The problem has three parts: (a) Given the set $A = [-7, 8) \cup [11, \infty)$, find the complement of $A$, denoted as $A'$. (b) Given the set $A = [-7, 8) \cup [11, \infty)$ and $B = [0, 20]$, find the intersection of $A$ and $B$, denoted as $A \cap B$. (c) Evaluate the integral $\int 7xe^{2x} dx$.

1. Problem Description

The problem has three parts:

(a) Given the set $A = [-7, 8) \cup [11, \infty)$, find the complement of $A$, denoted as $A'$.

(b) Given the set $A = [-7, 8) \cup [11, \infty)$ and $B = [0, 20]$, find the intersection of $A$ and $B$, denoted as $A \cap B$.

(c) Evaluate the integral $\int 7xe^{2x} dx$.

2. Solution Steps

(a) Finding $A'$:

The universal set is the set of real numbers, $\mathbb{R} = (-\infty, \infty)$. The complement of $A$ is the set of all elements in the universal set that are not in $A$.

$A = [-7, 8) \cup [11, \infty)$.

$A' = (-\infty, -7) \cup [8, 11)$.

(b) Finding $A \cap B$:

$A = [-7, 8) \cup [11, \infty)$ and $B = [0, 20]$.

$A \cap B = ([-7, 8) \cup [11, \infty)) \cap [0, 20] = ([-7, 8) \cap [0, 20]) \cup ([11, \infty) \cap [0, 20])$.

$[-7, 8) \cap [0, 20] = [0, 8)$.

$[11, \infty) \cap [0, 20] = [11, 20]$.

Therefore, $A \cap B = [0, 8) \cup [11, 20]$.

(c) Evaluating the integral $\int 7xe^{2x} dx$:

We will use integration by parts.

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

Let $u = x$ and $dv = 7e^{2x} dx$. Then $du = dx$ and $v = \int 7e^{2x} dx = \frac{7}{2}e^{2x}$.

$\int 7xe^{2x} dx = x(\frac{7}{2}e^{2x}) - \int \frac{7}{2}e^{2x} dx = \frac{7}{2}xe^{2x} - \frac{7}{2} \int e^{2x} dx = \frac{7}{2}xe^{2x} - \frac{7}{2} (\frac{1}{2}e^{2x}) + C = \frac{7}{2}xe^{2x} - \frac{7}{4}e^{2x} + C$.

3. Final Answer

(a) $A' = (-\infty, -7) \cup [8, 11)$

(b) $A \cap B = [0, 8) \cup [11, 20]$

(c) $\int 7xe^{2x} dx = \frac{7}{2}xe^{2x} - \frac{7}{4}e^{2x} + C$