Definite Integral: How to Solve

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

So 15/4 is the answer.

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

So 54 is the answer.

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

Integral of 1/x

Integral of f(ax + b)

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.