Simplify the following expression. \( \frac{d}{d x} \int_{3}^{x}\left(7 t^{2}+1+3\right) d t \) \( \frac{d}{d x} \int_{9}^{x}\left(7 t^{2}+1+3\right) d x=\square \)

Let's break this down step by step. We'll simplify the expression using the Fundamental Theorem of Calculus, which states that if \( F(t) \) is an antiderivative of \( f(t) \), then: \[ \frac{d}{dx} \int_a^{x} f(t) \, dt = f(x) \] Now, let's simplify the integrand: \[ 7t^2 + 1 + 3 = 7t^2 + 4 \] Applying the Fundamental Theorem of Calculus, the expression can be rewritten as: \[ \frac{d}{dx} \int_{3}^{x} (7t^2 + 4) \, dt = 7x^2 + 4 \] Thus, the answer is: \[ 7x^2 + 4 \] Now, for the second part we look at: \[ \frac{d}{dx} \int_{9}^{x} (7t^2 + 4) \, dt = 7x^2 + 4 \] So for both cases, \[ \frac{d}{d x} \int_{3}^{x}(7 t^{2}+4) dt = 7x^2 + 4 \quad \text{and} \quad \frac{d}{d x} \int_{9}^{x}(7 t^{2}+4) dt = 7x^2 + 4 \] The answer is: \[ 7x^2 + 4 \quad \text{for both cases!} \]