For what value of k, k > 0, does integral from 0 to k of 6 k x minus 2 k equals k cubed?

Question

For what value of k, k > 0, does ?

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Answer 1

∫_{(from 0 to k)} [6kx - 2k]dx = k³

(3kx² - 2kx)_{(from x = 0 to x = k)} = k³

3k³ - 2k² = k³

2k³ - 2k² = 0

k³ - k² = 0

k²(k - 1) = 0

k² = 0 or k = 1

k = 0 or 1

Since k > 0 by assumption, k = 1.