`int_(a+c)^(b+c) f(x) dx` is equal to

Options

A. `int_(a)^(b) f(x-c) dx`
B. `int_(a)^(b) f(x+c) dx`
C. `int_(a)^(b) f(x)dx`
D. `int_(a-c)^(b-c) f(x) dx`

Answer

To solve the integral \( \int_{a+c}^{b+c} f(x) \, dx \), we can use a substitution method. Here’s a step-by-step solution:

### Step 1: Substitution
Let’s perform the substitution \( t = x - c \). This means that \( x = t + c \).

### Step 2: Change of Limits
Now, we need to change the limits of integration according to our substitution:
- When \( x = a + c \), then \( t = (a + c) - c = a \).
- When \( x = b + c \), then \( t = (b + c) - c = b \).

### Step 3: Change of Differential
The differential \( dx \) will change as follows:
- Since \( t = x - c \), we have \( dt = dx \).

### Step 4: Rewrite the Integral
Now we can rewrite the integral in terms of \( t \):
\[ \int_{a+c}^{b+c} f(x) \, dx = \int_{a}^{b} f(t + c) \, dt \]

### Step 5: Final Result
Thus, the integral \( \int_{a+c}^{b+c} f(x) \, dx \) is equal to:
\[ \int_{a}^{b} f(t + c) \, dt \]

### Conclusion
The final answer is:
\[ \int_{a+c}^{b+c} f(x) \, dx = \int_{a}^{b} f(t + c) \, dt \]

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