Properties of Definite Integration

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

[inteval, from a to a, of f(x) dx] = 0

If the upper limit and the lower limit are the same,
then that integral is 0.

Proof (Property 1)

Properties of Definite Integration: Proof of Intervals of Length Zero

aa f(x) dx = F(a) - F(a)

Fundamental theorem of calculus (Part 2)

So the integral is 0.

Example 1

Find the value of a that makes the given equation true for all x. [inteval, from 1 to x, of f(t) dt] = x^2 + ax

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

[inteval, from a to b, of f(x) dx] +- [inteval, from a to b, of g(x) dx] = [inteval, from a to b, of [f(x) +- g(x)] dx]

If the integrals have the same interval,
then those integrals can be added (subtracted).

Proof (Property 2)

Properties of Definite Integration: Proof of Same Interval

ab f(x) dx = F(b) - F(a)
ab 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)] = ∫ab [f(x) ± g(x)] dx.

Example 2

Find the given integral. [inteval, from 1 to 3, of (x^2 + x - 1) dx] + [inteval, from 1 to 3, of 2y^2 dy] - [inteval, from 1 to 3, of z dz]

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

[inteval, from a to c, of f(x) dx] + [inteval, from c to b, of f(x) dx] = [inteval, from a to b, of f(x) dx]

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)

Properties of Definite Integration: Proof of Interval Addition

ac f(x) dx = F(c) - F(a)
cb f(x) dx = F(b) - F(c)

fundamental theorem of calculus (Part 2)

Cancel F(c).

Then F(b) - F(a) = ∫ab f(x) dx.

Example 3

Find the given integral. [integral, from 0 to 2, of (2x + 1) dx] + [integral, from 2 to 3, of (2x + 1) dx]

The given integrals' integrands are the same:
2x + 1.

And the integrals' intervals are adjacent:
[0, 2] and [2, 3].

So (given) = ∫03 (2x + 1) dx.

Solve the definite integral.

Definite integration of polynomials

Property 4: Reversing Limits

[inteval, from a to b, of f(x) dx] = -[inteval, from b to a, of f(x) dx]

If the upper limit and the lower limit are switched,
then (-) is added.

Proof (Property 4)

Properties of Definite Integration: Proof Reversing Limits

ab 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) = ∫ba f(x) dx

So -[F(a) - F(b)] = -∫ba f(x) dx.

Example 4

Find the given integral. [integral, from 0 to 16, of 5x^4 dx] - [integral, from 2 to 16, of 5x^4 dx]

Switch the second integral's limits.
And write (-) in front of the integral.

The given integrals' integrands are the same: 5x4.

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