Suppose for every integer `n, .int_(n)^(n+1) f(x)dx = n^(2)`. The value of `int_(-2)^(4) f(x)dx` is :

Answer

To solve the problem, we need to evaluate the integral \(\int_{-2}^{4} f(x) \, dx\) given that for every integer \(n\), \(\int_{n}^{n+1} f(x) \, dx = n^2\).

### Step-by-Step Solution:

1. **Define the Integral**:
Let \(I = \int_{-2}^{4} f(x) \, dx\).

2. **Split the Integral**:
We can split the integral from \(-2\) to \(4\) into smaller intervals:
\[ I = \int_{-2}^{-1} f(x) \, dx + \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx + \int_{2}^{3} f(x) \, dx + \int_{3}^{4} f(x) \, dx \]

3. **Apply the Given Condition**:
According to the problem, for each integer \(n\):
\[ \int_{n}^{n+1} f(x) \, dx = n^2 \]
We can evaluate each of the integrals:
- For \(n = -2\): \(\int_{-2}^{-1} f(x) \, dx = (-2)^2 = 4\)
- For \(n = -1\): \(\int_{-1}^{0} f(x) \, dx = (-1)^2 = 1\)
- For \(n = 0\): \(\int_{0}^{1} f(x) \, dx = 0^2 = 0\)
- For \(n = 1\): \(\int_{1}^{2} f(x) \, dx = 1^2 = 1\)
- For \(n = 2\): \(\int_{2}^{3} f(x) \, dx = 2^2 = 4\)
- For \(n = 3\): \(\int_{3}^{4} f(x) \, dx = 3^2 = 9\)

4. **Combine the Results**:
Now, we can add all these results together to find \(I\):
\[ I = 4 + 1 + 0 + 1 + 4 + 9 \]

5. **Calculate the Final Value**:
Performing the addition:
\[ I = 4 + 1 = 5 \]
\[ I = 5 + 0 = 5 \]
\[ I = 5 + 1 = 6 \]
\[ I = 6 + 4 = 10 \]
\[ I = 10 + 9 = 19 \]

Thus, the value of \(\int_{-2}^{4} f(x) \, dx\) is \(19\).

### Final Answer:
\[ \int_{-2}^{4} f(x) \, dx = 19 \]