evaluate the definite integral below. ∫₂⁴(3 + 2eˣ) dx

Explanation:

Step1: Use integral rules

We know that $\int(a + b)dx=\int a dx+\int b dx$ and $\int kdx=kx + C$ ($k$ is a constant), $\int e^{x}dx = e^{x}+C$. So $\int_{2}^{4}(3 + 2e^{x})dx=\int_{2}^{4}3dx+\int_{2}^{4}2e^{x}dx$.

Step2: Evaluate $\int_{2}^{4}3dx$

Using the rule $\int_{a}^{b}kdx=k(b - a)$ with $k = 3$, $a = 2$, $b = 4$, we get $\int_{2}^{4}3dx=3\times(4 - 2)=6$.

Step3: Evaluate $\int_{2}^{4}2e^{x}dx$

Using the rule $\int_{a}^{b}ke^{x}dx=k(e^{b}-e^{a})$ with $k = 2$, $a = 2$, $b = 4$, we get $\int_{2}^{4}2e^{x}dx=2(e^{4}-e^{2})$.

Step4: Combine results

$\int_{2}^{4}(3 + 2e^{x})dx=6+2(e^{4}-e^{2})=6 + 2e^{4}-2e^{2}$.

Answer:

$6 + 2e^{4}-2e^{2}$