If for every integer `n, int_(n)^(n+1) f(x) dx= n^(2)`, then the value of `int_(-2)^(4) f(x) dx` is

Options

A. 16
B. 14
C. 19
D. None of these

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

2. **Apply the Given Condition**:
According to the given condition, we can evaluate each of these 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 \( n = 3 \):
\[ \int_{3}^{4} f(x) \, dx = 3^2 = 9. \]

3. **Sum the Results**:
Now, we sum all the evaluated integrals:
\[ \int_{-2}^{4} f(x) \, dx = 4 + 1 + 0 + 1 + 4 + 9. \]

4. **Calculate the Total**:
Adding these values together:
\[ 4 + 1 = 5, \]
\[ 5 + 0 = 5, \]
\[ 5 + 1 = 6, \]
\[ 6 + 4 = 10, \]
\[ 10 + 9 = 19. \]

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

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