Posted by Rory on Sunday, December 4, 2011 at 8:45pm.
Let f(t) be an odd function
Using properties of the deﬁnite integral plus simple
substitution show that if f is continuous on [−a, a] for a positive number a, then
Z a
−a
f(t) dt = 0

Calculus  MathMate, Sunday, December 4, 2011 at 9:17pm
The definition of an odd function is
f(x)=f(x).
If f(x) is continuous on the interval [a,a], then
I=∫f(t)dt from a to a
= I1 + I2
where
I2 = ∫f(t)dt from 0 to a
and
I1 = ∫f(t)dt from a to 0
Using property of odd function, f(x)=f(x), and substitute u=t,dt=du
we get
I1 = ∫f(u)(du) from a to 0
=∫f(u)du from 0 to a
=I2
Thus
I=I1+I2=I2+I2=0
Incidentally, as exercise, you can use the above fact to solve:
∫[x^3 + x^4(tan^3(x))]dx from π/4 to π/4.
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