If `f(x)={(1-|x| ,, |x|lt=1),(0 ,, |x|>1):}` and `g(x)=f(x-1)+f(x+1),` then find the value of `int_(-3)^5g(x)dx`.

Explanation

To solve the problem, we need to find the value of the integral \(\int_{-3}^{5} g(x) \, dx\), where \(g(x) = f(x-1) + f(x+1)\) and \(f(x)\) is defined as follows:

\[ f(x) = \begin{cases} 1 - |x| & \text{if } |x| \leq 1 \\ 0 & \text{if } |x| > 1 \end{cases} \]

### Step 1: Analyze the function \(f(x)\)

The function \(f(x)\) is non-zero only in the interval \([-1, 1]\). Specifically:
- \(f(x) = 1 - |x|\) for \(-1 \leq x \leq 1\)
- \(f(x) = 0\) for \(x < -1\) and \(x > 1\)

### Step 2: Determine \(f(x-1)\) and \(f(x+1)\)

1. **For \(f(x-1)\)**:
- \(f(x-1) = 1 - |x-1|\) for \(0 \leq x \leq 2\) (since \(|x-1| \leq 1\) implies \(0 \leq x \leq 2\))
- \(f(x-1) = 0\) for \(x < 0\) and \(x > 2\)

2. **For \(f(x+1)\)**:
- \(f(x+1) = 1 - |x+1|\) for \(-2 \leq x \leq 0\) (since \(|x+1| \leq 1\) implies \(-2 \leq x \leq 0\))
- \(f(x+1) = 0\) for \(x < -2\) and \(x > 0\)

### Step 3: Combine \(f(x-1)\) and \(f(x+1)\) to find \(g(x)\)

Now, we can express \(g(x)\):
- For \(-2 \leq x < 0\):
\[ g(x) = f(x-1) + f(x+1) = 0 + (1 - |x+1|) = 1 - (x + 1) = 0 - x \]
- For \(0 \leq x \leq 2\):
\[ g(x) = f(x-1) + f(x+1) = (1 - |x-1|) + 0 = 1 - (x - 1) = 2 - x \]
- For \(x < -2\) and \(x > 2\):
\[ g(x) = 0 \]

### Step 4: Define \(g(x)\) piecewise

Thus, we can summarize \(g(x)\) as:
\[ g(x) = \begin{cases} -x & \text{if } -2 \leq x < 0 \\ 2 - x & \text{if } 0 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases} \]

### Step 5: Calculate the integral \(\int_{-3}^{5} g(x) \, dx\)

Since \(g(x) = 0\) for \(x < -2\) and \(x > 2\), we only need to integrate from \(-2\) to \(2\):
\[ \int_{-3}^{5} g(x) \, dx = \int_{-2}^{0} (-x) \, dx + \int_{0}^{2} (2 - x) \, dx \]

#### Step 5.1: Calculate \(\int_{-2}^{0} (-x) \, dx\)

\[ \int_{-2}^{0} (-x) \, dx = \left[-\frac{x^2}{2}\right]_{-2}^{0} = 0 - \left[-\frac{(-2)^2}{2}\right] = 0 - \left[-\frac{4}{2}\right] = 2 \]

#### Step 5.2: Calculate \(\int_{0}^{2} (2 - x) \, dx\)

\[ \int_{0}^{2} (2 - x) \, dx = \left[2x - \frac{x^2}{2}\right]_{0}^{2} = \left[2(2) - \frac{(2)^2}{2}\right] - \left[0\right] = 4 - 2 = 2 \]

### Step 6: Combine the results

Now, adding both areas:
\[ \int_{-3}^{5} g(x) \, dx = 2 + 2 = 4 \]

### Final Answer

Thus, the value of \(\int_{-3}^{5} g(x) \, dx\) is \(\boxed{4}\).