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September 30, 2014

September 30, 2014

Posted by **Rory** on Sunday, December 4, 2011 at 8:45pm.

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 9:17pmThe 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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