1. Interpret the region

We need the area of

$$\{(x,y): |x-5| \le y \le 4\sqrt{x}\}.$$

So the region lies between the curves

$$y=|x-5| \quad \text{and} \quad y=4\sqrt{x},$$

with the upper curve $y=4\sqrt{x}$ and lower curve $y=|x-5|$.

Hence, area exists for those $x$ where

$$|x-5| \le 4\sqrt{x}, \qquad x\ge 0.$$

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2. Find the points of intersection

Solve

$$|x-5|=4\sqrt{x}.$$

Let $t=\sqrt{x}$, so $x=t^2$ with $t\ge 0$. Then

$$|t^2-5|=4t.$$

We solve in two cases.

Case 1: $t^2\ge 5$

Then

$$t^2-5=4t$$ $$t^2-4t-5=0$$ $$(t-5)(t+1)=0.$$

Since $t\ge 0$, we get $t=5$, hence

$$x=t^2=25.$$

Case 2: $t^2<5$

Then

$$5-t^2=4t$$ $$t^2+4t-5=0$$ $$(t-1)(t+5)=0.$$

Since $t\ge 0$, we get $t=1$, hence

$$x=t^2=1.$$

So the curves intersect at $x=1$ and $x=25$.

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3. Set up the area integral

Thus,

$$A=\int_1^{25} \left(4\sqrt{x}-|x-5|\right)\,dx.$$

Because of the absolute value, split at $x=5$:

$$|x-5|= \begin{cases} 5-x, & 1\le x\le 5,\\ x-5, & 5\le x\le 25. \end{cases}$$

Hence

$$A=\int_1^5 \left(4\sqrt{x}-(5-x)\right)dx+\int_5^{25}\left(4\sqrt{x}-(x-5)\right)dx.$$

That is,

$$A=\int_1^5 (4\sqrt{x}+x-5)\,dx+\int_5^{25}(4\sqrt{x}-x+5)\,dx.$$

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4. Evaluate the first integral

$$I_1=\int_1^5 (4\sqrt{x}+x-5)\,dx.$$

Antiderivative:

$$\int 4\sqrt{x}\,dx=4\cdot \frac{2}{3}x^{3/2}=\frac{8}{3}x^{3/2},$$ $$\int x\,dx=\frac{x^2}{2},$$ $$\int (-5)\,dx=-5x.$$

So

$$I_1=\left[\frac{8}{3}x^{3/2}+\frac{x^2}{2}-5x\right]_1^5.$$

At $x=5$:

$$\frac{8}{3}(5\sqrt{5})+\frac{25}{2}-25=\frac{40\sqrt{5}}{3}-\frac{25}{2}.$$

At $x=1$:

$$\frac{8}{3}+\frac{1}{2}-5=\frac{8}{3}-\frac{9}{2}=-\frac{11}{6}.$$

Thus

$$I_1=\frac{40\sqrt{5}}{3}-\frac{25}{2}+\frac{11}{6} =\frac{40\sqrt{5}}{3}-\frac{32}{3}.$$

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5. Evaluate the second integral

$$I_2=\int_5^{25}(4\sqrt{x}-x+5)\,dx.$$

Antiderivative:

$$\frac{8}{3}x^{3/2}-\frac{x^2}{2}+5x.$$

So

$$I_2=\left[\frac{8}{3}x^{3/2}-\frac{x^2}{2}+5x\right]_5^{25}.$$

At $x=25$:

$$\frac{8}{3}(125)-\frac{625}{2}+125 =\frac{1000}{3}-\frac{375}{2} =\frac{875}{6}.$$

At $x=5$:

$$\frac{40\sqrt{5}}{3}-\frac{25}{2}+25 =\frac{40\sqrt{5}}{3}+\frac{25}{2}.$$

Thus

$$I_2=\frac{875}{6}-\frac{40\sqrt{5}}{3}-\frac{25}{2} =\frac{400}{3}-\frac{40\sqrt{5}}{3}.$$

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6. Compute total area

$$A=I_1+I_2 =\left(\frac{40\sqrt{5}}{3}-\frac{32}{3}\right)+\left(\frac{400}{3}-\frac{40\sqrt{5}}{3}\right).$$

The irrational terms cancel:

$$A=\frac{-32+400}{3}=\frac{368}{3}.$$

Therefore,

$$3A=368.$$

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7. Comparison with stored answer

Stored correct answer: $368$

Derived answer: $368$

So they agree.