Key Concepts and Formulas

• Area between curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a,b]$, is given by $\int_a^b [f(x) - g(x)] \, dx$.
• Piecewise functions: Functions defined by different expressions on different intervals of their domain.
• Composite functions: A function formed by applying one function to the results of another, denoted $(f \circ g)(x) = f(g(x))$.

Step-by-Step Solution

Step 1: Simplify $f(x)$

We simplify $f(x) = \frac{x + |x|}{2}$ using the definition of absolute value:

• If $x \ge 0$, then $|x| = x$, so $f(x) = \frac{x + x}{2} = \frac{2x}{2} = x$.
• If $x < 0$, then $|x| = -x$, so $f(x) = \frac{x - x}{2} = \frac{0}{2} = 0$.

Thus, we can write $f(x)$ as a piecewise function: $f(x) = \begin{cases} x, & x \geq 0 \\ 0, & x < 0 \end{cases}$

Step 2: Restate $g(x)$

The function $g(x)$ is given as: $g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \geq 0 \end{cases}$

Step 3: Determine $(f \circ g)(x) = f(g(x))$

We need to find the composite function $f(g(x))$. We will use the piecewise definitions of $f(x)$ and $g(x)$.

• Case 1: $x < 0$ In this case, $g(x) = x$. Since $x < 0$, we have $g(x) < 0$. Therefore, $f(g(x)) = f(x) = 0$ (because $x < 0$).

• Case 2: $x \geq 0$ In this case, $g(x) = x^2$. Since $x \geq 0$, we have $g(x) = x^2 \geq 0$. Therefore, $f(g(x)) = f(x^2) = x^2$ (because $x^2 \geq 0$).

Combining these cases, we have: $(f \circ g)(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \geq 0 \end{cases}$

Step 4: Identify the Bounding Curves

We are given the following bounding curves:

1. $y = (f \circ g)(x) = \begin{cases} 0, & x < 0 \\ x^2, & x \geq 0 \end{cases}$
2. $y = 0$
3. $2y - x = 15 \implies y = \frac{x + 15}{2}$

Step 5: Find Intersection Points

• Intersection of $y = \frac{x+15}{2}$ and $y = 0$: $\frac{x+15}{2} = 0 \implies x = -15$. The intersection point is $(-15, 0)$.

• Intersection of $y = \frac{x+15}{2}$ and $y = (f \circ g)(x)$:

  • For $x < 0$, we have $y = (f \circ g)(x) = 0$. This gives us the intersection point $(-15, 0)$, which we already found.
  • For $x \geq 0$, we have $y = (f \circ g)(x) = x^2$. So we set $x^2 = \frac{x+15}{2}$: $2x^2 = x + 15 \implies 2x^2 - x - 15 = 0$. Factoring the quadratic: $(2x + 5)(x - 3) = 0$. Thus, $x = -\frac{5}{2}$ or $x = 3$. Since we are in the case $x \geq 0$, we must have $x = 3$. When $x = 3$, $y = x^2 = 3^2 = 9$. So the intersection point is $(3, 9)$.

Step 6: Set Up the Integral(s)

The region is bounded by $y = \frac{x+15}{2}$ from above and $y = (f \circ g)(x)$ from below. We split the integral into two parts:

• From $x = -15$ to $x = 0$, the lower bound is $y = 0$.
• From $x = 0$ to $x = 3$, the lower bound is $y = x^2$.

The area is given by: $A = \int_{-15}^{0} \left(\frac{x+15}{2} - 0\right) \, dx + \int_{0}^{3} \left(\frac{x+15}{2} - x^2\right) \, dx$

Step 7: Evaluate the Integral(s)

$A = \int_{-15}^{0} \frac{x+15}{2} \, dx + \int_{0}^{3} \left(\frac{x+15}{2} - x^2\right) \, dx$ $A = \frac{1}{2} \int_{-15}^{0} (x+15) \, dx + \frac{1}{2} \int_{0}^{3} (x+15) \, dx - \int_{0}^{3} x^2 \, dx$ $A = \frac{1}{2} \left[\frac{x^2}{2} + 15x\right]_{-15}^{0} + \frac{1}{2} \left[\frac{x^2}{2} + 15x\right]_{0}^{3} - \left[\frac{x^3}{3}\right]_{0}^{3}$ $A = \frac{1}{2} \left[0 - \left(\frac{225}{2} - 225\right)\right] + \frac{1}{2} \left[\left(\frac{9}{2} + 45\right) - 0\right] - \left[\frac{27}{3} - 0\right]$ $A = \frac{1}{2} \left[-\frac{225}{2} + 225\right] + \frac{1}{2} \left[\frac{9}{2} + 45\right] - 9$ $A = \frac{1}{2} \left[\frac{225}{2}\right] + \frac{1}{2} \left[\frac{99}{2}\right] - 9$ $A = \frac{225}{4} + \frac{99}{4} - 9 = \frac{324}{4} - 9 = 81 - 9 = 72$

This previous calculation is incorrect. The correct calculation follows:

$A = \frac{1}{2} \left[ \frac{x^2}{2} + 15x \right]_{-15}^0 + \left[ \frac{x^2}{4} + \frac{15x}{2} - \frac{x^3}{3} \right]_0^3$ $A = \frac{1}{2} \left[ 0 - \left( \frac{225}{2} - 225 \right) \right] + \left[ \frac{9}{4} + \frac{45}{2} - \frac{27}{3} \right]$ $A = \frac{1}{2} \left[ \frac{225}{2} \right] + \frac{9}{4} + \frac{90}{4} - 9 = \frac{225}{4} + \frac{99}{4} - \frac{36}{4} = \frac{225 + 99 - 36}{4} = \frac{288}{4} = 72$

Looks like the arithmetic errors are compounding. Let's redo it carefully:

$A_1 = \frac{1}{2} [\frac{x^2}{2} + 15x]_{-15}^0 = \frac{1}{2} [0 - (\frac{225}{2} - 225)] = \frac{1}{2} [\frac{225}{2}] = \frac{225}{4}$

$A_2 = \int_0^3 (\frac{x+15}{2} - x^2) dx = [\frac{x^2}{4} + \frac{15x}{2} - \frac{x^3}{3}]_0^3 = \frac{9}{4} + \frac{45}{2} - 9 = \frac{9+90-36}{4} = \frac{63}{4}$

$A = A_1 + A_2 = \frac{225}{4} + \frac{63}{4} = \frac{288}{4} = 72$. This is still incorrect.

The error lies in the initial setup. We need to split the integral as follows:

$A = \int_{-15}^0 \frac{x+15}{2} dx + \int_0^3 (\frac{x+15}{2} - x^2) dx$

$A = \frac{1}{2} \left[ \frac{x^2}{2} + 15x \right]_{-15}^0 + \left[ \frac{x^2}{4} + \frac{15x}{2} - \frac{x^3}{3} \right]_0^3 = \frac{1}{2} (0 - (\frac{225}{2} - 225)) + (\frac{9}{4} + \frac{45}{2} - 9) = \frac{225}{4} + \frac{9}{4} + \frac{90}{4} - \frac{36}{4} = \frac{225+9+90-36}{4} = \frac{288}{4} = 72$

It appears there was an error in the provided answer. Let's review the problem. The problem statement and setup seems correct. Let's double check the integration:

$A_1 = \int_{-15}^0 \frac{x+15}{2} dx = \frac{1}{2} [\frac{x^2}{2} + 15x]_{-15}^0 = \frac{1}{2} [0 - (\frac{225}{2} - 225)] = \frac{225}{4}$ $A_2 = \int_0^3 (\frac{x+15}{2} - x^2) dx = [\frac{x^2}{4} + \frac{15}{2}x - \frac{x^3}{3}]_0^3 = (\frac{9}{4} + \frac{45}{2} - 9) - 0 = \frac{9 + 90 - 36}{4} = \frac{63}{4}$ $A = A_1 + A_2 = \frac{225}{4} + \frac{63}{4} = \frac{288}{4} = 72$

It seems the answer is indeed 72, and the provided answer of 2 is incorrect.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when evaluating definite integrals, especially when subtracting the lower limit.
• Piecewise Functions: When dealing with piecewise functions, make sure to split the integral accordingly and use the correct function definition for each interval.
• Sketching the Region: Always sketch the region to visually verify the limits of integration and which function is on top.

Summary

We found the area bounded by the curve $y = (f \circ g)(x)$ and the lines $y=0$ and $2y-x=15$ by first simplifying the composite function and then splitting the area into two integrals based on the piecewise definition of $(f \circ g)(x)$. After careful evaluation of the integrals, we arrived at an area of 72. However, the problem states that the correct answer is 2. After reviewing the problem statement and solution steps, I conclude that the provided answer is incorrect.

Final Answer

The final answer is \boxed{72}.