The area of the region bounded by $y-x=2$ and $x^2=y$ is equal to :
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To find the area between two curves, first find their points of intersection to determine the limits of integration. Then, integrate the difference between the upper curve and the lower curve over that interval.
We need to find the area of the region enclosed by the line $y = x+2$ and the parabola $y = x^2$.
First, find the points of intersection by setting the two equations equal to each other.
$x^2 = x+2$.
$x^2 - x - 2 = 0$.
Factor the quadratic equation:
$(x-2)(x+1) = 0$.
The points of intersection occur at $x=2$ and $x=-1$.
When $x=2$, $y = 2^2 = 4$. Point is (2, 4).
When $x=-1$, $y = (-1)^2 = 1$. Point is (-1, 1).
The area of the region bounded by two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is given by the integral:
Area = $\int_a^b (y_{upper} - y_{lower}) dx$.
In the interval $[-1, 2]$, we need to determine which function is on top. Let's test a point, say $x=0$.
For the line: $y = 0+2 = 2$.
For the parabola: $y = 0^2 = 0$.
Since $2>0$, the line $y=x+2$ is the upper curve and the parabola $y=x^2$ is the lower curve in this interval.
The limits of integration are from $a=-1$ to $b=2$.
Area = $\int_{-1}^2 ((x+2) - x^2) dx$.
Now, evaluate the integral:
Area = $\left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^2$.
Area = $\left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right) - \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right)$.
Area = $\left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 - \frac{-1}{3} \right)$.
Area = $\left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right)$.
Area = $\left( 6 - \frac{8}{3} \right) - \left( \frac{3-12+2}{6} \right)$.
Area = $\left( \frac{18-8}{3} \right) - \left( \frac{-7}{6} \right)$.
Area = $\frac{10}{3} + \frac{7}{6}$.
Area = $\frac{20}{6} + \frac{7}{6} = \frac{27}{6}$.
Simplify the fraction:
Area = $\frac{9 \times 3}{2 \times 3} = \frac{9}{2}$.
