Find the area of the curve enclosed between the curve x^2 = 4y and x = 4y - 2.

Steps to Solve

1. Identify the curves

We have two equations:

The parabola given by $x^2 = 4y$ can be rewritten as $y = \frac{x^2}{4}$.

The straight line given by $x = 4y - 2$ can be rearranged to find $y$: $$ y = \frac{x + 2}{4}. $$

2. Find points of intersection

To find the points of intersection, set the two expressions for $y$ equal to each other:

$$ \frac{x^2}{4} = \frac{x + 2}{4}. $$

Multiply through by 4 to eliminate the denominator:

$$ x^2 = x + 2. $$

Rearranging gives:

$$ x^2 - x - 2 = 0. $$

3. Solve the quadratic

We can use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -1$, and $c = -2$.

Calculating the discriminant:

$$ b^2 - 4ac = (-1)^2 - 4(1)(-2) = 1 + 8 = 9. $$

Thus, we have

$$ x = \frac{-(-1) \pm \sqrt{9}}{2(1)} = \frac{1 \pm 3}{2}. $$

This gives two solutions:

$$ x_1 = 2, ; x_2 = -1. $$

4. Find corresponding $y$ values

Using $y = \frac{x^2}{4}$, we compute $y$ for both $x$ values:

For $x = 2$: $$ y = \frac{2^2}{4} = 1. $$

For $x = -1$: $$ y = \frac{(-1)^2}{4} = \frac{1}{4}. $$

Thus, the points of intersection are $(2, 1)$ and $(-1, \frac{1}{4})$.

5. Set up the integral

The area $A$ between the curves from $x = -1$ to $x = 2$ is given by the integral:

$$ A = \int_{-1}^{2} \left( \frac{x + 2}{4} - \frac{x^2}{4} \right) , dx. $$

Simplifying the integrand:

$$ A = \int_{-1}^{2} \frac{(x + 2) - x^2}{4} , dx = \frac{1}{4} \int_{-1}^{2} (2 + x - x^2) , dx. $$

6. Evaluate the integral

First, find the antiderivative:

$$ \int (2 + x - x^2) , dx = 2x + \frac{x^2}{2} - \frac{x^3}{3} + C. $$

Now, evaluate from $-1$ to $2$:

$$ A = \frac{1}{4} \left[ \left( 2(2) + \frac{2^2}{2} - \frac{2^3}{3} \right) - \left( 2(-1) + \frac{(-1)^2}{2} - \frac{(-1)^3}{3} \right) \right]. $$

Calculating:

At $x = 2$: $$ = 4 + 2 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18 - 8}{3} = \frac{10}{3}. $$

At $x = -1$: $$ = -4 + \frac{1}{2} + \frac{1}{3} = -4 + \frac{3}{6} + \frac{2}{6} = -4 + \frac{5}{6} = -\frac{24}{6} + \frac{5}{6} = -\frac{19}{6}. $$

Now, substitute:

$$ A = \frac{1}{4} \left( \frac{10}{3} + \frac{19}{6} \right). $$

Finding a common denominator (6):

$$ = \frac{1}{4} \left( \frac{20}{6} + \frac{19}{6} \right) = \frac{1}{4} \left( \frac{39}{6} \right) = \frac{39}{24} = \frac{13}{8}. $$

7. Final calculation for the area

So the enclosed area between the curves is:

$$ A = \frac{13}{8}. $$