Unfortunately, this integral is not expressible in terms of elementary functions.
#3
islubio
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Ok my frene gave me the wrong info. int that interms of -1 to 1.
I know the answer is 0 but why?
#4
moxy
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Is the function f(x) = x^5 \sec{x} even, odd or neither? When you have decided that, it should become clear as to why
\int^{1}_{-1}{x^5 \sec{x}} dx = 0
#5
islubio
4
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I feel lost lol. How do i tell if it's even odd or neither?
#6
moxy
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A function f is even if f(x) = f(-x) \forall x \in domain
A function f is odd if -f(x) = f(-x) \forall x \in domain
For your function, check f(x) vs. f(-x) vs. -f(x).
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#7
islubio
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f(-x) = -x^5sec-x
-f(x) = -(x^5secx)
Is this the way?
#8
moxy
40
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Sorry, I initially missed a negative sign in my last post.
Yes, that's a way to check. And you know that sec(x) = 1/cos(x) and that cos(-x) = cos(x). Therefore, sec(-x) = sec(x). That implies that sec is an even function. However, (-x)^5 = -(x^5) implies that x^5 is an odd function. An even function times an odd function is itself an odd function.
One property of an odd function, g(x), is that
\int^{a}_{-a} g(x)dx = 0
#9
moxy
40
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On a related note, for an even function h(x),
\int^{a}_{-a} h(x)dx = 2\int^{a}_{0} h(x)dx
If you graph some simple even and odd functions, you will see why this is the case.