Given that k k\,k is a positive constant and ∫1k(32x+9)dx=8\displaystyle \int_1^k \left( \frac{3}{2\sqrt{x}} + 9 \right) dx = 8∫1k(2x3+9)dx=8
Show that 9k+3k−20=09k + 3\sqrt{k} - 20 = 09k+3k−20=0
Hence, using algebra, find any values of k k\,k such that ∫1k(32x+9)dx=8\displaystyle \int_1^k \left( \frac{3}{2\sqrt{x}} + 9 \right) dx = 8∫1k(2x3+9)dx=8
