Introduction to Indefinite Integrals, calculus lesson

So far you have started with a function and found its derivative. Often the situation is reversed. A speedometer gives you velocity and you want the distance covered, or a rate of growth is known and you want the size. Undoing differentiation is called antidifferentiation, and it opens the second half of calculus.

Key terms

Antiderivative

Definition Let $f$ be defined on an interval $I$ . A function $F$ is an antiderivative of $f$ on $I$ if $F$ is differentiable on $I$ and

F'(x) = f(x) \quad \text{for every } x \text{ in } I.

At an endpoint of $I$ that belongs to $I$ , the derivative is the one-sided derivative.

In plain words An antiderivative of $f$ is a function whose derivative is $f$ .

Example On $\mathbb{R}$ the function

F(x) = \frac{x^3}{3}

is an antiderivative of $f(x) = x^2$ , because $F'(x) = x^2$ . So is $x^3/3 + 7$ .

The theorems behind antiderivatives

Definition Both facts below follow from the mean value theorem.

If $F'(x) = 0$ for every $x$ in an interval $I$ , then $F$ is constant on $I$ .
If $F'(x) = G'(x)$ for every $x$ in an interval $I$ , then there is a constant $C$ with $F(x) = G(x) + C$ for every $x$ in $I$ .

Also, if $F$ and $G$ are antiderivatives of $f$ and $g$ on $I$ and $c$ is a constant, then $F + G$ is an antiderivative of $f + g$ and $cF$ is an antiderivative of $cf$ on $I$ .

In plain words On an interval, two functions with the same derivative can differ only by a constant. Antiderivatives of sums and constant multiples are built piece by piece.

Example For the first fact, suppose $F'(x) = 0$ on $I$ and $x_1 < x_2$ are in $I$ . Since $F$ is differentiable on $I$ , the mean value theorem gives some $c$ between them with

F(x_2) - F(x_1) = F'(c)(x_2 - x_1) = 0,

so $F(x_1) = F(x_2)$ . For the sum rule, $x^3/3$ and $-\cos x$ are antiderivatives of $x^2$ and $\sin x$ , so

\frac{x^3}{3} - \cos x

is an antiderivative of $x^2 + \sin x$ .

The family of antiderivatives

Definition If $F$ is an antiderivative of $f$ on an interval $I$ , then the antiderivatives of $f$ on $I$ are exactly the functions

F(x) + C, \qquad C \text{ a constant}.

This is the general antiderivative of $f$ on $I$ .

In plain words Once you have found one antiderivative, you have found them all. The rest are the same graph shifted up or down.

Example The general antiderivative of $\cos x$ on $\mathbb{R}$ is $\sin x + C$ . The graphs of $\sin x$ , $\sin x + 1$ and $\sin x - 2$ are vertical shifts of each other, and at every $x$ they have the same slope $\cos x$ .

The indefinite integral

Definition The indefinite integral of $f$ on an interval $I$ is the general antiderivative, written

\int f(x)\,dx = F(x) + C, \qquad \text{meaning } F'(x) = f(x) \text{ for every } x \text{ in } I.

The function $f$ is the integrand, $x$ is the variable of integration, and $C$ is the constant of integration.

In plain words The integral sign with no limits asks for every function whose derivative is the integrand. The $dx$ names the variable.

Example On $\mathbb{R}$

\int (4x^3 + 2)\,dx = x^4 + 2x + C,

because the derivative of $x^4 + 2x$ is $4x^3 + 2$ .

The constant of integration lives on an interval

Definition The statement that all antiderivatives differ by one constant is true on an interval. If the domain of $f$ is a union of several disjoint intervals, the most general antiderivative may use a different constant on each interval.

In plain words A gap in the domain lets each piece shift independently.

Example The function $f(x) = 1/x^2$ is defined for $x \neq 0$ . Its most general antiderivative is

F(x) = \begin{cases} -\dfrac{1}{x} + C_1, & x > 0, \\[1ex] -\dfrac{1}{x} + C_2, & x < 0, \end{cases}

with $C_1$ and $C_2$ chosen independently. For instance the function equal to $-1/x + 5$ for $x > 0$ and to $-1/x - 3$ for $x < 0$ has derivative $1/x^2$ everywhere on the domain, yet it is not $-1/x$ plus a single constant.

Initial conditions

Definition An initial value problem asks for a function $F$ on an interval $I$ with $F' = f$ on $I$ and $F(x_0) = y_0$ for a given $x_0$ in $I$ . Since every antiderivative has the form $G + C$ , the condition $G(x_0) + C = y_0$ fixes $C$ , and the solution on $I$ is unique.

In plain words One known value of the function picks one curve out of the whole family.

Example Find $f$ with $f'(x) = 6x^2 - 4$ and $f(1) = 3$ . The general antiderivative is $f(x) = 2x^3 - 4x + C$ , and $f(1) = 2 - 4 + C = 3$ gives $C = 5$ . So $f(x) = 2x^3 - 4x + 5$ . For a ball thrown upward at $15$ meters per second from $2$ meters with acceleration $-9.8$ , the same steps give $v(t) = -9.8t + 15$ and then $s(t) = -4.9t^2 + 15t + 2$ .

Checking by differentiation

Definition The equation $\int f(x)\,dx = F(x) + C$ on $I$ is correct exactly when $F'(x) = f(x)$ for every $x$ in $I$ . So any proposed antiderivative can be checked by differentiating it.

In plain words Integrals can be hard to find but are always easy to check. Differentiate your answer and compare with the integrand.

Example The claim $\int x\cos x\,dx = x\sin x + \cos x + C$ checks out, because

\frac{d}{dx}(x\sin x + \cos x) = \sin x + x\cos x - \sin x = x\cos x.

The guess $\tfrac{1}{2}x^2\sin x$ fails, since its derivative

x\sin x + \frac{x^2}{2}\cos x

is not $x\cos x$ .

Check yourself

Find $f$ with $f'(x) = 3x^2 + 2x$ and $f(0) = -1$ .

Show the answer

Every antiderivative has the form $f(x) = x^3 + x^2 + C$ , and $f(0) = C = -1$ . So $f(x) = x^3 + x^2 - 1$ .

Check whether $\int 2xe^{x^2}\,dx = e^{x^2} + C$ .

Show the answer

By the chain rule

\frac{d}{dx}e^{x^2} = e^{x^2} \cdot 2x,

which is the integrand. The formula is correct on $\mathbb{R}$ .

Find $f$ with $f''(x) = 6x$ , $f'(0) = 2$ and $f(0) = 1$ .

Show the answer

Integrating once gives $f'(x) = 3x^2 + C_1$ , and $f'(0) = 2$ gives $C_1 = 2$ . Then $f(x) = x^3 + 2x + C_2$ and $f(0) = 1$ gives $C_2 = 1$ . So $f(x) = x^3 + 2x + 1$ .

Introduction to Indefinite Integrals

Theory, part 89

Practice, part 90

Key terms

Antiderivative

The theorems behind antiderivatives

The family of antiderivatives

The indefinite integral

The constant of integration lives on an interval

Initial conditions

Checking by differentiation

Check yourself

How this lesson connects

Builds on

Used later in