- #1
flyers
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Homework Statement
The Attempt at a Solution
I have no idea how to start this integral. Can anybody give me a hint to start it off?
Thank you.
Why Is the Definite Integral Zero for Symmetric Functions?
- Thread starter flyers
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Homework Help Overview
The discussion centers around the evaluation of a definite integral for symmetric functions, particularly focusing on the properties of integrals over symmetric intervals and the implications of function symmetry.
Discussion Character
- Exploratory, Conceptual clarification, Assumption checking
Approaches and Questions Raised
- The original poster expresses uncertainty about how to begin the integral. Some participants suggest sketching a graph to explore the relationship between the function values for negative and positive inputs. Others observe that the function values for positive and negative inputs are reflections of each other, raising questions about the integral's value and the possibility of solving it.
Discussion Status
The discussion is ongoing, with participants exploring different aspects of the problem. Some guidance has been offered regarding the properties of definite integrals and symmetry, but there is no explicit consensus on the approach or final outcome.
Contextual Notes
Participants are considering the implications of function symmetry and the nature of the integral over a symmetric interval. There is an acknowledgment that finding an indefinite integral may not be the right approach for this problem.
I have no idea how to start this integral. Can anybody give me a hint to start it off?
Thank you.
- #2
Dick
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Sketch a graph. Anything strike you about the relation between the part of the graph for x<0 vs x>0?
- #3
flyers
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Well I noticed that x>0 is the reflection of x<0 except with negative values. Does this mean that the answer is 0? Also is it even possible to solve this integral?
- #4
Dick
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No, you aren't going to get far trying to find an indefinite integral. But the definite integral is zero. In general, if f(-x)=(-f(x)) then the integral over a symmetric interval around zero is zero. You can formally show this by splitting the integral into two parts and working out their relation by doing the substitution u=(-x).
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