We are given the functions $f$ and $g$ defined on the interval $[-1, 5]$ as follows: $f(x) = \begin{cases} x+3 & \text{if } -1 \leq x \leq 0 \\ -x+3 & \text{if } 0 < x \leq 3 \\ x-3 & \text{if } 3 < x \leq 5 \end{cases}$ and $g(x) = \begin{cases} -\frac{1}{2}x + \frac{3}{2} & \text{if } -1 \leq x \leq 1 \\ -\frac{1}{4}x + \frac{5}{4} & \text{if } 1 < x \leq 5 \end{cases}$ We need to: 1) Plot the functions $f$ and $g$ separately. 2) Calculate the integrals $I = \int_{-1}^{5} f(x) dx$ and $J = \int_{-1}^{5} g(x) dx$. 3) Deduce the integrals $\int_{-1}^{5} (f(x) + 4g(x)) dx$ and $\int_{-1}^{5} (5f(x) - 2g(x)) dx$.

1. Problem Description

We are given the functions $f$ and $g$ defined on the interval $[-1, 5]$ as follows:

$f(x) = \begin{cases} x+3 & \text{if } -1 \leq x \leq 0 \\ -x+3 & \text{if } 0 < x \leq 3 \\ x-3 & \text{if } 3 < x \leq 5 \end{cases}$

and

$g(x) = \begin{cases} -\frac{1}{2}x + \frac{3}{2} & \text{if } -1 \leq x \leq 1 \\ -\frac{1}{4}x + \frac{5}{4} & \text{if } 1 < x \leq 5 \end{cases}$

We need to:

1) Plot the functions $f$ and $g$ separately.

2) Calculate the integrals $I = \int_{-1}^{5} f(x) dx$ and $J = \int_{-1}^{5} g(x) dx$.

3) Deduce the integrals $\int_{-1}^{5} (f(x) + 4g(x)) dx$ and $\int_{-1}^{5} (5f(x) - 2g(x)) dx$.

2. Solution Steps

1) Plotting the functions $f$ and $g$ is not possible here.

2) Calculating the integrals $I$ and $J$:

$I = \int_{-1}^{5} f(x) dx = \int_{-1}^{0} (x+3) dx + \int_{0}^{3} (-x+3) dx + \int_{3}^{5} (x-3) dx$

$I = \left[\frac{x^2}{2} + 3x\right]_{-1}^{0} + \left[-\frac{x^2}{2} + 3x\right]_{0}^{3} + \left[\frac{x^2}{2} - 3x\right]_{3}^{5}$

$I = (0 - (\frac{1}{2} - 3)) + (-\frac{9}{2} + 9 - 0) + (\frac{25}{2} - 15 - (\frac{9}{2} - 9))$

$I = (0 - (-\frac{5}{2})) + (\frac{9}{2}) + (\frac{25}{2} - 15 - \frac{9}{2} + 9)$

$I = \frac{5}{2} + \frac{9}{2} + \frac{16}{2} - 6$

$I = \frac{30}{2} - 6 = 15 - 6 = 9$

$J = \int_{-1}^{5} g(x) dx = \int_{-1}^{1} (-\frac{1}{2}x + \frac{3}{2}) dx + \int_{1}^{5} (-\frac{1}{4}x + \frac{5}{4}) dx$

$J = \left[-\frac{1}{4}x^2 + \frac{3}{2}x\right]_{-1}^{1} + \left[-\frac{1}{8}x^2 + \frac{5}{4}x\right]_{1}^{5}$

$J = (-\frac{1}{4} + \frac{3}{2}) - (-\frac{1}{4} - \frac{3}{2}) + (-\frac{25}{8} + \frac{25}{4}) - (-\frac{1}{8} + \frac{5}{4})$

$J = -\frac{1}{4} + \frac{3}{2} + \frac{1}{4} + \frac{3}{2} - \frac{25}{8} + \frac{50}{8} + \frac{1}{8} - \frac{10}{8}$

$J = 3 + \frac{26}{8} - \frac{9}{8} = 3 + \frac{17}{8} = \frac{24 + 17}{8} = \frac{41}{8} = 5.125$

3) Deducing the integrals:

$\int_{-1}^{5} (f(x) + 4g(x)) dx = \int_{-1}^{5} f(x) dx + 4 \int_{-1}^{5} g(x) dx = I + 4J = 9 + 4(\frac{41}{8}) = 9 + \frac{41}{2} = \frac{18 + 41}{2} = \frac{59}{2} = 29.5$

$\int_{-1}^{5} (5f(x) - 2g(x)) dx = 5 \int_{-1}^{5} f(x) dx - 2 \int_{-1}^{5} g(x) dx = 5I - 2J = 5(9) - 2(\frac{41}{8}) = 45 - \frac{41}{4} = \frac{180 - 41}{4} = \frac{139}{4} = 34.75$

3. Final Answer

$I = \int_{-1}^{5} f(x) dx = 9$

$J = \int_{-1}^{5} g(x) dx = \frac{41}{8} = 5.125$

$\int_{-1}^{5} (f(x) + 4g(x)) dx = \frac{59}{2} = 29.5$

$\int_{-1}^{5} (5f(x) - 2g(x)) dx = \frac{139}{4} = 34.75$