The area enclosed between the graph of y = x³ and the lines x = 0, y = 1, y = 8 is

The region is bounded by:

• The curve $y = x^3$
• The y-axis (line $x = 0$)
• Horizontal line $y = 1$ (bottom boundary)
• Horizontal line $y = 8$ (top boundary)

When $y = 1$: $x^3 = 1$ → $x = 1$

When $y = 8$: $x^3 = 8$ → $x = 2$

The region is between the y-axis (left side) and the curve (right side), from $y = 1$ to $y = 8$.

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Since the boundaries are horizontal lines ($y = 1$ and $y = 8$), integrate with respect to $y$.

From $y = x^3$, taking the cube root:

$x = y^{1/3}$

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Area $= \int_1^8$ (width) $dy$

Width at any height $y$ is the distance from y-axis to curve $= x - 0 = y^{1/3}$

Area $= \int_1^8 y^{1/3} dy$

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Using the power rule:

$\int y^{1/3} dy = \dfrac{y^{1/3 + 1}}{1/3 + 1}$

$= \dfrac{y^{4/3}}{4/3}$

$= \dfrac{3}{4}y^{4/3}$

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Area $= \dfrac{3}{4}[y^{4/3}]_1^8$

$= \dfrac{3}{4}[8^{4/3} - 1^{4/3}]$

Since $8^{4/3} = (\sqrt[3]{8})^4 = 2^4 = 16$:

$= \dfrac{3}{4}[16 - 1]$

$= \dfrac{3}{4} \times 15$

$= \dfrac{45}{4}$ square units

Therefore, the area enclosed is $\dfrac{45}{4}$ square units.