Question

Let the area of the region enclosed by the curves $y = 3x$, $2y = 27 - 3x$, and $y = 3x - x\sqrt{x}$ be $A$. Then $10A$ is equal to:

• A. 184
• B. 154
• C. 172
• D. 162

Answer 1

Answer: (D)

162

Answer 2

The curves are $y = 3x$, $2y = 27 - 3x$, and $y = 3x - x\sqrt{x}$. The area $A$ is divided as:

$A = \int_0^3 \left(3x - (3x - x\sqrt{x})\right) dx + \int_3^9 \left(\frac{27 - 3x}{2} - (3x - x\sqrt{x})\right) dx.$

First Integral:

$\int_0^3 \left(3x - 3x + x\sqrt{x}\right) dx = \int_0^3 x\sqrt{x} dx = \int_0^3 x^{3/2} dx = \left[\frac{2x^{5/2}}{5}\right]_0^3.$ $= \frac{2}{5} \left(3^{5/2} - 0^{5/2}\right) = \frac{2}{5} \times 3^{5/2}.$

Second Integral:

$\int_3^9 \left(\frac{27}{2} - \frac{3x}{2} - 3x + x\sqrt{x}\right) dx = \int_3^9 \left(\frac{27}{2} - \frac{9x}{2} + x\sqrt{x}\right) dx.$

Evaluate term by term:

$\int_3^9 \frac{27}{2} dx = \frac{27}{2} \times (9 - 3) = 81.$ $\int_3^9 \frac{9x}{2} dx = \frac{9}{2} \int_3^9 x dx = \frac{9}{2} \times \left[\frac{x^2}{2}\right]_3^9 = \frac{9}{2} \times \frac{81 - 9}{2} = \frac{9}{2} \times 36 = 162.$ $\int_3^9 x\sqrt{x} dx = \int_3^9 x^{3/2} dx = \left[\frac{2x^{5/2}}{5}\right]_3^9 = \frac{2}{5} \left(9^{5/2} - 3^{5/2}\right).$

Combine results:

$A = \frac{2}{5} \times 3^{5/2} + 81 - 162 + \frac{2}{5} \times \left(9^{5/2} - 3^{5/2}\right).$ $A = \frac{2}{5} \times 9^{5/2} - \frac{2}{5} \times 3^{5/2} + 81 - 162.$ $A = \frac{2}{5} \times 9^{5/2} + 81 - 162.$ $A = \frac{486}{5} - 81 = \frac{81}{5}.$

Final result:

$10A = 162.$