The value of the integral `Sigma_(r=1)^(n) int_(0)^(1) f(r-1 +x) dx` is

Options

A. `int_(0)^(1) f(x) dx`
B. `int_(0)^(2) f(x) dx`
C. `int_(0)^(n) f(x) dx`
D. `n int_(0)^(1) f(x) dx`

Answer

To solve the integral \( \Sigma_{r=1}^{n} \int_{0}^{1} f(r-1+x) \, dx \), we will follow these steps:

### Step 1: Rewrite the summation and integral
We start with the expression:
\[ \Sigma_{r=1}^{n} \int_{0}^{1} f(r-1+x) \, dx \]
This means we need to evaluate the integral for each integer value of \( r \) from 1 to \( n \).

### Step 2: Change of variable in the integral
For each \( r \), we can perform a change of variable in the integral. Let:
\[ t = r - 1 + x \implies x = t - (r - 1) \implies dx = dt \]
When \( x = 0 \), \( t = r - 1 \) and when \( x = 1 \), \( t = r \).

Thus, the integral becomes:
\[ \int_{0}^{1} f(r-1+x) \, dx = \int_{r-1}^{r} f(t) \, dt \]

### Step 3: Substitute back into the summation
Now substituting this back into the summation:
\[ \Sigma_{r=1}^{n} \int_{r-1}^{r} f(t) \, dt \]

### Step 4: Combine the integrals
Now we can combine the integrals:
\[ \int_{0}^{1} f(t) \, dt + \int_{1}^{2} f(t) \, dt + \int_{2}^{3} f(t) \, dt + \ldots + \int_{n-1}^{n} f(t) \, dt \]
This can be written as:
\[ \int_{0}^{n} f(t) \, dt \]

### Final Result
Thus, the value of the integral is:
\[ \int_{0}^{n} f(t) \, dt \]