The problem asks us to evaluate two integrals: 1. $\int e^{2x} dx$

1. Problem Description

The problem asks us to evaluate two integrals:

1. $\int e^{2x} dx$

2. $\int xe^{-2x} dx$

2. Solution Steps

Problem 15: $\int e^{2x} dx$

We can use a simple substitution to solve this integral.

Let $u = 2x$. Then $du = 2 dx$, so $dx = \frac{1}{2} du$.

$\int e^{2x} dx = \int e^u \frac{1}{2} du = \frac{1}{2} \int e^u du = \frac{1}{2} e^u + C = \frac{1}{2} e^{2x} + C$

Problem 16: $\int xe^{-2x} dx$

This integral requires integration by parts. Recall the formula:

$\int u dv = uv - \int v du$

Let $u = x$ and $dv = e^{-2x} dx$. Then $du = dx$ and $v = \int e^{-2x} dx = -\frac{1}{2} e^{-2x}$.

So,

$\int xe^{-2x} dx = x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) dx$

$= -\frac{1}{2}xe^{-2x} + \frac{1}{2} \int e^{-2x} dx$

$= -\frac{1}{2}xe^{-2x} + \frac{1}{2} \left(-\frac{1}{2} e^{-2x}\right) + C$

$= -\frac{1}{2}xe^{-2x} - \frac{1}{4} e^{-2x} + C$

$= -\frac{1}{2} e^{-2x} \left(x + \frac{1}{2}\right) + C$

3. Final Answer

1. $\int e^{2x} dx = \frac{1}{2} e^{2x} + C$

2. $\int xe^{-2x} dx = -\frac{1}{2}xe^{-2x} - \frac{1}{4} e^{-2x} + C$