If the area bounded by the curve x^2 = by and | vedclass

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(A) The equation of the curve is $x^2 = by$, which implies $x = \sqrt{b} \sqrt{y}$ (since it is in the first quadrant).The area $A$ bounded by the cur

Vedclass · Accepted Answer

$36$

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(A) The equation of the curve is $x^2 = by$, which implies $x = \sqrt{b} \sqrt{y}$ (since it is in the first quadrant).The area $A$ bounded by the curve and the lines $y = 1$ and $y = 4$ is given by the integral:$A = \int_{1}^{4} x \, dy = \int_{1}^{4} \sqrt{b} \sqrt{y} \, dy$Given $A = 28$, we have:$28 = \sqrt{b} \int_{1}^{4} y^{1/2} \, dy$$28 = \sqrt{b} \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4}$$28 = \sqrt{b} \cdot \frac{2}{3} [4^{3/2} - 1^{3/2}]$$28 = \sqrt{b} \cdot \frac{2}{3} [8 - 1]$$28 = \sqrt{b} \cdot \frac{2}{3} \cdot 7$$28 = \sqrt{b} \cdot \frac{14}{3}$$\sqrt{b} = 28 \cdot \frac{3}{14} = 2 \cdot 3 = 6$$b = 6^2 = 36$.

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...

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