The area of the region bounded by `y = x^(2)` and y = 4x , for x between 0 and 1 is equal to

Options

A. 2 sq. units
B. `(5)/(3)` sq. units
C. 6 sq. units
D. 1 sq. units

Answer

To find the area of the region bounded by the curves \( y = x^2 \) and \( y = 4x \) for \( x \) between 0 and 1, we will follow these steps:

### Step 1: Find the points of intersection
To find the points where the two curves intersect, we set them equal to each other:

\[ x^2 = 4x \]

Rearranging gives:

\[ x^2 - 4x = 0 \]

Factoring out \( x \):

\[ x(x - 4) = 0 \]

This gives us the solutions:

\[ x = 0 \quad \text{and} \quad x = 4 \]

Since we are only interested in the interval from \( 0 \) to \( 1 \), we will consider \( x = 0 \) as the relevant intersection point.

### Step 2: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be found by integrating the difference of the two functions:

\[ A = \int_{0}^{1} (y_1 - y_2) \, dx \]

where \( y_1 = 4x \) (the line) and \( y_2 = x^2 \) (the parabola). Thus, we have:

\[ A = \int_{0}^{1} (4x - x^2) \, dx \]

### Step 3: Evaluate the integral
Now we will evaluate the integral:

\[ A = \int_{0}^{1} (4x - x^2) \, dx \]

We can split this into two separate integrals:

\[ A = \int_{0}^{1} 4x \, dx - \int_{0}^{1} x^2 \, dx \]

Calculating each integral separately:

1. For \( \int 4x \, dx \):

\[ \int 4x \, dx = 4 \cdot \frac{x^2}{2} = 2x^2 \]

Evaluating from 0 to 1:

\[ 2(1^2) - 2(0^2) = 2 - 0 = 2 \]

2. For \( \int x^2 \, dx \):

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

Evaluating from 0 to 1:

\[ \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3} \]

### Step 4: Combine the results
Now we can combine the results of the two integrals:

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

To combine these, we can express 2 as \( \frac{6}{3} \):

\[ A = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \]

### Final Answer
Thus, the area of the region bounded by the curves \( y = x^2 \) and \( y = 4x \) for \( x \) between 0 and 1 is:

\[ \boxed{\frac{5}{3}} \text{ square units} \]