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
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.
To prove that the definite integral of an odd function over a symmetric interval is equal to zero, we can use the properties of the definite integral and perform a simple substitution.
Let's start by considering the integral:
∫[−a,a] f(t) dt
Since f(t) is an odd function, we can observe the following property:
f(-t) = -f(t)
Now, let's perform a substitution by letting u = -t. As a result, both the limits of integration and the differential element change accordingly:
When t = -a, u = a
When t = a, u = -a
And dt = -du
The integral can now be rewritten as:
∫[−a,a] f(t) dt = -∫[a,-a] f(-u) (-du)
Using the property f(-t) = -f(t), we can simplify further:
∫[−a,a] f(t) dt = -∫[a,-a] -f(u) (-du)
∫[−a,a] f(t) dt = ∫[a,-a] f(u) du
Notice that the limits of integration are now switched, but since the integral is a definite integral, the value remains the same:
∫[−a,a] f(t) dt = ∫[a,-a] f(u) du
Since the limits of integration are symmetric, we can rewrite the integral as:
∫[a,-a] f(u) du = -∫[-a,a] f(u) du
Combining the two expressions obtained:
∫[−a,a] f(t) dt = ∫[a,-a] f(u) du = -∫[-a,a] f(u) du
From the properties of the definite integral, we know that changing the limits of integration does not affect the value of the integral:
∫[−a,a] f(t) dt = -∫[-a,a] f(u) du
Therefore, the definite integral of an odd function over a symmetric interval is equal to zero:
∫[−a,a] f(t) dt = 0