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

2 Answers
Aug 17, 2017

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}

Aug 17, 2017

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!