The area of the region (in sq. units) enclose | vedclass

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(C) The curve $y = 8x^3 - 1$ intersects the $x$-axis $(y=0)$ at $8x^3 - 1 = 0$,which gives $x^3 = \frac{1}{8}$,so $x = \frac{1}{2}$.For $x \in [-1, \f

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$\frac{19}{4}$

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(C) The curve $y = 8x^3 - 1$ intersects the $x$-axis $(y=0)$ at $8x^3 - 1 = 0$,which gives $x^3 = \frac{1}{8}$,so $x = \frac{1}{2}$.For $x \in [-1, \frac{1}{2}]$,$y \le 0$,so the area is $\int_{-1}^{\frac{1}{2}} -(8x^3 - 1) dx = \int_{-1}^{\frac{1}{2}} (1 - 8x^3) dx$.For $x \in [\frac{1}{2}, 1]$,$y \ge 0$,so the area is $\int_{\frac{1}{2}}^{1} (8x^3 - 1) dx$.Total Area = $\int_{-1}^{\frac{1}{2}} (1 - 8x^3) dx + \int_{\frac{1}{2}}^{1} (8x^3 - 1) dx$.$= [x - 2x^4]_{-1}^{\frac{1}{2}} + [2x^4 - x]_{\frac{1}{2}}^{1}$.$= ((\frac{1}{2} - 2(\frac{1}{16})) - (-1 - 2(1))) + ((2(1) - 1) - (2(\frac{1}{16}) - \frac{1}{2}))$.$= ((\frac{1}{2} - \frac{1}{8}) - (-3)) + (1 - (\frac{1}{8} - \frac{1}{2}))$.$= (\frac{3}{8} + 3) + (1 - (-\frac{3}{8}))$.$= \frac{27}{8} + \frac{11}{8} = \frac{38}{8} = \frac{19}{4}$ sq. units.

The area of the region (in sq. units) enclosed by the curves $y=8x^3-1$,$y=0$,$x=-1$ and $x=1$ is

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