Q94 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
If the area bounded by the curve $x^2 = by$ and the lines $y = 1, y = 4$ in the first quadrant is $28$ sq. units, then the value of b is...
- Correct answerA$36$
- B$6$
- C$9$
- D$3$
Solution
The equation of the curve is $x^2 = by$. In the first quadrant, $x = \sqrt{b}\sqrt{y}$.
The area bounded by the curve and the lines $y = 1$ and $y = 4$ in the first quadrant is given by:
$\int_{1}^{4} x \, dy = 28$
$\int_{1}^{4} \sqrt{b}\sqrt{y} \, dy = 28$
$\sqrt{b} \left[ \dfrac{2}{3} y^{3/2} \right]_{1}^{4} = 28$
$\dfrac{2}{3} \sqrt{b} (4^{3/2} - 1^{3/2}) = 28$
$\dfrac{2}{3} \sqrt{b} (8 - 1) = 28$
$\dfrac{14}{3} \sqrt{b} = 28$
$\sqrt{b} = 28 \times \dfrac{3}{14}$
$\sqrt{b} = 6$
$b = 36$
Answer: $36$
The area bounded by the curve and the lines $y = 1$ and $y = 4$ in the first quadrant is given by:
$\int_{1}^{4} x \, dy = 28$
$\int_{1}^{4} \sqrt{b}\sqrt{y} \, dy = 28$
$\sqrt{b} \left[ \dfrac{2}{3} y^{3/2} \right]_{1}^{4} = 28$
$\dfrac{2}{3} \sqrt{b} (4^{3/2} - 1^{3/2}) = 28$
$\dfrac{2}{3} \sqrt{b} (8 - 1) = 28$
$\dfrac{14}{3} \sqrt{b} = 28$
$\sqrt{b} = 28 \times \dfrac{3}{14}$
$\sqrt{b} = 6$
$b = 36$
Answer: $36$
Hence, the correct answer is option: 'A'.