Question:
The area of the region bounded by the curve \(y = x^3\) and the lines \(y = 8\) and \(x = 0\) is ... square units.
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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Integrate with respect to \(y\): \(x = y^{1/3}\) from \(0\) to \(8\).
Updated On: Oct 1, 2026
- \(8\)
- \(12\)
- \(16\)
- \(10\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The curve \(y = x^3\) meets \(y = 8\) at \(x = 2\). The region lies between the Y-axis (\(x=0\)), the curve and the line \(y=8\).
Step 2: Integrate along y:
\[ A = \int_0^8 y^{1/3}dy = \frac34\left[y^{4/3}\right]_0^8 = \frac34\cdot16 = 12 \]
Step 3: Check:
The area is the rectangle \(2\times8 = 16\) minus the area under the curve, \(\int_0^2x^3dx = 4\). So \(16 - 4 = 12\).
Final Answer:
The area is \(12\) square units, option (B). \[ \boxed{12} \]
The curve \(y = x^3\) meets \(y = 8\) at \(x = 2\). The region lies between the Y-axis (\(x=0\)), the curve and the line \(y=8\).
Step 2: Integrate along y:
\[ A = \int_0^8 y^{1/3}dy = \frac34\left[y^{4/3}\right]_0^8 = \frac34\cdot16 = 12 \]
Step 3: Check:
The area is the rectangle \(2\times8 = 16\) minus the area under the curve, \(\int_0^2x^3dx = 4\). So \(16 - 4 = 12\).
Final Answer:
The area is \(12\) square units, option (B). \[ \boxed{12} \]
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