. Evaluate the following definite integrals:a) ∫ (𝑒2𝑥 + 𝑒𝑥)𝑑𝑥32 (3)b) ∫ ( 1𝑥+2)𝑒−2−1 𝑑𝑥

1. Break Down the Problem

We have two definite integrals to evaluate: a) $\int_{2}^{3} (e^{2x} + e^{x}) \, dx$
b) $\int_{-1}^{-2} \left(\frac{1}{x+2}\right) e^{-2} \, dx$

2. Relevant Concepts

For both integrals, we need to find the antiderivative and then evaluate it at the given limits.

a) The integral of $(e^{2x} + e^{x})$ can be broken down as:
$\int (e^{2x} + e^{x}) \, dx = \int e^{2x} \, dx + \int e^{x} \, dx$

b) The integral of $\frac{1}{x+2} e^{-2}$ can be simplified since $e^{-2}$ is a constant multiplier. So: $\int \left(\frac{1}{x+2}\right) e^{-2} \, dx = e^{-2} \int \frac{1}{x+2} \, dx$

3. Analysis and Detail

a) Compute $\int_{2}^{3} (e^{2x} + e^{x}) \, dx$

1. Find the antiderivative:
   $\int e^{2x} \, dx = \frac{1}{2} e^{2x} + C$
   $\int e^{x} \, dx = e^{x} + C$
   Combine:
   $\int (e^{2x} + e^{x}) \, dx = \frac{1}{2} e^{2x} + e^{x} + C$

2. Evaluate from 2 to 3:
   $F(3) = \frac{1}{2} e^{6} + e^{3}$
   $F(2) = \frac{1}{2} e^{4} + e^{2}$
   Thus,
   $\int_{2}^{3} (e^{2x} + e^{x}) \, dx = F(3) - F(2) = \left( \frac{1}{2} e^{6} + e^{3} \right) - \left( \frac{1}{2} e^{4} + e^{2} \right)$

b) Compute $\int_{-1}^{-2} \left(\frac{1}{x+2}\right) e^{-2} \, dx$

1. Find the antiderivative: $\int \frac{1}{x+2} \, dx = \ln|x+2| + C$

2. Evaluate from -1 to -2 (note the limits are reversed so we will flip the sign): $F(-1) = \ln|1| = 0$
   $F(-2) = \ln|0| \text{ (undefined, hence cannot evaluate directly)}$

4. Verify and Summarize

For $b)$, because $F(-2)$ is not defined (as $\ln(0)$ is undefined), we note that the integral does not yield a well-defined value.

Final Answer

a) The final result for the first integral is:
$\int_{2}^{3} (e^{2x} + e^{x}) \, dx = \frac{1}{2} e^{6} + e^{3} - \frac{1}{2} e^{4} - e^{2}$

b) The integral $\int_{-1}^{-2} \left(\frac{1}{x+2}\right) e^{-2} \, dx$ is undefined due to the logarithmic singularity at $x = -2$.