6$ as $-6$ or as 0. $\int(-6)\,dx = -6x$ ⁠; a constant integrates to a multiple of $x$ ⁠.

What to take away

" $+\,C$ " describes a whole family of curves, all vertical shifts of one another.
One extra condition (an "initial value") fixes $C$ and gives a unique anti derivative.
Check both the derivative and the condition.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}x^n = nx^{n-1}$ ⁠; $\int x^n dx = \frac{x^{n+1}}{n+1} + C$	Step 1	Standard formula (i), Section 7.2
Functions with the same derivative differ by a constant	Step 2	Remark, Section 7.2 (Lesson: integration as anti-differentiation)
A condition fixes $C$ and gives a unique anti derivative	Steps 3, 4	Remark (i) after Example 4
$\int_a^b f\,dx = F(b) - F(a)$	Check box	Section 7.8, later in this chapter

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Example 4: The anti derivative of 4x3−64x^3 - 64x3−6 with F(0)=3F(0) = 3F(0)=3

Basics you need

Solution

Find one anti derivative

Write down every anti derivative

Use F(0)=3F(0) = 3F(0)=3 to find CCC