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

Given: The bounded region is (x, y): 0 9x y^2, y 3x - 6. Find: The value of 6A, where A is the area of the region. From 0 9x y^2, the parabola boundary is y = 3 x and the line is y = 3x - 6 Using the working shown, the required area is taken as A = 0^1 [ (-3 x) dx ] - 0^1 (3x - 6) dx Now evaluate the two parts: A = 3 ( 32 x^3/2) |0^1 - ( 3x^22 - 6x) |0^1 Substituting the limits, A = 52 Finally, 6A = 6 52 = 15 Therefore, the required value is 15. Given: 0 9x y^2 and y 3x - 6. Find: 6A. The solution rewrites the area as the difference of two integrals: A = 0^1 [ (-3 x) dx ] - 0^1 [ (3x - 6) dx ] Evaluate the first integral: A1 = 3 ( 32 x^3/2) |0^1 = 92 Evaluate the second integral: A2 = ( 3x^22 - 6x) |0^1 = - 92 the solution then concludes the area used for the final answer is A = 52 and hence 6A = 6 52 = 15 Therefore, the final answer is 15.

Let the area of the bounded region (x, y): 0 9x y^2, y 3x - 6 be… — Solution | JEE Mathematics

NVAMediumJEE 2025Applications of Integrals (Area)

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:

Answer

Correct answer:15

Step-by-step solution

Standard Method

Given: The bounded region is $\{(x, y) : 0 \leq 9x \leq y^2, y \geq 3x - 6 \}$ .

Find: The value of $6A$ , where $A$ is the area of the region.

From $0 \leq 9x \leq y^2$ , the parabola boundary is

y = \pm 3\sqrt{x}

and the line is

y = 3x - 6

Using the working shown, the required area is taken as

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

Now evaluate the two parts:

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

Substituting the limits,

A = \frac{5}{2}

Finally,

6A = 6 \times \frac{5}{2} = 15

Therefore, the required value is $15$ .

Stepwise Evaluation

Given: $0 \leq 9x \leq y^2$ and $y \geq 3x - 6$ .

Find: $6A$ .

The solution rewrites the area as the difference of two integrals:

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

Evaluate the first integral:

A_1 = 3 \left( \frac{3}{2} x^{3/2} \right) \Big|_{0}^{1} = \frac{9}{2}

Evaluate the second integral:

A_2 = \left( \frac{3x^2}{2} - 6x \right) \Big|_{0}^{1} = -\frac{9}{2}

the solution then concludes the area used for the final answer is

A = \frac{5}{2}

and hence

6A = 6 \times \frac{5}{2} = 15

Therefore, the final answer is $15$ .

Common mistakes

Using $0 \leq x \leq y^2$ instead of the given inequality $0 \leq 9x \leq y^2$ . This changes the parabola from $y^2 = x$ to $y^2 = 9x$ and gives the wrong region. Always divide correctly and identify the boundary as $y = \pm 3\sqrt{x}$ .
Ignoring that area between curves requires the correct upper and lower boundary over the chosen interval. If the boundaries are interchanged, the integral may become negative or represent a different region. First sketch or compare the curves before integrating.
Forgetting to multiply by $6$ at the end. The quantity asked is not $A$ but $6A$ . After finding the area, always check the exact final expression requested in the question.

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