# 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.

∫_{a}^{a} f(x) dx = 0

This is true because

∫_{a}^{a} 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 ∫_{1}^{1} f(t) dt = 1^{2} + a⋅1.

The upper limit and the lower limit

are the same: 1.

So the left side is 0.

Simplify the right side.

1^{2} + 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.

∫_{a}^{b} f(x) dx ± ∫_{a}^{b} g(x) dx = ∫_{a}^{b} [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.

+∫_{1}^{3} 2y^{2} dy = +∫_{1}^{3} 2x^{2} dx

-∫_{1}^{3} z dz = -∫_{1}^{3} x dx

These three integrals all have

the same upper limit (1) and the lower limit (3).

So ∫_{1}^{3} (x^{2} + x - 1) dx + ∫_{1}^{3} 2x^{2} dx - ∫_{1}^{3} x dx

= ∫_{1}^{3} (x^{2} + x - 1 + 2x^{2} - x) dx.

x^{2} + 2x^{2} = 3x^{2}

Cancel +x and -x.

Then ∫_{1}^{3} (3x^{2} - 1) dx.

Solve the integral.

Definite Integral: How to Solve

The integral of (3x^{2} - 1) is

3⋅[1/3]x^{3} - x.

Integral of a Polynomial

3⋅[1/3]x^{3} - x = x^{3} - x

Put 3 and 1

into x^{3} - x.

Then 3^{3} - 3 - (1^{3} - 1).

3^{3} = 27

1^{3} = 1

27 - 3 = 24

1 - 1 = 0

24 - 0 = 24

So 24 is the answer.

## Property 3: Connected Limits

### Formula

∫_{a}^{b} f(x) dx + ∫_{b}^{c} f(x) dx = ∫_{a}^{c} 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) = ∫_{0}^{4} f(x) dx.

Solve the integral.

The integral of (2x + 1) is

2⋅[1/2]x^{2} + x.

2⋅[1/2]x^{2} + x = x^{2} + x

Put 4 and 0

into x^{2} + x.

Then 4^{2} + 4 - (0^{2} + 0).

4^{2} = 16

-(0^{2} + 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.

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

## Example 4

### Example

### Solution

Write ∫_{0}^{8} 5x^{4} dx and -.

∫_{2}^{8} 5x^{4} dx = -∫_{8}^{2} 5x^{4} dx

So (given) = ∫_{0}^{8} 5x^{4} - (-∫_{8}^{2} 5x^{4} dx).

-(-∫_{8}^{2} 5x^{4} dx) = +∫_{8}^{2} 5x^{4} dx

The first integral is

from 0 to 8.

The second integral is

from 8 to 2.

And the inner functions of both integrals, 5x^{4},

are the same.

So (given) = ∫_{0}^{2} 5x^{4} dx

Property 3 is used.

Solve the integral.

The integral of 5x^{4} is

5⋅[1/5]x^{5}.

[1/5]x^{5} = x^{5}

Put 2 and 0

into x^{5}.

Then 2^{5} - 0^{5}.

2^{5} = 32

So 32 is the answer.