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Answer 1 · 2018-02-21T22:58:39

#k = 3#

We need to find the value for #k# that satisfies the equation, so we will solve for #k#.

However, first we need to integrate with respect to #x#:

#int_0^k (4kx - 5k) dx = k^2#
#=> 1/2*4kx^2 - 5kx |_0^(x=k) = k^2#
#=> 2kx^2 - 5kx |_0^(x=k) = k^2#

Now substitute in the upper and lower bounds:

#=> 2kk^2 - 5kk - (2k*0^2 - 5k*0) = k^2#
#=> 2k^3 - 5k^2 = k^2#

Subtract #k^2# from both sides:

#=> 2k^3 - 6k^2 = 0#

A solution is in sight, we just need to factor a bit:

#=> 2k^2*(k - 3) = 0#

We can see that for this entire thing to be zero, either #2k^2# must be zero, or #k-3# must be zero.

#2k^2 = 0# implies #k=0# so there is one solution for #k#. However, since we want a non-zero answer:

#=>k - 3 = 0#
#=>k = 3#

Thus the answer is #k=3#.