If `d/(dx){phi(x)}=f(x)," then "int_1^2f(x)dx` is equal to

Options

A. `f(1)-f(2)`
B. `phi(1)-phi(2)`
C. `f(2)-f(1)`
D. `phi(2)-phi(1)`

Answer

To solve the problem, we start with the given information and apply the Fundamental Theorem of Calculus.

### Step-by-Step Solution:

1. **Given Information**: We know that \( \frac{d}{dx} \phi(x) = f(x) \).

2. **Set Up the Integral**: We need to evaluate the integral \( \int_1^2 f(x) \, dx \).

3. **Apply the Fundamental Theorem of Calculus**: According to the Fundamental Theorem of Calculus, if \( F(x) \) is an antiderivative of \( f(x) \), then:
\[ \int_a^b f(x) \, dx = F(b) - F(a) \]
Here, since \( \frac{d}{dx} \phi(x) = f(x) \), we can say that \( \phi(x) \) is an antiderivative of \( f(x) \).

4. **Evaluate the Integral**: Therefore, we can write:
\[ \int_1^2 f(x) \, dx = \phi(2) - \phi(1) \]

5. **Final Result**: Thus, the value of the integral \( \int_1^2 f(x) \, dx \) is:
\[ \phi(2) - \phi(1) \]