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An in-depth exploration of odd and even functions, including their definitions, properties, and graphical representations. Students will learn how to identify odd and even functions, evaluate definite integrals, and sketch graphs. Topics covered include the definition of odd and even functions, properties of odd and even functions, odd functions, even functions, and definite integration of odd and even functions.
Typology: Lecture notes
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In this section we will focus on defining known functions as odd or even, and the implications of this, especially when sketching. By the end of this section, you should have the following skills:
We say a function f (x) is odd if f (−x) = −f (x) We say a function f (x) is even if f (−x) = f (x)
h(−x) = f (−x)g(−x) = −f (x)g(x) = −h(x)
Similarly for division of odd by an even function or vice versa. Examples include functions such as x^2 sin(x), x^3 cos(x)
f (g(−x)) = f (−g(x)) = −f (g(x))
So sin(x^3 − x) is odd.
The following functions are even:
h(−x) = f (−x)g(−x) = −(−f (x)g(x)) = h(x)
Also the product or division of two even functions is even. Examples of such even functions include the square f (x)^2 of any func- tion, x sin(x), x^2 cos(x) etc.
Sketch the graph for − 2 ≤ x ≤ 2 Solution
As f (x) is an odd function (it is an odd function of an odd func- tion) we first reflect in the x-axis to get
Reflect in x-axis
Then we reflect in the y-axis to obtain:
Reflect in y-axis
So putting the graphs together, the graph for − 2 ≤ x ≤ 2 is
Graph of sin(x^3 )
(x, f (x)) → (−x, f (x)) = (−x, f (−x))
and (−x, f (−x) is on the graph for negative values. Note we used the fact that f (−x) = f (x) here.
Example 2 Let f (x) = x^4 − 3 x^2 + 1. The graph of f (x) for 0 ≤ x ≤ 2 is
graph of x^4 − 3 x^2 + 1 for positive x
Sketch the graph for − 2 ≤ x ≤ 2 Solution
Proof Make the substitution u = −x ⇒ −du = dx in
I =
−a
f (x)dx
to get
a
f (−u)du
∫ (^) a
0
f (u)du
as f (−u) = f (u) Hence ∫ (^) a
−a
f (x)dx =
−a
f (x)dx +
∫ (^) a
0
f (x)dx
∫ (^) a
0
f (x)dx
27.5.2 Odd Functions
Let a ≥ 0. If f (x) is odd we have ∫ (^0)
−a
f (x)dx = −
∫ (^) a
0
f (x)dx ⇒ ∫ (^) a
−a
f (x)dx = 0
Proof Make the substitution u = −x ⇒ −du = dx in
−a
f (x)dx
to get
a
f (−u)du
∫ (^) a
0
f (u)du
as f (−u) = −f (u) Hence ∫ (^) a
−a
f (x)dx =
−a
f (x)dx +
∫ (^) a
0
f (x)dx
= 0
27.5.3 Examples
Example 3 Evaluate the following definite integrals
(a)
I =
− 1
x^4 sin(x)dx
(b)
I =
− 2
(x^2 sin(3x) + x^5 + 3x^2 )dx
(c)
I =
− 1
arctan(x^3 + x) cos(x) x^4 + x^2 + 1
dx
Solution
(a) x^4 sin(x) is odd as it is the product of an odd and an even function. Hence I = 0
(b) x^2 sin(3x)+x^5 is an odd function as x^2 sin(3x) is odd as is x^5. Hence
− 2
3 x^2 dx = 2
0
3 x^2 dx = 2[x^3 ]^20 = 16
as 3 x^2 is an even function.
(c) arctan(x^3 + x) cos(x) x^4 + x^2 + 1
(c) arcsin(tan(x)) sin(x) is an even function, x^3 + x is an odd function, hence arcsin(tan(x)) sin(x) x^3 + x is an odd function. It follows that
− 2
arcsin(tan(x)) sin(x) x^3 + x
− 2
3 x^2 + 5 dx
= [x^3 + 5x]^2 − 2 = 36