Question

The area enclosed between the graph of \( y = x^3 \) { and the lines} \[ x = 0, \, y = 1, \, y = 8 { is:} \]

• A. \( \frac{45}{4} \)
• B. 14
• C. 7
• D. None of these

Hint:

To find the area between curves, integrate the difference of the functions over the given interval.

Answer 1

The Correct Option is A

Step 1: The given curve is \( y = x^3 \), and we need to find the area enclosed by the curve between the lines \( y = 1 \) and \( y = 8 \). We first find the points of intersection of the curve and the lines \( y = 1 \) and \( y = 8 \). For \( y = 1 \), we have: \[ 1 = x^3 \quad \Rightarrow \quad x = 1 \] For \( y = 8 \), we have: \[ 8 = x^3 \quad \Rightarrow \quad x = 2 \] Step 2: The required area is the integral of \( x \) with respect to \( y \) from \( y = 1 \) to \( y = 8 \). Since \( y = x^3 \), we can write \( x = y^{1/3} \). Thus, the area \( A \) is: \[ A = \int_{1}^{8} y^{1/3} \, dy \] Step 3: Solve the integral: \[ A = \left[ \frac{3}{4} y^{4/3} \right]_{1}^{8} = \frac{3}{4} \left( 8^{4/3} - 1^{4/3} \right) \] Since \( 8^{4/3} = 16 \) and \( 1^{4/3} = 1 \), we get: \[ A = \frac{3}{4} (16 - 1) = \frac{3}{4} \times 15 = \frac{45}{4} \] Thus, the area enclosed is \( \frac{45}{4} \).