The area of the region bounded by the curve $y=x^3$, and the lines, $y=8$, and $x=0$, is

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Aicharya · Accepted Answer

Required Area $=\int_{y=0}^8 y^{1 / 3} d y$

$ \begin{aligned} & =\left.\frac{y^{1 / 3+1}}{\frac{1}{3}+1}\right|_0 ^8=\left.\frac{3}{4}\left(y^{4 / 3}\right)\right|_0 ^8 \\ & =\frac{3}{4}\left[(8)^{4 / 3}-0\right]=\frac{3}{4}\left[2^4\right] \\ & =\frac{3}{4} \times 16=12 \text { sq. unit. } \end{aligned} $

The area of the region bounded by the curve $y=x^3$, and the lines, $y=8$, and $x=0$, is

Solution