Answer to Question #239400 in Calculus for sam

Question

True or false:

If the function  f is  odd, then $\displaystyle{\int_{-1}^{1}f(x)dx=0}$

Answer 1

Function x , is equal to minus function (-x) i.e. f(x) = -f(-x) for all x values in the domain like x€D_f

D_f = domain for the function f

Figure shows that it is an odd function because a straight line passing through the origin which also called function x i.e f(x)

Now consider odd function f(x)= x

These function is an odd function because the power of x is equal to 1 so it is called an odd function

Now we have to prove the odd function over the limit -1 to 1 is equal to 0

In the figure , function f(x) = x,

function f(1)=1, and then f(-1) = -1

So it is full fill the definition f(1)=-f(-1)

$\int_{-1}^{1}f(x)dx=0$

$\int_{-1}^{0}xdx + \int_{0}^{1}xdx=0$

$\int_{-1}^{1}xdx=0$

First integrate the function (x)dx over the limit area from -1 to 0 getting the same value but negative sign

When integrate the function (x)dx over the limit area from 0 to -1 getting the same area, same value but the opposite sign, so on integrating the f(x) on whole limit area from -1 to 1 getting zero (0) value because cancel on each other.

$\int_{-1}^{1}xdx=0$

Hence it is proved