The area bounded by `y=x^(2), y=sqrt(x), 0 le x le 4` is

Options

A. `(20)/(3)`
B. `(10)/(3)`
C. 10
D. `(50)/(3)`

Answer

To find the area bounded by the curves \( y = x^2 \), \( y = \sqrt{x} \), and the lines \( x = 0 \) and \( x = 4 \), we can follow these steps:

### Step 1: Find the Points of Intersection
To find the points where the curves intersect, we set \( y = x^2 \) equal to \( y = \sqrt{x} \):

\[ x^2 = \sqrt{x} \]

Squaring both sides gives:

\[ x^4 = x \]

Rearranging this, we have:

\[ x^4 - x = 0 \]

Factoring out \( x \):

\[ x(x^3 - 1) = 0 \]

This gives us:

\[ x = 0 \quad \text{or} \quad x^3 - 1 = 0 \implies x = 1 \]

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

### Step 2: Set Up the Integral for the Area
The area between the curves from \( x = 0 \) to \( x = 1 \) can be found using the integral:

\[ \text{Area}_1 = \int_0^1 (\sqrt{x} - x^2) \, dx \]

For the area from \( x = 1 \) to \( x = 4 \):

\[ \text{Area}_2 = \int_1^4 (x^2 - \sqrt{x}) \, dx \]

### Step 3: Calculate the First Area
Now we calculate \( \text{Area}_1 \):

\[ \text{Area}_1 = \int_0^1 (\sqrt{x} - x^2) \, dx \]

Calculating the integral:

\[ = \int_0^1 x^{1/2} \, dx - \int_0^1 x^2 \, dx \]

Calculating each part:

\[ \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3} \]

\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]

Thus,

\[ \text{Area}_1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

### Step 4: Calculate the Second Area
Now we calculate \( \text{Area}_2 \):

\[ \text{Area}_2 = \int_1^4 (x^2 - \sqrt{x}) \, dx \]

Calculating the integral:

\[ = \int_1^4 x^2 \, dx - \int_1^4 x^{1/2} \, dx \]

Calculating each part:

\[ \int_1^4 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^4 = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21 \]

\[ \int_1^4 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_1^4 = \frac{2}{3} (8 - 1) = \frac{14}{3} \]

Thus,

\[ \text{Area}_2 = 21 - \frac{14}{3} = \frac{63}{3} - \frac{14}{3} = \frac{49}{3} \]

### Step 5: Total Area
Now, we add both areas to get the total area:

\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{1}{3} + \frac{49}{3} = \frac{50}{3} \]

### Final Answer
Thus, the area bounded by the curves is:

\[ \boxed{\frac{50}{3}} \]