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

  • Thread starter tandoorichicken
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In summary, the integral \int_{-a}^{a} h(x) \,dx is equal to 0 because h(x) is an odd function, meaning it is symmetric about the origin and the areas above and below the x-axis will cancel each other out. This is represented by the property h(x) = -h(-x) and can be seen by drawing a picture or using a specific example, such as h(x) = x^3.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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]
 
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  • #5
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.
 
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  • #6
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
 
  • #7
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
 

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

What does the equation h(x)=-h(-x) mean?

The equation h(x)=-h(-x) means that the function h is an odd function, where the output values for negative inputs are the negative of the output values for positive inputs. This can also be written as h(-x)=-h(x).

How do you solve h(x)=-h(-x)?

To solve h(x)=-h(-x), you can use the property that states the integral of an odd function from -a to a is equal to zero. This means that the area under the curve for positive x values will be equal to the negative of the area under the curve for negative x values.

What is the significance of solving h(x)=-h(-x)?

Solving h(x)=-h(-x) can be useful in finding the area under a curve or the average value of a function over a symmetric interval. It can also help in simplifying integrals and solving differential equations.

Can h(x)=-h(-x) be solved for any value of a?

Yes, h(x)=-h(-x) can be solved for any value of a since it is a general property of odd functions. However, the specific method of solving may vary depending on the function or interval given.

What other applications does solving h(x)=-h(-x) have?

In addition to finding the area and average value of a function, solving h(x)=-h(-x) can be used in physics and engineering to analyze systems with symmetric properties, such as electric fields or mechanical systems. It can also be applied in signal processing and image analysis.

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