Let `f(x)` be a contiuous function such a `int_n^(n+1) f(x)dx=n^3, n in Z.` Then, the value of the intergral `int_-3^3 f(x) dx` (A) 9 (B) `-27` (C) `-9` (D) none of these

Options

A. 9
B. `-27`
C. `-9`
D. 27

Answer

To solve the problem, we need to evaluate the integral \( \int_{-3}^{3} f(x) \, dx \) given that \( \int_{n}^{n+1} f(x) \, dx = n^3 \) for \( n \in \mathbb{Z} \).

### Step 1: Break down the integral from -3 to 3
We can express the integral \( \int_{-3}^{3} f(x) \, dx \) as a sum of integrals over smaller intervals:
\[ \int_{-3}^{3} f(x) \, dx = \int_{-3}^{-2} f(x) \, dx + \int_{-2}^{-1} f(x) \, dx + \int_{-1}^{0} f(x) \, dx + \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx + \int_{2}^{3} f(x) \, dx \]

### Step 2: Apply the given condition for each interval
According to the problem, for each integer \( n \):
- For \( n = -3 \):
\[ \int_{-3}^{-2} f(x) \, dx = (-3)^3 = -27 \]
- For \( n = -2 \):
\[ \int_{-2}^{-1} f(x) \, dx = (-2)^3 = -8 \]
- For \( n = -1 \):
\[ \int_{-1}^{0} f(x) \, dx = (-1)^3 = -1 \]
- For \( n = 0 \):
\[ \int_{0}^{1} f(x) \, dx = 0^3 = 0 \]
- For \( n = 1 \):
\[ \int_{1}^{2} f(x) \, dx = 1^3 = 1 \]
- For \( n = 2 \):
\[ \int_{2}^{3} f(x) \, dx = 2^3 = 8 \]

### Step 3: Combine all the results
Now we can combine all these results:
\[ \int_{-3}^{3} f(x) \, dx = (-27) + (-8) + (-1) + 0 + 1 + 8 \]

### Step 4: Simplify the expression
Calculating this step by step:
- First, combine the negative values:
\[ -27 - 8 - 1 = -36 \]
- Now add the positive values:
\[ -36 + 0 + 1 + 8 = -36 + 9 = -27 \]

### Final Answer
Thus, the value of the integral \( \int_{-3}^{3} f(x) \, dx \) is:
\[ \boxed{-27} \]