Area lying between the curves `y^2=4x` and `y=2x` is

Options

A. `2/3`
B. `1/2`
C. `1/4`
D. `3/4`

Answer

To find the area lying between the curves \( y^2 = 4x \) and \( y = 2x \), we will follow these steps:

### Step 1: Find Points of Intersection
To find the points where the curves intersect, we need to set \( y^2 = 4x \) equal to \( y = 2x \).

Substituting \( y = 2x \) into \( y^2 = 4x \):

\[ (2x)^2 = 4x \]
\[ 4x^2 = 4x \]
\[ 4x^2 - 4x = 0 \]
Factoring out \( 4x \):

\[ 4x(x - 1) = 0 \]

This gives us the solutions:

\[ x = 0 \quad \text{or} \quad x = 1 \]

Now, substituting these values back to find the corresponding \( y \) values:

- For \( x = 0 \):
\[ y = 2(0) = 0 \quad \Rightarrow \quad (0, 0) \]

- For \( x = 1 \):
\[ y = 2(1) = 2 \quad \Rightarrow \quad (1, 2) \]

Thus, the points of intersection are \( (0, 0) \) and \( (1, 2) \).

### Step 2: Set Up the Integral
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be found by integrating the difference of the upper curve and the lower curve.

The upper curve is \( y = 2x \) and the lower curve is \( y = 2\sqrt{x} \) (from \( y^2 = 4x \)).

Thus, the area \( A \) is given by:

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

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

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

Calculating each integral separately:

1. For \( \int_{0}^{1} 2x \, dx \):
\[ = 2 \left[ \frac{x^2}{2} \right]_{0}^{1} = \left[ x^2 \right]_{0}^{1} = 1^2 - 0^2 = 1 \]

2. For \( \int_{0}^{1} 2\sqrt{x} \, dx \):
\[ = 2 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{4}{3} \left[ x^{3/2} \right]_{0}^{1} = \frac{4}{3}(1^{3/2} - 0^{3/2}) = \frac{4}{3} \]

Putting it all together:

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

Since area cannot be negative, we take the absolute value:

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

### Final Answer
The area lying between the curves \( y^2 = 4x \) and \( y = 2x \) is:

\[ \boxed{\frac{1}{3}} \text{ square units.} \]