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

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The integral of an odd function h(x) defined by h(x) = -h(-x) over the interval from -a to a equals zero. This is due to the symmetry of odd functions, where the area above the x-axis cancels out the area below it. The discussion emphasizes that for any odd function, the integral from -a to a will always yield a result of zero. Examples provided illustrate that while even functions yield non-zero integrals, odd functions consistently result in zero. Thus, the conclusion is that \int_{-a}^{a} h(x) \,dx = 0 for any odd function h(x).
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Ummmm...

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

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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
\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
Suppose H(x)+C=\int h(x) \,dx
\int_{-a}^{0}-h(-x)dx+\int_{0}^{a}h(x)dx=\int_{0}^{-a}h(-x)dx+\int_{0}^{a}h(x)dx
Using u=-x, du=-dx
\int_{0}^{-a}h(-x)dx+\int_{0}^{a}h(x)dx=-\int_{0}^{a}h(u)du+\int_{0}^{a}h(x)dx
=-H(a)+H(0)+H(a)-H(0)=0
 
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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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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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