The area of the region (in sq. units), in the first quadrant bounded by the parabola $y=9 x^2$ and the lines $x=0, y=1$ and $y=4$, is :

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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { Required area }=\int_{y=1}^4 \sqrt{\frac{y}{9}} d y \ & =\frac{1}{3} \int_{y=1}^4 y^{1 / 2} d y=\frac{1}{3} 	imes\left.\frac{2}{3}\left(y^{3 / 2}ight)ight|_1 ^4 \ & =\frac{2}{9}\left[\left(4^{1 / 2}ight)^3-\left(1^{1 / 2}ight)^3ight]=\frac{2}{9}[8-1] \end{aligned} $ $ =\frac{2}{9} 	imes 7=\frac{14}{9} 	ext { sq. units. } $

The area of the region (in sq. units), in the first quadrant bounded by the parabola $y=9 x^2$ and the lines…

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