If `f(x)` is continuous for all real values of `x ,` then `sum_(r=1)^nint_0^1f(r-1+x)dx `is equal to (a)`int_0^nf(x)dx` (b) `int_0^1f(x)dx` (c)`int_0^1f(x)dx` (d) `(n-1)int_0^1f(x)dx`

Options

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

Explanation

To solve the problem, we need to evaluate the sum

\[ \sum_{r=1}^{n} \int_0^1 f(r-1+x) \, dx. \]

### Step 1: Rewrite the sum
We can express the sum as follows:

\[ \sum_{r=1}^{n} \int_0^1 f(r-1+x) \, dx = \int_0^1 f(0+x) \, dx + \int_0^1 f(1+x) \, dx + \int_0^1 f(2+x) \, dx + \ldots + \int_0^1 f((n-1)+x) \, dx. \]

### Step 2: Change of variable
For each integral, we can perform a change of variable. Let \( u = r - 1 + x \). Then, when \( x = 0 \), \( u = r - 1 \) and when \( x = 1 \), \( u = r \). Thus, we can rewrite each integral:

\[ \int_0^1 f(r-1+x) \, dx = \int_{r-1}^{r} f(u) \, du. \]

### Step 3: Combine the integrals
Now, we can combine all these integrals:

\[ \sum_{r=1}^{n} \int_{r-1}^{r} f(u) \, du = \int_0^1 f(u) \, du + \int_1^2 f(u) \, du + \int_2^3 f(u) \, du + \ldots + \int_{n-1}^{n} f(u) \, du. \]

### Step 4: Evaluate the total integral
The total integral from \( 0 \) to \( n \) can be expressed as:

\[ \int_0^n f(u) \, du. \]

### Conclusion
Thus, we conclude that

\[ \sum_{r=1}^{n} \int_0^1 f(r-1+x) \, dx = \int_0^n f(x) \, dx. \]

Therefore, the answer is:

\[ \text{(a) } \int_0^n f(x) \, dx. \]

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