# Question 6bd6c

May 12, 2016

0

#### Explanation:

$f \left(x\right) = {x}^{3} - x$ is an odd function. It verifies $f \left(x\right) = - f \left(- x\right)$
so ${\int}_{-} {1}^{1} f \left(x\right) \mathrm{dx} = {\int}_{-} {1}^{0} f \left(x\right) \mathrm{dx} + {\int}_{0}^{1} f \left(x\right) \mathrm{dx} = {\int}_{0}^{1} f \left(- x\right) \mathrm{dx} + {\int}_{0}^{1} f \left(x\right) \mathrm{dx} = {\int}_{0}^{1} \left(f \left(x\right) + f \left(- x\right)\right) \mathrm{dx} = 0$

May 13, 2016

${\int}_{-} {1}^{1} \left({x}^{3} - x\right) \mathrm{dx} = 0$

It could be the area, but the function does not maintain a constant sign between $x \in \left[- 1 , 1\right]$. Also, because of symmetry in $x = 0$ which cuts by half this interval, the areas cancel out each other and nulify the area.

#### Explanation:

Geometrically, the integral of a function of only one variable equals to an area. However, the geometry suggests the that smaller valued function is substracted from the bigger valued function in order for the area to not be negative. More specifically, for two functions $f \left(x\right)$ and $g \left(x\right)$ the area between the two graphs in $\left[a , b\right]$ is:

${\int}_{a}^{b} | f \left(x\right) - g \left(x\right) | \mathrm{dx}$

That is, one must know which one of the following cases actually hold true:

$f \left(x\right) > g \left(x\right)$

$f \left(x\right) < g \left(x\right)$

Now considering your function, the find the sign of the difference between these functions:

${x}^{3} - x = 0$

$x \left({x}^{2} - 1\right) = 0$

$x \left(x - 1\right) \left(x + 1\right) = 0$

We see that for the given area of $\left[- 1 , 1\right]$ that the exercise gives you, the sign actually changes from positive to negative at $x = 0$. Therefore, geometrically this definite integral does NOT represent the area. The actual area is:

$A = {\int}_{-} {1}^{0} \left({x}^{3} - x\right) \mathrm{dx} - {\int}_{0}^{1} \left({x}^{3} - x\right) \mathrm{dx}$

Since the area from 0 to 1 would be negative, we just add a minus sign so it adds up. If you solve the integrals:

$A = {\left[{x}^{4} / 4 - {x}^{2} / 2\right]}_{-} {1}^{0} - {\left[{x}^{4} / 4 - {x}^{2} / 2\right]}_{0}^{1}$

$A = \frac{1}{4} - \left(- \frac{1}{4}\right)$

Α=2/4#

Notice that the two integrals yield the same value? That's because of the symmetry of the function, which causes your integral to be negative.

To sum up:

${\int}_{-} {1}^{1} \left({x}^{3} - x\right) \mathrm{dx} = {\left[{x}^{4} / 4 - {x}^{2} / 2\right]}_{-} {1}^{1} = \frac{1}{4} - \frac{1}{4} = 0$
$A = {\int}_{-} {1}^{0} \left({x}^{3} - x\right) \mathrm{dx} - {\int}_{0}^{1} \left({x}^{3} - x\right) \mathrm{dx} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4}$
Therefore, it may remind of area, but the integral you are given does NOT represent area (you could know this from the beginning, since an area can't be 0). The only geometrical result that could be obtained would be the symmetry of the function. For axis of symmetry $x = 0$ the symmetrical values of $x$ $- 1$ and $+ 1$ yield equal areas, so the function is most likely symmetrical. Graphing the two functions in the same sheet, you can see is actually is symmetrical: