Animated Solution for Mathematics - Definite Integration: Sketch the region bounded by the curves $y=\sqrt{5-x^2}$ and $y=|x-1|$ and find its area.

Welcome, future engineer. Today, we are not just solving a problem; we are choreographing a dance between two distinct mathematical entities: the smooth, elegant curve of a semi-circle and the sharp, angular precision of a modulus function.

When you look at the equations $y=\sqrt{5-x^2}$ and $y=|x-1|$, see them as actors on a stage. One is a graceful arc, the other a rigid V-shape. Our goal is to find the area of the stage they enclose together.

The equation $y=\sqrt{5-x^2}$ represents the upper half of a circle. If you square both sides, you get $x^2+y^2=5$, which is a circle centered at the origin with a radius of $\sqrt{5}$. Because of the square root, we are restricted to $y\ge 0$.

The equation $y=|x-1|$ is the classic modulus function, shifted to the right by one unit. It creates a sharp vertex at $(1,0)$. This function acts as the floor of our region, while the semi-circle acts as the roof.

To find the boundaries of our region, we set the two equations equal:

Squaring both sides yields $5-x^2=(x-1{)}^2$. Expanding the right side, we get $5-x^2=x^2-2x+1$.

Rearranging the terms to one side, we arrive at the quadratic equation:

Dividing by $2$, we obtain $x^2-x-2=0$, which factors into $(x-2)(x+1)=0$. Thus, our intersection points are $x=-1$ and $x=2$.

The area $A$ is defined by the integral of the upper curve minus the lower curve:

We split this into two parts: the integral of the semi-circle and the integral of the modulus function. The semi-circle integral is evaluated using the standard formula:

Substituting $a^2=5$, we evaluate this from $-1$ to $2$.

When substituting the limits, we encounter terms like ${\sin }^{-1}(\frac{2}{\sqrt{5}})$ and ${\sin }^{-1}(\frac{1}{\sqrt{5}})$. If we let $\alpha ={\sin }^{-1}(\frac{1}{\sqrt{5}})$, then $\sin (\alpha )=\frac{1}{\sqrt{5}}$.

By constructing a right triangle, we find that $\cos (\alpha )=\frac{2}{\sqrt{5}}$, which implies $\alpha ={\cos }^{-1}(\frac{2}{\sqrt{5}})$. Consequently:

The area under the semi-circle simplifies to $2+\frac{5\pi }{4}$.

Finally, we subtract the area under the modulus function. The integral of $|x-1|$ from $-1$ to $2$ represents the sum of the areas of two triangles, which is $\frac{5}{2}$.

Subtracting this from our semi-circle area:

Combining these terms, we arrive at our final answer: