1. **Problem:** State the fundamental theorem of calculus, and hence evaluate the definite integral $$\int (3x^2 + 6) \, dx$$. 2. **Fundamental Theorem of Calculus:** It states that if $F$ is an antiderivative of $f$ on an interval $[a,b]$, then $$\int_a^b f(x) \, dx = F(b) - F(a).$$ This means the definite integral of a function can be found by evaluating its antiderivative at the bounds and subtracting. 3. **Find the antiderivative:** For the integrand $3x^2 + 6$, integrate term-by-term: $$\int (3x^2 + 6) \, dx = \int 3x^2 \, dx + \int 6 \, dx = 3 \int x^2 \, dx + 6x + C.$$ 4. **Calculate each integral:** $$3 \int x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3,$$ so the antiderivative is $$F(x) = x^3 + 6x + C.$$ 5. **Evaluate the definite integral:** Since no limits are given, the integral is indefinite. If limits $a$ and $b$ were given, we would compute $$\int_a^b (3x^2 + 6) \, dx = F(b) - F(a) = (b^3 + 6b) - (a^3 + 6a).$$ **Final answer:** $$\int (3x^2 + 6) \, dx = x^3 + 6x + C.$$