1. Write both curves in a convenient form

The parabola is $x^2=4y \implies y=\frac{x^2}{4}.$

The line is $x=4y-2 \implies 4y=x+2 \implies y=\frac{x+2}{4}.$

So we need the area between $y=\frac{x^2}{4} \quad \text{and} \quad y=\frac{x+2}{4}.$

2. Find points of intersection

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

Multiply by $4$: $x^2=x+2.$

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

Factorizing, $(x-2)(x+1)=0.$

Hence, $x=2 \quad \text{or} \quad x=-1.$

Now corresponding $y$-values:

• For $x=2$: $y=\frac{2^2}{4}=1.$
• For $x=-1$: $y=\frac{(-1)^2}{4}=\frac14.$

Thus the curves intersect at $(-1,\tfrac14) \quad \text{and} \quad (2,1).$

3. Determine which curve is above the other

Take a test value, say $x=0$.

For the line: $y=\frac{0+2}{4}=\frac12.$

For the parabola: $y=\frac{0^2}{4}=0.$

So in the interval $[-1,2]$, the line lies above the parabola.

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

4. Evaluate the integral

$A=\frac14\int_{-1}^{2}(x+2-x^2)\,dx.$

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

Therefore, $A=\frac14\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}.$

Now compute:

At $x=2$, $\frac{2^2}{2}+2(2)-\frac{2^3}{3}=2+4-\frac{8}{3}=6-\frac{8}{3}=\frac{10}{3}.$

At $x=-1$, $\frac{(-1)^2}{2}+2(-1)-\frac{(-1)^3}{3}=\frac12-2+\frac13=\frac{5}{6}-2=-\frac{7}{6}.$

So,