Let the area of the bounded region (x,y):0 9x y^2,y 3x-6 be A.

Question

Let the area of the bounded region ${(x,y):0\le 9x\le {y}^{2},y\ge 3x-6}$ be $A$. Then $6A$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 15

Accepted Answer

From $0\le9x\le y^2$: $x \ge 0$ and $x \le y^2/9$, i.e. $y \ge 3\sqrt{x}$ (taking $y \ge 0$ branch relevant here as $-3\sqrt x$). Combined with $y \ge 3x-6$, integrating with respect to $x$:

$$A = \int_0^1\left[(-3\sqrt{x}) - (3x-6)\right]dx = -3\left[\frac{x^{3/2}}{3/2}\right]_0^1 - \left[\frac{3x^2}{2}-6x\right]_0^1$$

$$= -2 - \left(\frac32-6\right) = -2+\frac92 = \frac52$$

$$6A = 6\times\frac52 = 15$$

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