`int_(-a)^(a) f (x) dx` is equal to

Options

A. `int_(0)^(a) [f(x) + f(-x)]dx`
B. `int_(0)^(a) [f(x)- f(-x)] dx`
C. `2int_(0)^(a) f(x) dx`
D. 0

Answer

To solve the integral \( \int_{-a}^{a} f(x) \, dx \), we will break it down step by step.

### Step 1: Break the Integral
We start by splitting the integral into two parts:
\[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx \]

### Step 2: Change of Variable for the First Integral
For the first integral \( I_1 = \int_{-a}^{0} f(x) \, dx \), we will use the substitution \( x = -t \). Then, we have:
- When \( x = -a \), \( t = a \)
- When \( x = 0 \), \( t = 0 \)
- The differential changes as \( dx = -dt \)

Thus, we can rewrite \( I_1 \):
\[ I_1 = \int_{-a}^{0} f(x) \, dx = \int_{a}^{0} f(-t)(-dt) = \int_{0}^{a} f(-t) \, dt \]

### Step 3: Combine the Integrals
Now we can express the original integral as:
\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(-t) \, dt + \int_{0}^{a} f(x) \, dx \]
Since we are changing the variable \( t \) back to \( x \) in the second integral, we have:
\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(-x) \, dx + \int_{0}^{a} f(x) \, dx \]

### Step 4: Final Expression
Combining both integrals gives us:
\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} (f(x) + f(-x)) \, dx \]

### Conclusion
Thus, the final result for the integral \( \int_{-a}^{a} f(x) \, dx \) is:
\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} (f(x) + f(-x)) \, dx \]