Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by the curves $y=x^2-4x+4$ and $y^2=16-8x$ is:

Imagine you are standing at the point $(2,0)$ on a Cartesian plane. You look up and see a graceful curve, a parabola opening upwards, defined by $y=(x-2{)}^2$.

You turn your head and see another curve, a parabola opening to the left, defined by $y^2=-8(x-2)$. These two curves are partners in a dance, enclosing a beautiful, finite region. Our goal is to find the area of this region.

To find the area, we first need to know where these curves meet. We set up the intersection equation by substituting $y=(x-2{)}^2$ into $y^2=-8(x-2)$.

This gives us:

This simplifies to $(x-2{)}^4=-8(x-2)$. A common mistake here is to expand the fourth power; instead, bring everything to one side:

Now, factor out $(x-2)$ to get $(x-2)((x-2{)}^3+8)=0$. This reveals our intersection points: $x=2$ and $(x-2{)}^3=-8$, which means $x-2=-2$, so $x=0$. These are the boundaries of our integral.

Now that we have our limits, we need to set up the integral for the area. The area $A$ is the integral of the upper curve minus the lower curve from $x=0$ to $x=2$.

The upper boundary is the horizontal parabola $y=\sqrt{-8(x-2)}=\sqrt{16-8x}$, and the lower boundary is the vertical parabola $y=(x-2{)}^2$. So, our integral is:

We can split this into two parts: $I_1={\int }_0^2\sqrt{16-8x}dx$ and $I_2={\int }_0^2(x-2{)}^{2}dx$.

For $I_1$, we use substitution. Let $u=16-8x$, then $du=-8dx$. The limits change from $x=0$ to $u=16$ and $x=2$ to $u=0$.

This gives:

For $I_2$, it is a straightforward integration:

Finally, the area is $I_1-I_2=\frac{16}{3}-\frac{8}{3}=\frac{8}{3}$. The area of the region is $\frac{8}{3}$ square units.