Definite Integral: Property

How to use the properties of definite integral to solve the given integrals: 4 properties, 4 examples, and their solutions.

Property 1: Same Limits

Formula

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

aa f(x) dx = 0

This is true because
aa f(x) dx
= F(a) - F(a)
= 0.

Example 1

Example

Solution

It says
the given equation is true for all x.

So, if x = 1,
the equation is true.

So put 1 into the equation.
Then ∫11 f(t) dt = 12 + a⋅1.

The upper limit and the lower limit
are the same: 1.

So the left side is 0.

Simplify the right side.
12 + a⋅1 = 1 + a

So 0 = 1 + a.

Solve this equation.

Then a = -1.

So a = -1 is the answer.

Property 2: Same Limit Integrals

Formula

For two integrals,
if the upper limit and the lower limit
are the same,
you can add and subtract the inner functions.

ab f(x) dx ± ∫ab g(x) dx = ∫ab [f(x) ± g(x)] dx

Example 2

Example

Solution

A definite integral is a value,
not a function.

So you can change
the variable of a definite integral.

+∫13 2y2 dy = +∫13 2x2 dx

-∫13 z dz = -∫13 x dx

These three integrals all have
the same upper limit (1) and the lower limit (3).

So ∫13 (x2 + x - 1) dx + ∫13 2x2 dx - ∫13 x dx
= ∫13 (x2 + x - 1 + 2x2 - x) dx.

x2 + 2x2 = 3x2

Cancel +x and -x.

Then ∫13 (3x2 - 1) dx.

Solve the integral.

Definite Integral: How to Solve

The integral of (3x2 - 1) is
3⋅[1/3]x3 - x.

Integral of a Polynomial

3⋅[1/3]x3 - x = x3 - x

Put 3 and 1
into x3 - x.

Then 33 - 3 - (13 - 1).

33 = 27

13 = 1

27 - 3 = 24

1 - 1 = 0

24 - 0 = 24

So 24 is the answer.

Property 3: Connected Limits

Formula

ab f(x) dx + ∫bc f(x) dx = ∫ac f(x) dx

Example 3

Example

Solution

The first integral is
from 0 to 1.

The second integral is
from 1 to 4.

And the inner functions of both integrals, (2x + 1),
are the same.

Then (given) = ∫04 f(x) dx.

Solve the integral.

The integral of (2x + 1) is
2⋅[1/2]x2 + x.

2⋅[1/2]x2 + x = x2 + x

Put 4 and 0
into x2 + x.

Then 42 + 4 - (02 + 0).

42 = 16

-(02 + 0) = 0

16 + 4 = 20

So 20 is the answer.

Property 4: Switching Limits

Formula

If the upper limit and the lower limit are switched,
then the sign of the integral is changed.

ab f(x) dx = -∫ba f(x) dx

Example 4

Example

Solution

Write ∫08 5x4 dx and -.

28 5x4 dx = -∫82 5x4 dx

So (given) = ∫08 5x4 - (-∫82 5x4 dx).

-(-∫82 5x4 dx) = +∫82 5x4 dx

The first integral is
from 0 to 8.

The second integral is
from 8 to 2.

And the inner functions of both integrals, 5x4,
are the same.

So (given) = ∫02 5x4 dx

Property 3 is used.

Solve the integral.

The integral of 5x4 is
5⋅[1/5]x5.

[1/5]x5 = x5

Put 2 and 0
into x5.

Then 25 - 05.

25 = 32

So 32 is the answer.