The integration of #x^4-1# is simple. You can integrate it piece by piece, so #intx^4dx = x^(4+1)/(4+1) = x^5/5 + C# and #int-1dx = -x + C#. We will drop
#C# as it is unneeded for finding #a#.
So what we have is #x^5/5-x# where the upper-bound is #a# and the lower-bound is #-a#. If we plug in #a# and #-a# into our function as per with every definite integral, we will have:
#a^5/5-a-(-a^5/5+a) = 2a^5/5-2a#.
Realize that our goal is to have that function equate to zero, so rewriting it, we will get:
#2a^5/5-2a = 0#.
So, adding #2a# to both sides, we get #2a^5/5 = 2a#. Then we multiply by #5# and we get #2a^5 = 10a#. Then dividing by #2#, we get #a^5 = 5a# then continuing we finally get #a^4 = 5#.
Then we fourth-root both sides to get #a = +-root4(5)#.
