Question

The area enclosed by the curves \( y = x^3 \) and \( y = \sqrt{x} \) is:

• A. \( \frac{5}{3} \) sq. units
• B. \( \frac{5}{4} \) sq. units
• C. \( \frac{5}{12} \) sq. units
• D. \( \frac{12}{5} \) sq. units

Hint:

To find the area enclosed by two curves, first identify the points of intersection. Then integrate the difference between the functions over the range defined by the intersection points. Make sure to subtract the lower function from the upper function before integrating.

Answer 1

The Correct Option is C

Given curves \( y = x^3 \) and \( y = \sqrt{x} \). The curves intersect at \( x = 0 \) and \( x = 1 \). The area enclosed by these curves is given by the integral: \[ A = \int_0^1 \left( \sqrt{x} - x^3 \right) dx = \left[ \frac{x^{3/2}}{3/2} - \frac{x^4}{4} \right]_0^1 = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}. \] Thus, the area is \( \frac{5}{12} \) sq. units.

Answer 2

Step 1: Given Curves
The given curves are: \[ y = x^3 \quad \text{and} \quad y = \sqrt{x} \]
Step 2: Find the Points of Intersection
To find the points of intersection, set the equations equal to each other: \[ x^3 = \sqrt{x} \] Square both sides to eliminate the square root: \[ x^6 = x \] Rearrange the equation: \[ x^6 - x = 0 \] Factor the equation: \[ x(x^5 - 1) = 0 \] Thus, \( x = 0 \) or \( x^5 = 1 \). Solving \( x^5 = 1 \) gives \( x = 1 \). The points of intersection are \( x = 0 \) and \( x = 1 \).
Step 3: Set up the Integral for the Area
The area between the curves is given by: \[ \text{Area} = \int_{x_1}^{x_2} \left( y_{\text{right}} - y_{\text{left}} \right) dx \] Here, the right curve is \( y = \sqrt{x} \), and the left curve is \( y = x^3 \). Thus, the area is: \[ \text{Area} = \int_{0}^{1} \left( \sqrt{x} - x^3 \right) dx \]
Step 4: Compute the Integral
Now, compute the integral: \[ \text{Area} = \int_{0}^{1} \left( x^{1/2} - x^3 \right) dx \] 1. For \( \int_{0}^{1} x^{1/2} \, dx \): \[ \int_{0}^{1} x^{1/2} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3}(1^{3/2} - 0^{3/2}) = \frac{2}{3} \] 2. For \( \int_{0}^{1} x^3 \, dx \): \[ \int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1^4}{4} - 0 = \frac{1}{4} \]
Step 5: Subtract the Results
Now, subtract the results of the integrals: \[ \text{Area} = \frac{2}{3} - \frac{1}{4} \] Find a common denominator: \[ \text{Area} = \frac{8}{12} - \frac{3}{12} = \frac{5}{12} \]
Step 6: Conclusion
The area enclosed by the curves is: \[ \boxed{\frac{5}{12}} \, \text{sq. units} \]