`int_(-1)^1 g'(x) dx=`

Options

A. `2g(-1)`
B. `g(1)-g(-1)`
C. `-2g(1)`
D. `2g(1)`

Answer

To solve the integral \( \int_{-1}^{1} g'(x) \, dx \), we can follow these steps:

### Step 1: Understand the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, the integral of the derivative of a function gives us the original function evaluated at the boundaries of the integral. Specifically, we have:
\[ \int g'(x) \, dx = g(x) + C \]
where \( C \) is the constant of integration.

### Step 2: Set Up the Integral with Limits
We need to evaluate the definite integral from \(-1\) to \(1\):
\[ \int_{-1}^{1} g'(x) \, dx = \left[ g(x) \right]_{-1}^{1} \]

### Step 3: Evaluate the Integral at the Limits
Now we will substitute the limits into the expression:
\[ = g(1) - g(-1) \]

### Step 4: Conclusion
Thus, the value of the integral \( \int_{-1}^{1} g'(x) \, dx \) is:
\[ g(1) - g(-1) \]

### Final Answer
The final answer is:
\[ g(1) - g(-1) \]

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