Definite Integral: Odd and Even Functions

How to solve the integral of odd and even functions: definitions, formulas, 2 examples, and their solutions.

Odd Function

Definition

An odd function is a function
whose graph is symmetric about the origin.

Examples:
x1, x3, x5, x7, ... (Odd exponent polynomial functions)
sin x, tan x
sinh x

Sine: Graph

Tangent: Graph

Even Function

Definition

An even function is a function
whose graph is symmetric about the y-axis.

Examples:
C (constant), x2, x4, x6, ... (Even exponent polynomial functions)
cos x
cosh x

Cosine: Graph

Formulas

Integral of an Odd Function

-aa (odd function) dx = 0

The left colored region and the right colored region
have the same shape
but have different signs.

So the two areas are cancelled.

So the integral of an odd function is 0.
(from -a to a)

Integral of an Even Function

-aa (even function) dx = 2⋅∫0a (even function) dx

The left colored region and the right colored region
have the same shape
and have the same sign.

So the integral of an even function
is the double of the right colored region.

Example 1

Example

Solution

The upper limit (1) and the lower limit (-1)
are the opposites.

5x4, -6x2, and +7 are even functions.
+8x3 and -2x are odd functions.
The terms of the inner function
are either odd functions or even functions.

So you can use the above formulas.

First write 2.

Write the integral from 0 to 1.

Write the even functions:
5x4 - 6x2 + 7.

And write dx.

The integral of the odd functions are 0.
So the odd functions, +8x3 and -2x, are removed.

Solve the integral.

Definite Integral: How to Solve

The integral of (5x4 - 6x2 + 7) is
5⋅[1/5]x5 - 6⋅[1/3]x3 + 7x.

Integral of a Polynomial

5⋅[1/5]x5 - 6⋅[1/3]x3 + 7x
= x5 - 2x3 + 7x

Put 1 and 0
into x5 - 2x3 + 7x.

Then 2[15 - 2⋅13 + 7⋅1 - (05 - 2⋅03 + 7⋅0)]

15 = 1
-2⋅13 = -2
+7⋅1 = +7

-(05 - 2⋅03 + 7⋅0) = -0

-2 + 7 = +5

1 + 5 = 6

2⋅6 = 12

So 12 is the answer.

Example 2

Example

Solution

The upper limit (π/3) and the lower limit (-π/3)
are the opposites.

sin θ is an odd function.
cos θ is an odd function.
The terms of the inner function
are either odd function or even function.

So you can use the above formulas.

First write 2.

Write the integral from 0 to π/3.

Write the even function:
cos θ.

And write dθ.

The integral of the odd function is 0.
So the odd function, sin θ, is removed.

Solve the integral.

The integral of cos θ is
sin θ.

Put π/3 and 0
into sin θ.

Then [sin π/3 - sin 0].

sin π/3 = √3/2

sin 0 = 0

Sine Values of Commonly Used Angles

2⋅[√3/2 - 0] = 2⋅[√3/2]

2⋅[√3/2] = √3

So √3 is the answer.