The area of the region bounded by the curve y | vedclass

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(B) The region is bounded by $y = x^3$, $y = 8$, and $x = 0$.Expressing $x$ in terms of $y$, we get $x = y^{1/3}$.The intersection of $y = x^3$ and $y

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The region is bounded by $y = x^3$, $y = 8$, and $x = 0$.Expressing $x$ in terms of $y$, we get $x = y^{1/3}$.The intersection of $y = x^3$ and $y = 8$ occurs at $x^3 = 8$, which gives $x = 2$.The area $A$ is given by the integral of $x$ with respect to $y$ from $y = 0$ to $y = 8$:$A = \int_{0}^{8} y^{1/3} \, dy$$A = \left[ \frac{y^{4/3}}{4/3} ight]_{0}^{8}$$A = \frac{3}{4} \left[ y^{4/3} ight]_{0}^{8}$$A = \frac{3}{4} \left( 8^{4/3} - 0^{4/3} ight)$$A = \frac{3}{4} \left( (2^3)^{4/3} ight) = \frac{3}{4} 	imes 2^4 = \frac{3}{4} 	imes 16 = 12 	ext{ square units.}$

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

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

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