Animated Solution for Mathematics - Definite Integration: If the area of the region $\{(x,y):|x-5|\le y\le 4\sqrt{x}\}$ is $A$, then $3A$ is equal to _____ .

Imagine you are standing on a coordinate plane, looking at two distinct paths. One is a graceful, sweeping curve, the parabola defined by $y=4\sqrt{x}$, which opens to the right, starting from the origin and climbing steadily.

The other is a sharp, angular V-shape, the absolute value function $y=|x-5|$, which has its vertex firmly planted at $x=5$. The problem asks us to find the area trapped between these two curves.

This is not just a calculation; it is a dance between the smooth and the sharp. To find the area $A$, we must first understand where these two paths cross.

Before we can integrate, we need to know our boundaries. We set the two functions equal to each other:

Because of the absolute value, we must split our investigation into two distinct worlds. In the first world, where $x<5$, the expression $|x-5|$ becomes $5-x$.

Setting $5-x=4\sqrt{x}$ gives us a quadratic equation in terms of $\sqrt{x}$. By substituting $u=\sqrt{x}$, we get:

Factoring this, we find $(u+5)(u-1)=0$. Since $\sqrt{x}$ cannot be negative, we discard $u=-5$ and keep $u=1$, which gives us $x=1$.

In the second world, where $x\ge 5$, the expression $|x-5|$ becomes $x-5$. Setting $x-5=4\sqrt{x}$ leads to:

Factoring this, we get $(u-5)(u+1)=0$. Again, we keep the positive root, $u=5$, which gives us $x=25$. Our boundaries are set: we are integrating from $x=1$ to $x=25$.

Now, we calculate the area $A$. The total area is the integral of the upper curve minus the lower curve:

Let us tackle the parabola first. The integral of $4\sqrt{x}$ is $4\int x^{\frac{1}{2}}dx$, which becomes:

Evaluating this from $1$ to $25$, we get:

This is the total area under the parabola. Now, for the V-shape, we use geometry. The area under $|x-5|$ from $1$ to $25$ is composed of two triangles.

The first triangle, from $x=1$ to $x=5$, has a base of $4$ and a height of $4$ (since $|1-5|=4$). Its area is:

The second triangle, from $x=5$ to $x=25$, has a base of $20$ and a height of $20$ (since $|25-5|=20$). Its area is:

The total area under the V-shape is $8+200=208$.

We are almost there. The net area $A$ is the difference between the area under the parabola and the area under the V-shape:

Converting $208$ to a fraction with a denominator of $3$, we get $208=\frac{624}{3}$. Thus:

The question asks for $3A$. We multiply our result by $3$:

The elegance of the final cancellation is the reward for our careful work. You have successfully navigated the curves and the algebra to reach the final answer of 368.