Find the area enclosed between the curves
`y^(2)=2x,y=4x-1`

Answer

To find the area enclosed between the curves \( y^2 = 2x \) and \( y = 4x - 1 \), we will follow these steps:

### Step 1: Find the points of intersection
We start by substituting \( y = 4x - 1 \) into the equation \( y^2 = 2x \).

\[ (4x - 1)^2 = 2x \]

Expanding the left side:

\[ 16x^2 - 8x + 1 = 2x \]

Rearranging the equation gives:

\[ 16x^2 - 10x + 1 = 0 \]

### Step 2: Solve the quadratic equation
Now we will solve the quadratic equation \( 16x^2 - 10x + 1 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 16 \), \( b = -10 \), and \( c = 1 \):

\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 16 \cdot 1}}{2 \cdot 16} \]
\[ x = \frac{10 \pm \sqrt{100 - 64}}{32} \]
\[ x = \frac{10 \pm \sqrt{36}}{32} \]
\[ x = \frac{10 \pm 6}{32} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{16}{32} = \frac{1}{2} \)
2. \( x = \frac{4}{32} = \frac{1}{8} \)

### Step 3: Set up the integral for the area
The area \( A \) between the curves can be found using the integral:

\[ A = \int_{x_1}^{x_2} (f(x) - g(x)) \, dx \]

Where \( f(x) = 4x - 1 \) (the line) and \( g(x) = \sqrt{2x} \) (the upper part of the parabola). The limits of integration are from \( x = \frac{1}{8} \) to \( x = \frac{1}{2} \).

Thus, we have:

\[ A = \int_{\frac{1}{8}}^{\frac{1}{2}} ((4x - 1) - \sqrt{2x}) \, dx \]

### Step 4: Compute the integral
Now we compute the integral:

\[ A = \int_{\frac{1}{8}}^{\frac{1}{2}} (4x - 1 - \sqrt{2x}) \, dx \]

Calculating each part separately:

1. The integral of \( 4x \) is \( 2x^2 \).
2. The integral of \( -1 \) is \( -x \).
3. The integral of \( -\sqrt{2x} \) can be computed as \( -\frac{2}{3} (2x)^{3/2} \).

Putting it all together:

\[ A = \left[ 2x^2 - x - \frac{2}{3} (2x)^{3/2} \right]_{\frac{1}{8}}^{\frac{1}{2}} \]

### Step 5: Evaluate the definite integral
Now we evaluate this expression at the limits \( \frac{1}{2} \) and \( \frac{1}{8} \):

1. At \( x = \frac{1}{2} \):

\[ 2\left(\frac{1}{2}\right)^2 - \frac{1}{2} - \frac{2}{3} (2 \cdot \frac{1}{2})^{3/2} = 2 \cdot \frac{1}{4} - \frac{1}{2} - \frac{2}{3} \cdot 1 = \frac{1}{2} - \frac{1}{2} - \frac{2}{3} = -\frac{2}{3} \]

2. At \( x = \frac{1}{8} \):

\[ 2\left(\frac{1}{8}\right)^2 - \frac{1}{8} - \frac{2}{3} (2 \cdot \frac{1}{8})^{3/2} = 2 \cdot \frac{1}{64} - \frac{1}{8} - \frac{2}{3} \cdot \left(\frac{1}{4}\right) = \frac{1}{32} - \frac{1}{8} - \frac{1}{6} \]

Finding a common denominator (which is 96):

\[ = \frac{3}{96} - \frac{12}{96} - \frac{16}{96} = -\frac{25}{96} \]

### Step 6: Combine results
Now, we combine the results:

\[ A = \left(-\frac{2}{3}\right) - \left(-\frac{25}{96}\right) = -\frac{2}{3} + \frac{25}{96} \]

Finding a common denominator (which is 96):

\[ = -\frac{64}{96} + \frac{25}{96} = -\frac{39}{96} = \frac{39}{96} = \frac{13}{32} \]

Thus, the area enclosed between the curves is:

\[ \boxed{\frac{13}{32}} \]