Solving h(x)=-h(-x): Integral from -a to a

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Homework Help Overview

The discussion revolves around the integral of a function \( h(x) \) that satisfies the property \( h(x) = -h(-x) \), specifically evaluating \( \int_{-a}^{a} h(x) \,dx \). The subject area includes properties of odd functions and definite integrals.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the property \( h(x) = -h(-x) \) and its geometric interpretation. Some suggest visualizing the function to understand the integral's behavior. Others question the generality of the problem, considering whether a specific answer is sought or if a broader understanding is needed.

Discussion Status

The discussion is active, with various interpretations being explored. Some participants have provided reasoning related to the integral's value being zero due to the symmetry of odd functions, while others express skepticism about the generality of the problem and the adequacy of the examples provided.

Contextual Notes

There are mentions of specific functions and their properties, such as even and odd functions, which may influence the evaluation of the integral. Participants also reference the behavior of definite integrals concerning areas above and below the x-axis.

tandoorichicken
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Ummmm...

If h(x)= -h(-x) for all x, what is [tex]\int_{-a}^{a} h(x) \,dx[/tex]

?
 
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Draw a picture! It should be clear. What does h(x)=-h(-x) mean? What does a definite integral represent if h(x) is positive?

Think of h(x)=x3 for instance. This should make the answer clear.
 
It means it's flipped around the origin.

That question must want a really really generic answer because if it wanted anything specific, you could just make up an answer very easily.
 
Originally posted by ShawnD

That question must want a really really generic answer because if it wanted anything specific, you could just make up an answer very easily.
There's only one possible answer.
0
[tex]\int_{-a}^{a} h(x) \,dx =\int_{-a}^{0} h(x)dx + \int_{0}^{a}h(x)dx=\int_{-a}^{0}-h(-x)dx+\int_{0}^{a}h(x)dx[/tex]
Suppose [tex]H(x)+C=\int h(x) \,dx[/tex]
[tex]\int_{-a}^{0}-h(-x)dx+\int_{0}^{a}h(x)dx=\int_{0}^{-a}h(-x)dx+\int_{0}^{a}h(x)dx[/tex]
Using u=-x, du=-dx
[tex]\int_{0}^{-a}h(-x)dx+\int_{0}^{a}h(x)dx=-\int_{0}^{a}h(u)du+\int_{0}^{a}h(x)dx[/tex]
[tex]=-H(a)+H(0)+H(a)-H(0)=0[/tex]
 
Last edited:
What you just wrote is true with EVERY function that has a real domain and is in the form a^b where b is an integer and NOT a variable. Let's look at the integration of X^2 between -1 and 1 then between -1 and 0 then 0 and 1.
from -1 to 1 = 2/3
from -1 to 0 and 0 to 1 = 1/3 + 1/3 = 2/3

Lets try another equation, this time something like (5 - 2x)^4 - 10x +7 between -10 and 10
from -10 to 10 = 1052640
from -10 to 0 and 0 to 10 = 976820 + 75820 = 1052640

That's just way too generic to be the answer he's looking for.
 
Last edited:
f(x)=x^2 does not satisfy the hypothesis f(x)=-f(-x)
f(x)=x^2 is EVEN
f(x)=-f(-x) means f is ODD
 
Definte Integrals includes + as well as - signs
area below the x-axis is -
area above the x-axis is +

Now the function is odd therefore it issymmetric with I & III quadrant

Hence if one area is positve the other will be negative.
result the integral will be zero under the limits i repeat under the limits -a to +a
 

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