We are given that $f(x) \ge 0$ for $a \le x \le b$, and that $\int_a^b f(x) \, dx \ge (k+4+3)$. We want to find the value of $k$.

1. Problem Description

We are given that $f(x) \ge 0$ for $a \le x \le b$, and that $\int_a^b f(x) \, dx \ge (k+4+3)$. We want to find the value of $k$.

2. Solution Steps

First, simplify the inequality on the right side of the $\ge$ sign.

$k + 4 + 3 = k + 7$

So we have:

$\int_a^b f(x) \, dx \ge k + 7$

Subtract 7 from both sides of the inequality:

$\int_a^b f(x) \, dx - 7 \ge k$

Or:

$k \le \int_a^b f(x) \, dx - 7$

Since we are not given a specific function $f(x)$, or the limits of integration $a$ and $b$, we cannot evaluate the integral $\int_a^b f(x) \, dx$. However, we are given that $f(x) \ge 0$ on $[a, b]$. This means that $\int_a^b f(x) \, dx \ge 0$.

Since the problem asks for "the value of k", there might be an additional constraint that has not been clearly mentioned in the statement of the problem. If $\int_a^b f(x) dx = 0$, then we have $k \le -7$.

The question asks for *the* value of k, implying a single possible answer. However, since we only have an inequality for k, any value of k that satisfies the inequality $k \le \int_a^b f(x) \, dx - 7$ is a possible value.

I will assume that the question has an implicit assumption that the integral attains its minimum value, i.e. 0, since $f(x) \ge 0$.

So $\int_a^b f(x) dx \ge 0$, giving $k+7 \le \int_a^b f(x) dx \ge 0$. Therefore, $k+7 \le 0$, hence $k \le -7$.

Also, the integral can be 0 (e.g., if $f(x) = 0$).

Now, we are looking for "the value of $k$", which means there must be a particular value of $k$ that satisfies the condition. This must mean that $k$ can achieve its maximum possible value.

Let's assume the integral is equal to $k+7$. Since $f(x) \ge 0$, the minimum value of the integral is $0$. Then the smallest value of $k+7$ is also $0$. If $k+7=0$, $k=-7$. Then we have the inequality $0 \ge k+7$, i.e. $k \le -7$.

If the integral were to be exactly $k+7$, and the smallest possible value the integral could be is zero, then we require $k+7=0$, which implies $k=-7$.

However, if the question assumes $f(x)>0$, then the integral will be greater than zero. Hence the minimum value of the integral is a small value which is greater than

0. Therefore, $k>-7$.

Since we are looking for *the* value of k and the integral is greater than or equal to $k+7$, this suggests we must use the lower bound of the integral. Since $f(x) \ge 0$, the integral is greater than or equal to zero. Therefore, $\int_a^b f(x) dx \ge 0$.

Thus, $k+7 \le \int_a^b f(x) dx \ge 0$. Hence, $k+7 \le 0$, so $k \le -7$.

If we select the maximum possible value of $k$, which is -7, then the inequality holds.

3. Final Answer

-7