Let `f(x)={1-|x|,|x| leq 1 and 0,|x| lt 1 and g(x)=f(x-)+f(x + 1)`, for all `x in R`.Then,the value of `int_-3^3 g(x)dx` is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to evaluate the integral \( \int_{-3}^{3} g(x) \, dx \), 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: Understand the function \( f(x) \)

The function \( f(x) \) is a triangular function that peaks at \( x = 0 \) and is equal to 1 at that point. It decreases linearly to 0 at \( x = -1 \) and \( x = 1 \). For values of \( x \) outside the interval \([-1, 1]\), \( f(x) = 0 \).

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

1. **For \( f(x-1) \)**:
- \( f(x-1) = 1 - |x-1| \) if \( |x-1| \leq 1 \) (which means \( 0 \leq x \leq 2 \))
- \( f(x-1) = 0 \) if \( |x-1| > 1 \) (which means \( x < 0 \) or \( x > 2 \))

2. **For \( f(x+1) \)**:
- \( f(x+1) = 1 - |x+1| \) if \( |x+1| \leq 1 \) (which means \( -2 \leq x \leq 0 \))
- \( f(x+1) = 0 \) if \( |x+1| > 1 \) (which means \( x < -2 \) or \( x > 0 \))

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

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

- For \( 0 \leq x \leq 2 \):
- \( g(x) = f(x-1) + f(x+1) = (1 - |x-1|) + 0 = 1 - (x-1) = 2 - x \)

- For \( x < -2 \) or \( x > 2 \):
- \( g(x) = 0 \)

### Step 4: Determine the intervals for \( g(x) \)

Thus, we have:
- \( g(x) = x + 2 \) for \( -2 \leq x < 0 \)
- \( g(x) = 2 - x \) for \( 0 \leq x \leq 2 \)
- \( g(x) = 0 \) for \( x < -2 \) or \( x > 2 \)

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

We can break the integral into three parts:
1. From \( -3 \) to \( -2 \): \( g(x) = 0 \)
2. From \( -2 \) to \( 0 \): \( g(x) = x + 2 \)
3. From \( 0 \) to \( 2 \): \( g(x) = 2 - x \)
4. From \( 2 \) to \( 3 \): \( g(x) = 0 \)

Calculating the integral:

\[ \int_{-3}^{3} g(x) \, dx = \int_{-2}^{0} (x + 2) \, dx + \int_{0}^{2} (2 - x) \, dx \]

### Step 6: Evaluate the integrals

1. **For \( \int_{-2}^{0} (x + 2) \, dx \)**:
\[ = \left[ \frac{x^2}{2} + 2x \right]_{-2}^{0} = \left( 0 + 0 \right) - \left( \frac{(-2)^2}{2} + 2(-2) \right) = 0 - (2 - 4) = 2 \]

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

### Step 7: Combine the results

Thus, the total integral is:
\[ \int_{-3}^{3} g(x) \, dx = 2 + 2 = 4 \]

### Final Answer
The value of \( \int_{-3}^{3} g(x) \, dx \) is \( 4 \).

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