Question

The area of the region (in square units) enclosed by the curves \( y = 8x^3 - 1 \), \( y = 0 \), \( x = -1 \), and \( x = 1 \) is:

• A. \( \frac{15}{4} \)
• B. \( \frac{15}{8} \)
• C. \( \frac{19}{4} \)
• D. \( \frac{19}{8} \)

Hint:

For areas enclosed by curves, split the integral where the function changes sign, take the absolute value, and sum the definite integrals over the appropriate limits.

Answer 1

The Correct Option is C

Step 1: Set Up the Integral for the Enclosed Area The enclosed area is given by the definite integral: \[ A = \int_{-1}^{1} |(8x^3 - 1) - 0| \,dx \] Since \( 8x^3 - 1 \) changes sign at \( x = \frac{1}{2} \), we split the integral into two parts: \[ A = \int_{-1}^{\frac{1}{2}} |(8x^3 - 1)| \,dx + \int_{\frac{1}{2}}^{1} (8x^3 - 1) \,dx \] For \( x<\frac{1}{2} \), \( 8x^3 - 1 \) is negative, so we take the absolute value: \[ A = \int_{-1}^{\frac{1}{2}} (1 - 8x^3) \,dx + \int_{\frac{1}{2}}^{1} (8x^3 - 1) \,dx \] 

Step 2: Evaluate Each Integral Computing: \[ \int x^3 \,dx = \frac{x^4}{4}, \quad \int dx = x \] Evaluating both integrals: \[ \int_{-1}^{\frac{1}{2}} (1 - 8x^3) \,dx = \left[ x - 2x^4 \right]_{-1}^{\frac{1}{2}} \] \[ = \left( \frac{1}{2} - 2 \left( \frac{1}{16} \right) \right) - \left( -1 - 2(1) \right) \] \[ = \left( \frac{1}{2} - \frac{1}{8} \right) - (-3) \] \[ = \frac{4}{8} - \frac{1}{8} + 3 = \frac{3}{8} + 3 = \frac{27}{8} \] Similarly, for the second integral: \[ \int_{\frac{1}{2}}^{1} (8x^3 - 1) \,dx = \left[ 2x^4 - x \right]_{\frac{1}{2}}^{1} \] \[ = \left( 2(1) - 1 \right) - \left( 2 \times \frac{1}{16} - \frac{1}{2} \right) \] \[ = (2 - 1) - \left( \frac{2}{16} - \frac{8}{16} \right) \] \[ = 1 - (-\frac{6}{16}) = 1 + \frac{3}{8} = \frac{11}{8} \] 

Step 3: Compute Total Area \[ A = \frac{27}{8} + \frac{11}{8} = \frac{38}{8} = \frac{19}{4} \] Thus, the enclosed area is: \[ \frac{19}{4} \]