The area enclosed between the graph of $y = x^3$ and the lines $x = 0, y = 1,y= 8$ is

$\frac{45}{4}$ Sq. units

The correct answer is Option (3) → $\frac{45}{4}$ Sq. units

Given:

Curve: $y = x^3$

Lines: $x = 0$, $y = 1$, $y = 8$

Convert $y = x^3$ to $x = y^{\frac{1}{3}}$ since limits are in terms of $y$.

Area enclosed between $x = 0$ and $x = y^{\frac{1}{3}}$ from $y = 1$ to $y = 8$ is:

$\displaystyle A = \int_1^8 y^{\frac{1}{3}}\,dy$

$= \left[ \frac{y^{\frac{4}{3}}}{\frac{4}{3}} \right]_1^8$

$= \left[ \frac{3}{4} y^{\frac{4}{3}} \right]_1^8$

$= \frac{3}{4} \left( 8^{\frac{4}{3}} - 1^{\frac{4}{3}} \right)$

$= \frac{3}{4} \left( (2^3)^{\frac{4}{3}} - 1 \right)$

$= \frac{3}{4} \left( 2^4 - 1 \right) = \frac{3}{4} (16 - 1)$

$= \frac{3}{4} \cdot 15 = \frac{45}{4}$