Question

R is the 4th quadrant region enclosed by the x-axis and the curve y = x^2 - 2kx, where k>0. find the value of k so that the area of the region r is 36 square units.
a. 2
b. 3
c. 4
d. 6
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Answer 1

Observe that

y = x² - 2kx = x (x - 2k)

has roots at x = 0 and x = 2k. Since the leading coefficient is 1 > 0, the parabola opens upward, and in particular this means the x-axis lies above the curve.

We can compute the area of R by integrating:

$\displaystyle \int_0^{2k} |x^2-2kx| \, dx = - \int_0^{2k} (x^2-2kx) \, dx \\\\ = kx^2 -\frac{x^3}3 \bigg|_{0}^{2k} \\\\ = \frac{4k^3}3$

Solve for k :

$\dfrac{4k^3}3 = 36 \implies k^3 = 27 \implies \boxed{k=3} ~~~~ (B)$