If `int_(-2)^(5) f(x) dx = 4` and `int_(0)^(5) {1+f(x)}dx = 7`, then what is `int_(-2)^(0) f(x) dx` equal to ?

Options

A. `-3`
B. `2`
C. `3`
D. `5`

Explanation

To solve the problem, we need to find the value of the integral \(\int_{-2}^{0} f(x) \, dx\) given the following information:

1. \(\int_{-2}^{5} f(x) \, dx = 4\)
2. \(\int_{0}^{5} (1 + f(x)) \, dx = 7\)

Let's break this down step by step.

### Step 1: Simplify the second integral
We can rewrite the second integral:
\[ \int_{0}^{5} (1 + f(x)) \, dx = \int_{0}^{5} 1 \, dx + \int_{0}^{5} f(x) \, dx \]
Calculating \(\int_{0}^{5} 1 \, dx\):
\[ \int_{0}^{5} 1 \, dx = [x]_{0}^{5} = 5 - 0 = 5 \]
Thus, we have:
\[ \int_{0}^{5} (1 + f(x)) \, dx = 5 + \int_{0}^{5} f(x) \, dx \]
Setting this equal to 7:
\[ 5 + \int_{0}^{5} f(x) \, dx = 7 \]
Subtracting 5 from both sides gives:
\[ \int_{0}^{5} f(x) \, dx = 2 \]

### Step 2: Use the property of definite integrals
Now we can relate the integrals from \(-2\) to \(5\) and from \(-2\) to \(0\):
\[ \int_{-2}^{5} f(x) \, dx = \int_{-2}^{0} f(x) \, dx + \int_{0}^{5} f(x) \, dx \]
We know:
\[ \int_{-2}^{5} f(x) \, dx = 4 \quad \text{and} \quad \int_{0}^{5} f(x) \, dx = 2 \]
Substituting these values into the equation gives:
\[ 4 = \int_{-2}^{0} f(x) \, dx + 2 \]

### Step 3: Solve for the unknown integral
Now, we can solve for \(\int_{-2}^{0} f(x) \, dx\):
\[ \int_{-2}^{0} f(x) \, dx = 4 - 2 = 2 \]

### Final Answer
Thus, the value of \(\int_{-2}^{0} f(x) \, dx\) is:
\[ \boxed{2} \]