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You may use the following properties of functions and definite integrals, such that:
`int_(-a)^a f(x)dx = 0` if f(x) is odd function
`int_(-a)^a f(x)dx = 2int_0^a f(x)dx` if f(x) even function
Hence, you need to test if the function is odd or even, or a combination of both.
You need to use the definitions of odd and even functions, such that:
`- if f(x) = f(-x) => f(x)` is even
`- if f(-x) = -f(x) => f(x)` is odd
Reasoning by analogy, you need to replace -x for x in equation of the function, such that:
`f(-x) = (-x)^(2013)*e^(-(-x)^2) => f(-x) = -x^2013*e^(-x^2)`
Since `f(x) = x^2013*e^(-x^2)` yields that `f(-x) = -f(x)` , hence the given function is odd.
Since the function is odd, then you do not need to evaluate the definite integral, anymore because, by definition, `int_(-1)^1 x^2013*e^(-x^2)dx = 0` .
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