The area inside the parabola y^2 = 4ax,betwee | vedclass

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(C) The area $A$ bounded by the parabola $y^2 = 4ax$ and the lines $x = a$ and $x = 4a$ is given by the integral:$A = 2 \int_{a}^{4a} y \, dx$Since $y

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$\frac{56a^2}{3}$ sq. units

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(C) The area $A$ bounded by the parabola $y^2 = 4ax$ and the lines $x = a$ and $x = 4a$ is given by the integral:$A = 2 \int_{a}^{4a} y \, dx$Since $y^2 = 4ax$,we have $y = 2\sqrt{a}\sqrt{x}$ (considering the area above the $x$-axis and multiplying by $2$ for symmetry).$A = 2 \int_{a}^{4a} 2\sqrt{a} \sqrt{x} \, dx = 4\sqrt{a} \int_{a}^{4a} x^{1/2} \, dx$$A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{a}^{4a} = 4\sqrt{a} \cdot \frac{2}{3} \left[ x^{3/2} \right]_{a}^{4a}$$A = \frac{8\sqrt{a}}{3} \left[ (4a)^{3/2} - a^{3/2} \right]$$A = \frac{8\sqrt{a}}{3} \left[ 8a\sqrt{a} - a\sqrt{a} \right]$$A = \frac{8\sqrt{a}}{3} \left[ 7a\sqrt{a} \right] = \frac{56a^2}{3}$ sq. units.

The area inside the parabola $y^2 = 4ax$,between the lines $x = a$ and $x = 4a$ is equal to

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