The area of the region in first quadrant bounded by the curves `y = x^(3)` and `y = sqrtx` is equal to

Options

A. `(12)/(5)` sq. units
B. `(5)/(12)` sq. units
C. `(5)/(3)` sq. units
D. `(3)/(5)` units

Answer

To find the area of the region in the first quadrant bounded by the curves \( y = x^3 \) and \( y = \sqrt{x} \), we will follow these steps:

### Step 1: Find the Points of Intersection
To determine the area between the curves, we first need to find the points where they intersect. We set the equations equal to each other:

\[ x^3 = \sqrt{x} \]

Squaring both sides to eliminate the square root gives:

\[ x^6 = x \]

Rearranging this, we have:

\[ x^6 - x = 0 \]

Factoring out \( x \):

\[ x(x^5 - 1) = 0 \]

This gives us two solutions:

1. \( x = 0 \)
2. \( x^5 - 1 = 0 \) which implies \( x = 1 \)

Thus, the points of intersection are \( (0, 0) \) and \( (1, 1) \).

### Step 2: Set Up the Integral
Next, we need to set up the integral to find the area between the curves from \( x = 0 \) to \( x = 1 \). The area \( A \) is given by the integral of the upper curve minus the lower curve:

\[ A = \int_{0}^{1} (\sqrt{x} - x^3) \, dx \]

### Step 3: Calculate the Integral
Now we will compute the integral:

\[ A = \int_{0}^{1} \sqrt{x} \, dx - \int_{0}^{1} x^3 \, dx \]

Calculating the first integral:

\[ \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]

Evaluating from 0 to 1:

\[ \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3} (1^{3/2}) - \frac{2}{3} (0^{3/2}) = \frac{2}{3} - 0 = \frac{2}{3} \]

Now for the second integral:

\[ \int x^3 \, dx = \frac{x^4}{4} \]

Evaluating from 0 to 1:

\[ \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} - 0 = \frac{1}{4} \]

### Step 4: Combine the Results
Now we can find the area:

\[ A = \frac{2}{3} - \frac{1}{4} \]

To combine these fractions, we find a common denominator, which is 12:

\[ A = \frac{8}{12} - \frac{3}{12} = \frac{5}{12} \]

Thus, the area of the region bounded by the curves in the first quadrant is:

\[ \boxed{\frac{5}{12}} \text{ square units} \]