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

[tandoorichicken]

Ummmm...

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

?

 

[StephenPrivitera]

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.

 

[ShawnD]

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.

 

[StephenPrivitera]

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]

 

[ShawnD]

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.

 

[StephenPrivitera]

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

 

[himanshu121]

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