Animated Solution for Mathematics - Definite Integration: The area (in square units) of the region bounded by the parabola $y^2=4(x-2)$ and the line $y=2x-8$

Imagine you are standing on a coordinate plane. You see a parabola, $y^2=4(x-2)$, stretching out to the right, its vertex anchored firmly at $(2,0)$.

Then, you see a line, $y=2x-8$, cutting across the plane like a blade. The region trapped between these two curves is our target.

If we were to use vertical strips (integrating with respect to $x$), we would face a dilemma: the parabola's upper and lower branches are different functions of $x$, meaning we would have to split our integral at the vertex. Instead, let us be strategic and use horizontal strips.

To make our lives easier, let us introduce a coordinate shift. Let $X=x-2$.

By shifting our origin to the vertex of the parabola, the equation $y^2=4(x-2)$ transforms into the much friendlier:

Similarly, the line $y=2x-8$ becomes $y=2(X+2)-8$, which simplifies to $y=2X-4$. The landscape is now clear and ready to be tamed.

To find the area, we must know where these curves meet. We set the expressions for $X$ equal to each other.

From the line, $X=\frac{y+4}{2}$. From the parabola, $X=\frac{y^2}{4}$. Setting them equal, we get:

Multiplying by $4$, we get $2(y+4)=y^2$, or $y^2-2y-8=0$. Factoring this quadratic, we find $(y-4)(y+2)=0$.

Thus, our intersection points occur at $y=4$ and $y=-2$. These are the boundaries of our integration.

We are integrating with respect to $y$ from $-2$ to $4$. The area is the integral of the right curve minus the left curve:

Substituting our expressions, we have:

The antiderivative of $\frac{y}{2}$ is $\frac{y^2}{4}$, the antiderivative of $\frac{4}{2}$ is $2y$, and the antiderivative of $-\frac{y^2}{4}$ is $-\frac{y^3}{12}$. We evaluate the expression:

Plugging in the upper limit $y=4$:

Plugging in the lower limit $y=-2$:

Subtracting the lower limit from the upper limit:

The area is exactly 9 square units.