A handful of properties let you take a complicated integral apart into pieces you already know how to handle. Three of them split an integral up; one common-looking move is not allowed at all.

Definition The constant multiple rule

\int k\,f(x)\,dx = k\int f(x)\,dx

A constant factor can be lifted out in front of the integral, left there while you integrate what remains, and multiplied back in at the end. It never interferes with the integration itself.

The added constant absorbs it: $k$ times $C$ is still just an unknown constant, so it is written $C$ again.

Example Pulling the constant out

Take the factor outside, integrate the rest, then multiply back.

\displaystyle\int 3x\,dx = 3\int x\,dx = 3\cdot\dfrac{x^{2}}{2} = \dfrac{3x^{2}}{2} + C

\displaystyle\int \tfrac{1}{2}x\,dx = \tfrac{1}{2}\cdot\dfrac{x^{2}}{2} = \dfrac{x^{2}}{4} + C

\displaystyle\int 2e^{x}\,dx = 2\int e^{x}\,dx = 2e^{x} + C

\displaystyle\int 3\cdot 2^{x}\,dx = 3\cdot\dfrac{2^{x}}{\ln 2} + C

In every case the constant waits outside until the integration is done.

Note What a constant does to the area

An integral is the area between the curve and the $x$ -axis. Multiplying the function by a constant stretches that area by the same factor: $2$ doubles it, a factor between $0$ and $1$ shrinks it, and a negative factor flips it to the other side of the axis.

Across the same width, the region under

2f

is exactly twice the height of the region under

f

, so it holds twice the area. That is the whole content of the rule, read as a picture.

Definition The sum and difference rule

\[ \int \bigl[\,f(x) \pm g(x)\,\bigr]\,dx = \int f(x)\,dx \pm \int g(x)\,dx \]

Integrate each term on its own and combine the results with the sign that joined them. Nothing needs to be rearranged first.

Example Integrating term by term

\displaystyle\int (x + 1)\,dx = \int x\,dx + \int 1\,dx = \dfrac{x^{2}}{2} + x + C

\displaystyle\int (x - 1)\,dx = \int x\,dx - \int 1\,dx = \dfrac{x^{2}}{2} - x + C

Each term brings its own $C$ , but the sum of unknown constants is a single unknown constant, so only one $C$ is written.

As areas: adding two functions stacks their heights, so the region under $f + g$ is the region under $f$ plus the region under $g$ .

Note Splitting the interval

A definite integral over an interval equals the sum of the integrals over its pieces. Pick any point inside and cut there.

\int_{0}^{5} f(x)\,dx = \int_{0}^{3} f(x)\,dx + \int_{3}^{5} f(x)\,dx

The area from

0

5

is the area from

0

3

placed next to the area from

3

5

. The cut point is yours to choose.

Note The product does not split

The rules above cover a constant times a function, and a sum or difference of functions. A product of two functions is where the pattern stops.

\int f(x)\,g(x)\,dx \;\neq\; \left(\int f(x)\,dx\right)\!\left(\int g(x)\,dx\right)

A quick check shows why: $\displaystyle\int x\cdot x\,dx = \int x^{2}\,dx = \dfrac{x^{3}}{3} + C$ , but multiplying the separate integrals gives $\dfrac{x^{2}}{2}\cdot\dfrac{x^{2}}{2} = \dfrac{x^{4}}{4}$ — not the same expression.

Integrating a product needs its own method, covered in a later lesson.

Summary

$\displaystyle\int k\,f(x)\,dx = k\int f(x)\,dx$ : a constant factor comes out and goes back in.
A constant scales the area — enlarging, shrinking for $0 < k < 1$ , or flipping it when $k < 0$ .
$\displaystyle\int \bigl[f(x) \pm g(x)\bigr]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$ : integrate term by term.
Adding functions adds their areas.
$\displaystyle\int_{0}^{5} = \int_{0}^{3} + \int_{3}^{5}$ : an interval can be cut at any interior point.
A product does not split into a product of integrals; it has a separate method.

Properties of Integration