Question 42

Mathematics Area Under Curve Hard

The area enclosed between the graphs of \( y=x^3 \) and the lines \( x=0, y=1, y=8 \) is

(A) 7
(B) 12
(C) \( \frac{45}{4} \)
(D) \( \frac{21}{8} \)
View Dynamic Solution & Explanation
Correct Solution: Option C

Step-by-step Solution:

Step 1: Identify the Region The region is bounded: - On the left by the line \( x = 0 \), - On the bottom by the line \( y = 1 \), - On the top by the line \( y = 8 \), - On the right by the curve \( y = x^3 \). Step 2: Solve for \( x \) in Terms of \( y \) From \( y = x^3 \), we solve for \( x \): \[ x = y^{1/3} \] Step 3: Set Up the Integral The area \( A \) can be calculated by integrating with respect to \( y \) from \( y = 1 \) to \( y = 8 \): \[ A = \int_{1}^{8} x \, dy = \int_{1}^{8} y^{1/3} \, dy \] Step 4: Evaluate the Integral Compute the integral: \[ A = \int_{1}^{8} y^{1/3} \, dy = \left[ \frac{3}{4} y^{4/3} \right]_{1}^{8} \] \[ = \frac{3}{4} \left( 8^{4/3} - 1^{4/3} \right) \] \[ = \frac{3}{4} \left( (2^3)^{4/3} - 1 \right) \] \[ = \frac{3}{4} \left( 2^4 - 1 \right) \] \[ = \frac{3}{4} \left( 16 - 1 \right) \] \[ = \frac{3}{4} \times 15 \] \[ = \frac{45}{4} \] Step 5: Conclusion The area enclosed between the graphs is: \[ \boxed{\frac{45}{4}} \] Correct Answer: \(\boxed{C}\)