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 $A$ is given by the in

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$36$

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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 $A$ is given by the integral of $x$ with respect to $y$ from $y = 1$ to $y = 4$:$A = \int_{1}^{4} \sqrt{b} \sqrt{y} \, dy = 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} = \frac{28 \cdot 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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