# Properties of Definite Integration

How to use the properties of definite integration to solve definite integral problems: properties, proofs, examples, and their solutions.

## Property 1: Intervals of Length Zero

If the upper limit and the lower limit are the same,

then that integral is 0.

## Proof (Property 1)

∫_{a}^{a} *f*(*x*) *dx* = *F*(*a*) - *F*(*a*)

Fundamental theorem of calculus (Part 2)

So the integral is 0.

## Example 1

It says the given equation is true for all *x*.

So this equation is true when *x* = 1.

So put 1 into the equation.

See the left side's integral.

The upper limit and the lower limit are both 1.

So the left side integral is 0.

So 0 = 1 + *a*.

## Property 2: Same Interval

If the integrals have the same interval,

then those integrals can be added (subtracted).

## Proof (Property 2)

∫_{a}^{b} *f*(*x*) *dx* = *F*(*b*) - *F*(*a*)

∫_{a}^{b} *g*(*x*) *dx* = *G*(*b*) - *G*(*a*)

fundamental theorem of calculus (Part 2)

Arrange the terms to make

[*F*(*b*) ± *G*(*b*)] - [*F*(*a*) ± *G*(*a*)].

Then [*F*(*b*) ± *G*(*b*)] - [*F*(*a*) ± *G*(*a*)] = ∫_{a}^{b} [*f*(*x*) ± *g*(*x*)] *dx*.

## Example 2

The result of a definite integral is a number.

So even when you change the variables

the result is the same.

So change the variables *y* and *z* into *x*.

The integrals all have the same interval: [1, 3].

So you can combine the integrals like this.

Solve the definite integral.

Definite integration of polynomials

## Property 3: Interval Addition

If the integrals have the same integrand, *f*(*x*),

and whose intervals are adjacent, [*a*, *c*] and [*c*, *b*],

then the adjacent intervals can be added.

## Proof (Property 3)

∫_{a}^{c} *f*(*x*) *dx* = *F*(*c*) - *F*(*a*)

∫_{c}^{b} *f*(*x*) *dx* = *F*(*b*) - *F*(*c*)

fundamental theorem of calculus (Part 2)

Cancel *F*(*c*).

Then *F*(*b*) - *F*(*a*) = ∫_{a}^{b} *f*(*x*) *dx*.

## Example 3

The given integrals' integrands are the same:

2*x* + 1.

And the integrals' intervals are adjacent:

[0, 2] and [2, 3].

So (given) = ∫_{0}^{3} (2*x* + 1) *dx*.

Solve the definite integral.

Definite integration of polynomials

## Property 4: Reversing Limits

If the upper limit and the lower limit are switched,

then (-) is added.

## Proof (Property 4)

∫_{a}^{b} *f*(*x*) *dx* = *F*(*b*) - *F*(*a*)

fundamental theorem of calculus (Part 2)

Switch *F*(*b*) and *F*(*a*)

and write (-) in front of them.

*F*(*a*) - *F*(*b*) = ∫_{b}^{a} *f*(*x*) *dx*

So -[*F*(*a*) - *F*(*b*)] = -∫_{b}^{a} *f*(*x*) *dx*.

## Example 4

Switch the second integral's limits.

And write (-) in front of the integral.

The given integrals' integrands are the same: 5*x*^{4}.

And the integrals' intervals are adjacent:

[0, 16] and [16, 2].

So use the property 3 (interval addition)

you've just learned.

Solve the definite integral.

Definite integration of polynomials