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

Step 1: Identify the bounding curves

The region is bounded below by the function $y_1 = |x-5|$ and above by the function $y_2 = 4\sqrt{x}$. We need to find the area between these two curves.

Step 2: Find intersection points

To find the limits of integration, we need to determine where the two curves intersect. We set $y_1 = y_2$ and solve for $x$. We consider two cases for $|x-5|$: $x-5$ for $x \ge 5$ and $-(x-5)$ for $x < 5$.

Step 3: Solve for intersection points (Case 1: $x \ge 5$)

For $x \ge 5$, we have $x-5 = 4\sqrt{x}$. Squaring both sides gives $(x-5)^2 = (4\sqrt{x})^2$, which simplifies to $x^2 - 10x + 25 = 16x$. Rearranging gives $x^2 - 26x + 25 = 0$. Factoring this quadratic equation yields $(x-1)(x-25) = 0$. Since we assumed $x \ge 5$, the valid solution is $x=25$.

Step 4: Solve for intersection points (Case 2: $x < 5$)

For $x < 5$, we have $-(x-5) = 4\sqrt{x}$, which is $5-x = 4\sqrt{x}$. Squaring both sides gives $(5-x)^2 = (4\sqrt{x})^2$, which simplifies to $25 - 10x + x^2 = 16x$. Rearranging gives $x^2 - 26x + 25 = 0$. Factoring this quadratic equation yields $(x-1)(x-25) = 0$. Since we assumed $x < 5$, the valid solution is $x=1$.

Step 5: Set up the integral for area A

The area $A$ is found by integrating the difference between the upper curve $y_2 = 4\sqrt{x}$ and the lower curve $y_1 = |x-5|$. Since $|x-5|$ changes its definition at $x=5$, we split the integral into two parts: from $x=1$ to $x=5$ (where $y_1 = 5-x$) and from $x=5$ to $x=25$ (where $y_1 = x-5$).

Step 6: Evaluate the integrals

We evaluate each integral separately. The antiderivative of $4\sqrt{x} = 4x^{1/2}$ is $4 \cdot \frac{2}{3}x^{3/2} = \frac{8}{3}x^{3/2}$. The antiderivative of $5-x$ is $5x - \frac{x^2}{2}$. The antiderivative of $x-5$ is $\frac{x^2}{2} - 5x$.

Step 7: Calculate the definite integrals and find A

Substitute the limits of integration into the antiderivatives and perform the arithmetic. After careful calculation, the total area $A$ is found to be $\frac{200}{3}$.

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