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

The area bounded by the curves \( y = x^3 \) and \( y = \sqrt{x} \) between their intersection points at \( x = 0 \) and \( x = 1 \) is calculated as follows:\[A = \int_0^1 \left( \sqrt{x} - x^3 \right) dx = \left[ \frac{2}{3}x^{3/2} - \frac{1}{4}x^4 \right]_0^1 = \left( \frac{2}{3}(1)^{3/2} - \frac{1}{4}(1)^4 \right) - \left( \frac{2}{3}(0)^{3/2} - \frac{1}{4}(0)^4 \right) = \frac{2}{3} - \frac{1}{4} = \frac{8-3}{12} = \frac{5}{12}.\]The resulting area is \( \frac{5}{12} \) square units.