We are given that $(2 + 4x^2) \le f(x)$. We need to find a correct estimate for the integral $\int_0^2 f(x) \, dx$.

1. Problem Description

We are given that $(2 + 4x^2) \le f(x)$. We need to find a correct estimate for the integral $\int_0^2 f(x) \, dx$.

2. Solution Steps

Since $(2 + 4x^2) \le f(x)$, we can integrate both sides of the inequality from 0 to 2:

$\int_0^2 (2 + 4x^2) \, dx \le \int_0^2 f(x) \, dx$

Now, let's evaluate the integral on the left side:

$\int_0^2 (2 + 4x^2) \, dx = \int_0^2 2 \, dx + \int_0^2 4x^2 \, dx$

$= 2 \int_0^2 dx + 4 \int_0^2 x^2 \, dx$

$= 2[x]_0^2 + 4 \left[ \frac{x^3}{3} \right]_0^2$

$= 2(2 - 0) + 4 \left( \frac{2^3}{3} - \frac{0^3}{3} \right)$

$= 2(2) + 4 \left( \frac{8}{3} \right)$

$= 4 + \frac{32}{3}$

$= \frac{12}{3} + \frac{32}{3}$

$= \frac{44}{3}$

Since $\frac{44}{3} \approx 14.6667$, we have

$\frac{44}{3} \le \int_0^2 f(x) \, dx$

Therefore, $\int_0^2 f(x) \, dx \ge \frac{44}{3} \approx 14.6667$.

Comparing this with the given options:

A. $\int_0^2 f(x) \, dx \ge 14.67$

B. $\int_0^2 f(x) \, dx \le 14.67$

C. $\int_0^2 f(x) \, dx \le 2$

D. $\int_0^2 f(x) \, dx \ge 0$

Since $\frac{44}{3} \approx 14.6667$, and $14.6667 < 14.67$, it follows that $\int_0^2 f(x) dx \ge 14.6667$, which is very close to 14.

6

7. Also, $f(x) \ge 2 + 4x^2 \ge 2$, therefore $\int_0^2 f(x) dx \ge \int_0^2 2 dx = 4$. Therefore, option C can be ruled out. Option D is too weak of a bound, as $f(x)$ must be positive on the interval $[0, 2]$. Option B is not true, as $\int_0^2 f(x) dx \ge \frac{44}{3}$. Since $\frac{44}{3} \approx 14.6667 < 14.67$, it follows that $\int_0^2 f(x) dx \ge 14.6667$. We also know that $\int_0^2 f(x) dx \ge 14.67$ can be false, but the problem asks which of the following is a correct estimate. The integral is greater than or equal to $\frac{44}{3}$, which is about 14.67, so option A is correct.

3. Final Answer

A. $\int_0^2 f(x) \, dx \ge 14.67$