Question

What are all the values for `k` for which `int_2^kx^5dx=0`?

Answer 1

See below.

Explanation:

`int_2^kx^5 dx = 1/6(k^6-2^6)`

and

`k^6-2^6=(k^3+2^3)(k^3-2^3)` but

`k^3+2^3 = (k +2)(k^2-2k+2^2)` and

`k^3-2^3 = (k-2)(k^2+2k+2^2)` so

`k^6-2^6=(k +2)(k^2-2k+2^2) (k-2)(k^2+2k+2^2)`

or

`{(k+2=0),(k^2-2k+2^2=0),(k-2=0),(k^2+2k+2^2=0):}`

then finally

real values `k = {-2,2}`
complex values `k = {-1pm i sqrt3,1pm i sqrt3}`

Answer 2

`k = +- 2`

Explanation:

We require:

`int_2^k x^5 \ dx = 0`

Integrating we get:

`[ x^6/6 ]_2^k = 0`

`:. 1/6 [ color(white)(""/"") x^6 ]_2^k = 0`

`:. 1/6(k^6-2^6) = 0`

`:. (k^3)^2-(2^3)^2 = 0`

`:. k^3 = +- 2^3`

`:. \ \k = +- 2`,

Assuming that `k in RR` (there are actually `6` roots, `4` of which are complex)

Now, depending upon the context of the problem, one could argue that `k<2` (ie `k=-2`) is invalid as `k>=2` to make the internal "proper" thus excluding that solution, but without any context it is reasonable to include both solutions.

Also, note that `k=+-2` could be shown to be solutions without actually performing any integration.

Firstly, a property of definite integrals is that:

`int_a^a f(x) = 0`

so we can immediately establish `k=2` is a solution.

Secondly, `x^5` is an odd function, and odd functions satisfy:

`f(-x) = f(x)`

and have rotational symmetry about the origin. as such, if `f(x)` is odd then:

`int_(a)^a \ f(x) = 0`

so we can immediately establish `k=-2` is a solution.

The integration and subsequent calculations do however prove that these are the only solutions!