# Definite Integral: How to Solve

How to solve a definite integral: formula, 3 examples, and their solutions.

## Formula

### Formula

ab f(x) dx = F(b) - F(a)

First write the integral of f(x), F(x).

Next, find the value of F(b) - F(a).
(Put b and a into F(x).)

## Example 1

### Solution

The integral of x3 is
[1/4]x4.
This is F(x).

Integral of a Polynomial

Draw a bracket.

Write 2 and -1
on the right side of the bracket.

To find F(2),
put the upper limit 2
into [1/4]x4.

Then [1/4]⋅24.

Write -.

To find F(-1),
put the lower limit -1
into [1/4]x4.

Then [1/4]⋅(-1)4.

So [1/4]⋅24 - [1/4]⋅(-1)4.

24 = 16
(-1)4 = 1

[1/4]⋅16 - [1/4]⋅1 = [1/4]⋅(16 - 1)

16 - 1 = 15

## Example 2

### Solution

The integral of 6x2 - 2x + 5 is
6⋅[1/3]x3 - 2⋅[1/2]x2 + 5x.

Draw a bracket.

Write 3 and 1
on the right side of the bracket.

6⋅[1/3]x3 = 2x3
-2⋅[1/2]x2 = -x2

Put 3 into 2x3 - x2 + 5x.

Then 2⋅33 - 32 + 5⋅3.

Write -.

Put 1 into 2x3 - x2 + 5x.

Then (2⋅13 - 12 + 5⋅1).

So 2⋅33 - 32 + 5⋅3 - (2⋅13 - 12 + 5⋅1).

2⋅33 = 2⋅27
-32 = -9
+5⋅3 = +15

2⋅13 = 2
-12 = -1
+5⋅1 = +5

2⋅27 = 54

-9 + 15 = +6

-(2 - 1 + 5) = -2 + 1 - 5

+6 - 2 = +4
+1 - 5 = -4

+4 - 4 = 0

## Example 3

### Solution

The integral of 1/[3x + 1] is
[1/3]⋅(ln |3x + 1|).

Integral of 1/x

Linear Change of Variable Rule

Draw a bracket.

Write 1 and 0
on the right side of the bracket.

Put 1 into [1/3]⋅(ln |3x + 1|).

Then [1/3]⋅(ln |3⋅1 + 1|).

Write -.

Put 0 into [1/3]⋅(ln |3x + 1|).

Then [1/3]⋅(ln |3⋅0 + 1|).

So [1/3]⋅(ln |3⋅1 + 1|) - [1/3]⋅(ln |3⋅0 + 1|).

|3⋅1 + 1| = |4|

|3⋅0 + 1| = |1|

ln |4| = ln 4 = ln 22

ln |1| = 0
So [1/3]⋅(ln |1|) = [1/3]⋅0 = 0.

[1/3]⋅[ln 22]
= [1/3]⋅2⋅[ln 2]
= [2/3]⋅[ln 2]

Logarithm of a Power

So [2/3]⋅[ln 2] is the answer.