Math Core

The Fundamental Theorem turns every definite integral into a question about antiderivatives: find a function whose derivative is the integrand. This lesson collects the basic antiderivative rules, shows how to rewrite a function so the rules apply, and uses an initial condition to pick out one specific antiderivative.

Antiderivatives and the constant C

An antiderivative of $f$ is a function $F$ with $F'(x) = f(x)$ . For example, $x^2$ is an antiderivative of $2x$ . But so are $x^2 + 5$ and $x^2 - 17$ , because the derivative of a constant is 0.

In fact, if two functions have the same derivative on an interval, they differ by a constant (this follows from the Mean Value Theorem). So once you find one antiderivative, you have them all: add an arbitrary constant $C$ .

Definition

Indefinite integral

The indefinite integral of $f$ is the family of all antiderivatives of $f$ :

\int f(x)\,dx = F(x) + C, \quad \text{where } F'(x) = f(x).

$C$ is called the constant of integration.

A definite integral $\int_a^b f(x)\,dx$ is a number. An indefinite integral $\int f(x)\,dx$ is a family of functions. Graphically, the family is a stack of vertical shifts of the same curve: they all have the same slope at each $x$ .

Three antiderivatives of 2x: y = x² − 2, y = x² and y = x² + 2. At every x the three curves have the same slope.Open in grapher →

The basic rules

Each rule comes from reading a derivative rule backward. You can always check an antiderivative by differentiating it.

Basic antiderivatives

\begin{aligned} &\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \ne -1) &&\int \frac{1}{x}\,dx = \ln\lvert x\rvert + C \\ &\int e^x\,dx = e^x + C &&\int a^x\,dx = \frac{a^x}{\ln a} + C \\ &\int \cos x\,dx = \sin x + C &&\int \sin x\,dx = -\cos x + C \\ &\int \sec^2 x\,dx = \tan x + C &&\int \csc^2 x\,dx = -\cot x + C \\ &\int \sec x \tan x\,dx = \sec x + C &&\int \csc x \cot x\,dx = -\csc x + C \\ &\int \frac{1}{1 + x^2}\,dx = \arctan x + C &&\int \frac{1}{\sqrt{1 - x^2}}\,dx = \arcsin x + C \end{aligned}

Constants factor out, and sums split: $\displaystyle\int \big(k f(x) \pm g(x)\big)\,dx = k\int f(x)\,dx \pm \int g(x)\,dx$ .

The power rule in words: raise the exponent by one, then divide by the new exponent. It works for negative and fractional exponents too, as long as the exponent isn't $-1$ . That one case is covered by $\ln\lvert x\rvert$ , which is why $\int x^{-1}\,dx$ has its own rule.

Worked example: A polynomial

Find $\displaystyle\int (4x^3 - 6x + 5)\,dx$ .

Integrate term by term with the power rule. The constant 5 is $5x^0$ , whose antiderivative is $5x$ .

\int (4x^3 - 6x + 5)\,dx = 4 \cdot \frac{x^4}{4} - 6 \cdot \frac{x^2}{2} + 5x + C = x^4 - 3x^2 + 5x + C.

Check: $\dfrac{d}{dx}(x^4 - 3x^2 + 5x + C) = 4x^3 - 6x + 5$ , the original integrand.

Rewrite before you integrate

There's no quotient rule or product rule for antiderivatives. When the integrand is a fraction, a product or a radical, rewrite it as a sum of powers first.

Radicals become fractional exponents: $\sqrt{x} = x^{1/2}$ , $\dfrac{1}{\sqrt[3]{x}} = x^{-1/3}$ .
A fraction with a single-term denominator splits: $\dfrac{x^2 + 1}{x} = x + \dfrac{1}{x}$ .
Products of polynomials can be expanded: $(x + 2)^2 = x^2 + 4x + 4$ .

Worked example: Splitting a fraction

Find $\displaystyle\int \frac{x^2 + 1}{\sqrt{x}}\,dx$ .

Divide each term by $x^{1/2}$ :

\frac{x^2 + 1}{x^{1/2}} = x^{3/2} + x^{-1/2}.

Now use the power rule:

\int \left(x^{3/2} + x^{-1/2}\right) dx = \frac{x^{5/2}}{5/2} + \frac{x^{1/2}}{1/2} + C = \frac{2}{5}x^{5/2} + 2x^{1/2} + C.

Worked example: Mixing function types

Find $\displaystyle\int \left(3\cos x - 2e^x + \frac{4}{x}\right) dx$ .

Use a different basic rule for each term:

\int \left(3\cos x - 2e^x + \frac{4}{x}\right) dx = 3\sin x - 2e^x + 4\ln\lvert x\rvert + C.

Common mistake

Don't integrate a product or quotient piece by piece. $\displaystyle\int x \cdot x^2\,dx$ is not $\dfrac{x^2}{2} \cdot \dfrac{x^3}{3}$ . Multiply first: $\displaystyle\int x^3\,dx = \frac{x^4}{4} + C$ . And don't forget $+\,C$ on every indefinite integral; on the AP exam, leaving it off costs points.

Initial conditions: finding one antiderivative

Often you know the rate $f'(x)$ and one value of $f$ , such as $f(1) = 5$ . That extra fact, called an initial condition, pins down $C$ and picks one curve out of the family.

Worked example: Using an initial condition

Find $f(x)$ if $f'(x) = 6x^2 - 2$ and $f(1) = 5$ .

First, find the general antiderivative:

f(x) = 2x^3 - 2x + C.

Then use $f(1) = 5$ : $2(1)^3 - 2(1) + C = 5$ , so $0 + C = 5$ and $C = 5$ .

f(x) = 2x^3 - 2x + 5.

The same method works twice for a second derivative. Given $f''$ , integrate once and use a value of $f'$ to find the first constant; integrate again and use a value of $f$ to find the second. In motion problems this is how you go from acceleration to velocity to position.

Tip

Always verify with a quick derivative. If $\dfrac{d}{dx}$ of your answer doesn't give back the integrand exactly, something went wrong: often a missing coefficient, like forgetting to divide by the new exponent.

Practice

Practice 1

Find the antiderivative $F$ of $f(x) = 3x^2 + 4x$ that satisfies $F(0) = 2$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

Find the antiderivative $F$ of $f(x) = \cos x$ that satisfies $F(0) = 3$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 3

For $x \gt 0$ , $f'(x) = \dfrac{2}{x}$ and $f(1) = 4$ . Find $f(x)$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

$\displaystyle\int \frac{x^3 - 1}{x^2}\,dx =$

Practice 5

$\displaystyle\int \left(\frac{2}{1 + x^2} - \sin x\right) dx =$

Practice 6

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

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 7

A particle moves along a line with velocity $v(t) = 4t - 3$ . Its position at $t = 0$ is $s(0) = 5$ . Find $s(3)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.