`f(x)` is a continuous function for all real values of `x` and satisfies `int_n^(n+1)f(x)dx=(n^2)/2AAn in Idot` Then `int_(-3)^5f(|x|)dx` is equal to `(19)/2` (b) `(35)/2` (c) `(17)/2` (d) none of these

Options

A. `19//2`
B. `35//2`
C. `17//2`
D. none of these

Explanation

To solve the problem, we need to evaluate the integral \( \int_{-3}^{5} f(|x|) \, dx \) given that \( f(x) \) is a continuous function satisfying the condition:

\[ \int_{n}^{n+1} f(x) \, dx = \frac{n^2}{2} \quad \text{for all } n \in \mathbb{Z}. \]

### Step-by-step Solution:

1. **Understanding the Integral**:
We start with the integral \( \int_{-3}^{5} f(|x|) \, dx \). Since \( |x| \) is an even function, we can split the integral into two parts:
\[ \int_{-3}^{5} f(|x|) \, dx = \int_{-3}^{0} f(-x) \, dx + \int_{0}^{5} f(x) \, dx. \]

2. **Changing Variables**:
For the first integral, we can make a substitution \( u = -x \), which gives \( du = -dx \). The limits change from \( x = -3 \) to \( x = 0 \) into \( u = 3 \) to \( u = 0 \). Thus:
\[ \int_{-3}^{0} f(-x) \, dx = \int_{3}^{0} f(u) (-du) = \int_{0}^{3} f(u) \, du. \]

3. **Combining Integrals**:
Now we can combine the two integrals:
\[ \int_{-3}^{5} f(|x|) \, dx = \int_{0}^{3} f(x) \, dx + \int_{0}^{5} f(x) \, dx. \]

4. **Splitting the Integral**:
We can split \( \int_{0}^{5} f(x) \, dx \) into two parts:
\[ \int_{0}^{5} f(x) \, dx = \int_{0}^{3} f(x) \, dx + \int_{3}^{5} f(x) \, dx. \]
Therefore:
\[ \int_{-3}^{5} f(|x|) \, dx = \int_{0}^{3} f(x) \, dx + \left( \int_{0}^{3} f(x) \, dx + \int_{3}^{5} f(x) \, dx \right) = 2 \int_{0}^{3} f(x) \, dx + \int_{3}^{5} f(x) \, dx. \]

5. **Using the Given Condition**:
Now we can evaluate the integrals using the given condition:
- For \( n = 0 \):
\[ \int_{0}^{1} f(x) \, dx = \frac{0^2}{2} = 0. \]
- For \( n = 1 \):
\[ \int_{1}^{2} f(x) \, dx = \frac{1^2}{2} = \frac{1}{2}. \]
- For \( n = 2 \):
\[ \int_{2}^{3} f(x) \, dx = \frac{2^2}{2} = 2. \]
- For \( n = 3 \):
\[ \int_{3}^{4} f(x) \, dx = \frac{3^2}{2} = \frac{9}{2}. \]
- For \( n = 4 \):
\[ \int_{4}^{5} f(x) \, dx = \frac{4^2}{2} = 8. \]

6. **Calculating the Total**:
Now we can calculate:
\[ \int_{0}^{3} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx + \int_{2}^{3} f(x) \, dx = 0 + \frac{1}{2} + 2 = \frac{5}{2}. \]
Therefore:
\[ 2 \int_{0}^{3} f(x) \, dx = 2 \cdot \frac{5}{2} = 5. \]
And:
\[ \int_{3}^{5} f(x) \, dx = \int_{3}^{4} f(x) \, dx + \int_{4}^{5} f(x) \, dx = \frac{9}{2} + 8 = \frac{25}{2}. \]

7. **Final Calculation**:
Now we can combine everything:
\[ \int_{-3}^{5} f(|x|) \, dx = 5 + \frac{25}{2} = \frac{10}{2} + \frac{25}{2} = \frac{35}{2}. \]

### Conclusion:
Thus, the value of \( \int_{-3}^{5} f(|x|) \, dx \) is \( \frac{35}{2} \).