# For what non zero value of a does the definite integral int_-a^ax^4-1dx have a value equal to 0? Help!?

Aug 23, 2017

See below.

#### Explanation:

Integrating and equating to zero

${\int}_{- a}^{a} \left({x}^{4} - 1\right) \mathrm{dx} = {\left(\frac{1}{5} {x}^{5} - x\right)}_{- a}^{a} = 2 \left({a}^{5} / 5 - a\right) = 0$ or

$a \left({a}^{4} / 5 - 1\right) = 0$ giving

$a = \left\{\begin{matrix}0 \\ - \sqrt[4]{5} \\ \sqrt[4]{5}\end{matrix}\right. \Rightarrow a = \pm \sqrt[4]{5}$

Aug 23, 2017

If you have no idea what an imaginary number is, then the answers you are looking for are $a = \pm {5}^{\frac{1}{4}}$

Otherwise, upon finding the other two solutions we have
$a = {5}^{\frac{1}{4}} {e}^{\frac{i \pi k}{2}} , k = 0 , 1 , 2 , 3$

ie $a = {5}^{\frac{1}{4}} , - {5}^{\frac{1}{4}} , {5}^{\frac{1}{4}} i , - {5}^{\frac{1}{4}} i$

#### Explanation:

${\int}_{-} {a}^{a} {x}^{4} - 1 \mathrm{dx} = {\left[\frac{1}{5} {x}^{5} - x\right]}_{-} {a}^{a}$

$= \frac{2}{5} {a}^{5} - 2 a$

To find the values of a that make this integral zero we set this equal to zero, ie

$\frac{2}{5} {a}^{5} = 2 a$

$\implies a \left({a}^{4} - 5\right) = 0$

We are looking for non-zero solutions so

$\therefore {a}^{4} = 5$

How you deal with the remainder of this question depends on if you have met complex numbers yet. If not, take the two real roots and you are done.

If you have heard of them, or are interested then read on - it's a pretty cool topic! So how come we have a polynomial of degree 4 but there's only two solutions? That doesn't seem quite right!

And it isn't - there are two entirely real solutions. The other two lie somewhere in the complex plane (in this case it turns out they are entirely imaginary).

We are going to make use something known as Roots of Unity (fancy name eh?).
Convert to polar form via $a = r {e}^{i \theta}$ to obtain

${r}^{4} {e}^{4 i \theta} = 5$

To end up with the right magnitude we need to have $r = {5}^{\frac{1}{4}}$ hence we have

${e}^{4 i \theta} = 1 = {e}^{2 \pi i k} \forall k \in \mathbb{Z}$

Take natural logs of both sides, giving

$4 i \theta = 2 \pi i k$

$\implies \theta = \frac{\pi}{2} k$

Notice that $k = 0 , 1 , 2 , 3$ give unique solutions however $k = 4$ gives $\theta = 2 \pi$ which, by the periodicity of the complex plane is the same as $\theta = 0$. Hence we have obtained four roots, as we would expect from a polynomial of degree 4. These are:

$a = {5}^{\frac{1}{4}} {e}^{\frac{i \pi k}{2}} , k = 0 , 1 , 2 , 3$