Question

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

Hint:

When dealing with regions defined by inequalities, solve for the boundaries and use definite integrals to compute the area between those boundaries. Remember to evaluate carefully.

Answer 1

Correct Answer: 15

Given \( 0 \leq x \leq y^2 \) and \( y \geq 3x - 6 \) 

Required Area \( A \) is given by:

\( A = \int_{0}^{1} \left[ (-3\sqrt{x}) \, dx - \int_{0}^{1} (3x - 6) \, dx \right] \)

Now, solving for the area:

\( A = 3 \left( \frac{3}{2} x^{3/2} \right) \Big|_{0}^{1} - \left( \frac{3x^2}{2} - 6x \right) \Big|_{0}^{1} \)

\( A = -2 \left[ 1 - 0 \right] \left[ \frac{3}{2} - 6 \right] \)

Thus, the area becomes:

\( A = -2 \times [1] \times \left( \frac{3}{2} - 6 \right) = \frac{5}{2} \, \text{square units} \)

Finally, multiplying by 6:

\( 6A = 6 \times \frac{5}{2} = 15 \, \text{square units} \)

Answer 2

Given the inequalities: \( 0 \leq x \leq y^2 \) and \( y \geq 3x - 6 \), we need to find the required area \( A \). 

The area \( A \) is expressed as the difference of two integrals:

\( A = \int_{0}^{1} \left[ (-3\sqrt{x}) \, dx \right] - \int_{0}^{1} \left[ (3x - 6) \, dx \right] \)

Now, solving each integral separately:

For the first integral:

\( A_1 = 3 \left( \frac{3}{2} x^{3/2} \right) \Big|_{0}^{1} \)

For the second integral:

\( A_2 = \left( \frac{3x^2}{2} - 6x \right) \Big|_{0}^{1} \)

Evaluating the integrals at the limits:

\( A_1 = 3 \times \left( \frac{3}{2} \times 1^{3/2} - \frac{3}{2} \times 0^{3/2} \right) = 3 \times \frac{3}{2} = \frac{9}{2} \)

\( A_2 = \left( \frac{3 \times 1^2}{2} - 6 \times 1 \right) - \left( \frac{3 \times 0^2}{2} - 6 \times 0 \right) = \frac{3}{2} - 6 = -\frac{9}{2} \)

Now, combining both parts:

\( A = A_1 - A_2 = \frac{9}{2} - \left( -\frac{9}{2} \right) = \frac{9}{2} + \frac{9}{2} = 9 \, \text{square units} \)

Thus, the total area is:

\( 6A = 6 \times \frac{5}{2} = 15 \, \text{square units} \)