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

1. Break Down the Problem

We need to evaluate the following definite integrals: a) $\int (e^{2x} + e^x) \, dx$ from 2 to 3
b) $\int \left(\frac{1}{x+2}\right)e^{-2} \, dx$ from -1 to 3

2. Relevant Concepts

To evaluate definite integrals, we need to find the antiderivative of the function within the integral and then use the Fundamental Theorem of Calculus to compute the values at the given bounds.

3. Analysis and Detail

Part a:

Evaluate the integral $\int (e^{2x} + e^x) \, dx$

1. Find the antiderivative:

   • The antiderivative of $e^{2x}$ is $\frac{1}{2}e^{2x}$.
   • The antiderivative of $e^x$ is $e^x$.

   Thus, $\int (e^{2x} + e^x) \, dx = \frac{1}{2}e^{2x} + e^x + C$

2. Evaluate from 2 to 3: $\left[\frac{1}{2}e^{2x} + e^x\right]_2^3 = \left(\frac{1}{2}e^{6} + e^{3}\right) - \left(\frac{1}{2}e^{4} + e^{2}\right)$ Simplifying this: $= \left(\frac{1}{2}e^{6} + e^{3} - \frac{1}{2}e^{4} - e^{2}\right)$

Part b:

Evaluate the integral $\int \left(\frac{1}{x+2}\right)e^{-2} \, dx$

1. Factor out the constant: Since $e^{-2}$ is a constant, we can factor it out: $e^{-2} \int \frac{1}{x+2} \, dx$

2. Find the antiderivative: The antiderivative of $\frac{1}{x+2}$ is $\ln|x+2|$.

   Thus, $\int \left(\frac{1}{x+2}\right)e^{-2} \, dx = e^{-2} \ln|x+2| + C$

3. Evaluate from -1 to 3: $e^{-2} \left[\ln|x+2|\right]_{-1}^{3} = e^{-2} \left(\ln|5| - \ln|1|\right)$ Since $\ln|1| = 0$, $= e^{-2} \ln(5)$

4. Verify and Summarize

After checking the antiderivatives and computation steps for any arithmetic errors, the evaluations of both integrals are accurate.

Final Answer

a) $\left(\frac{1}{2}e^{6} + e^{3} - \frac{1}{2}e^{4} - e^{2}\right)$

b) $e^{-2} \ln(5)$