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

The Correct Option is D

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:

\[ 10A = 162. \]