Question: top-left: Consider the function $F$ such that $F'(x) = 4e^x$. If $F(-3) = 10$, determine $F(2)$.
bottom-right: (5 Marks)
1. **State the problem:** We are given the derivative of a function $F$, which is $F'(x) = 4e^x$, and a value of the function at $x = -3$, $F(-3) = 10$. We need to find the value of $F(2)$.
2. **Recall the formula:** Since $F'(x)$ is the derivative of $F(x)$, we can find $F(x)$ by integrating $F'(x)$:
$$F(x) = \int F'(x) \, dx + C$$
where $C$ is the constant of integration.
3. **Integrate $F'(x)$:**
$$F(x) = \int 4e^x \, dx + C = 4 \int e^x \, dx + C = 4e^x + C$$
4. **Use the initial condition $F(-3) = 10$ to find $C$:**
$$10 = F(-3) = 4e^{-3} + C$$
Solve for $C$:
$$C = 10 - 4e^{-3}$$
5. **Write the full function:**
$$F(x) = 4e^x + 10 - 4e^{-3}$$
6. **Find $F(2)$:**
$$F(2) = 4e^2 + 10 - 4e^{-3}$$
This is the exact value of $F(2)$.
**Final answer:**
$$\boxed{F(2) = 4e^2 + 10 - 4e^{-3}}$$