The value of integral ` sum _(k=1)^(n) int _(0)^(1) f(k - 1+x) dx ` is

Options

A. `underset (0)overset(1) intf(x)dx`
B. `underset (0)overset(2) intf(x)dx`
C. `underset (0)overset(2) intf(x)dx`
D. `n underset (0)overset(2) intf(x)dx`

Answer

To solve the integral \( \sum_{k=1}^{n} \int_{0}^{1} f(k - 1 + x) \, dx \), we will follow a systematic approach.

### Step 1: Change of Variables
Let’s perform a change of variables in the integral. We set:
\[ t = k - 1 + x \]
Then, differentiating both sides gives:
\[ dx = dt \]
Now, we need to change the limits of integration. When \( x = 0 \):
\[ t = k - 1 + 0 = k - 1 \]
When \( x = 1 \):
\[ t = k - 1 + 1 = k \]
Thus, the integral becomes:
\[ \int_{0}^{1} f(k - 1 + x) \, dx = \int_{k - 1}^{k} f(t) \, dt \]

### Step 2: Substitute Back into the Summation
Now we substitute this back into the summation:
\[ \sum_{k=1}^{n} \int_{0}^{1} f(k - 1 + x) \, dx = \sum_{k=1}^{n} \int_{k - 1}^{k} f(t) \, dt \]

### Step 3: Combine the Integrals
Next, we can combine the integrals:
\[ \sum_{k=1}^{n} \int_{k - 1}^{k} f(t) \, dt = \int_{0}^{n} f(t) \, dt \]
This is because the intervals \([k-1, k]\) for \(k = 1\) to \(n\) cover the entire interval \([0, n]\).

### Step 4: Final Result
Thus, we conclude:
\[ \sum_{k=1}^{n} \int_{0}^{1} f(k - 1 + x) \, dx = \int_{0}^{n} f(t) \, dt \]

### Conclusion
The value of the integral \( \sum_{k=1}^{n} \int_{0}^{1} f(k - 1 + x) \, dx \) is:
\[ \int_{0}^{n} f(t) \, dt \]