Let's apply the Trapezoidal Rule with the given values step by step.
#### Step 1: Determine [tex]\(\Delta x\)[/tex]
For [tex]\(n=4\)[/tex], the interval [tex]\([0, 1]\)[/tex] is divided into 4 subintervals.
[tex]\[
\Delta x = \frac{1-0}{4} = 0.25
\][/tex]
#### Step 2: Calculate [tex]\(x\)[/tex] values
The [tex]\(x\)[/tex] values are calculated as follows:
[tex]\[
\begin{array}{l}
x_0 = 0 \cdot 0.25 = 0 \\
x_1 = 1 \cdot 0.25 = 0.25 \\
x_2 = 2 \cdot 0.25 = 0.5 \\
x_3 = 3 \cdot 0.25 = 0.75 \\
x_4 = 4 \cdot 0.25 = 1 \\
\end{array}
\][/tex]
#### Step 3: Formulate the Trapezoidal Rule
Given the function [tex]\( f(x) = \frac{2}{(x+2)^2} \)[/tex], we can apply the Trapezoidal Rule:
[tex]\[
\int_0^1 \frac{2}{(x+2)^2} \, dx \approx \frac{\Delta x}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4) \right]
\][/tex]
Substituting the [tex]\(x\)[/tex] values:
[tex]\[
\begin{array}{l}
\int_0^1 \frac{2}{(x+2)^2} \, dx \approx \frac{0.25}{2} \left[ \frac{2}{(0+2)^2} \right. \\
+ 2 \left( \frac{2}{(0.25+2)^2} \right) \\
+ 2 \left( \frac{2}{(0.5+2)^2} \right) \\
+ 2 \left( \frac{2}{(0.75+2)^2} \right) \\
\left. + \frac{2}{(1+2)^2} \right]
\end{array}
\][/tex]
#### Step 4: Evaluate the function at the [tex]\(x\)[/tex] values
[tex]\[
\begin{array}{l}
f(x_0) = \frac{2}{(0+2)^2} = \frac{2}{4} = 0.5 \\
f(x_1) = \frac{2}{(0.25+2)^2} = \frac{2}{5.0625} \approx 0.394 \\
f(x_2) = \frac{2}{(0.5+2)^2} = \frac{2}{6.25} = 0.32 \\
f(x_3) = \frac{2}{(0.75+2)^2} = \frac{2}{7.5625} \approx 0.264 \\
f(x_4) = \frac{2}{(1+2)^2} = \frac{2}{9} \approx 0.222 \\
\end{array}
\][/tex]
#### Step 5: Substitute and compute
[tex]\[
\begin{array}{l}
\int_0^1 \frac{2}{(x+2)^2} \, dx \approx \frac{0.25}{2} \left[0.5 \right.\\
+ 2 \left(0.394\right) + 2 \left(0.32\right) + 2 \left(0.264\right) + 0.222 \\
\left.\right] \\
\approx \frac{0.25}{2} \left[0.5 + 0.788 + 0.64 + 0.528 + 0.222\right] \\
= \frac{0.25}{2} \left[2.678\right] \\
= 0.125 \cdot 2.678 \\
\approx 0.335
\end{array}
\][/tex]
Hence, we successfully calculated the integral [tex]\(\int_0^1 \frac{2}{(x+2)^2} \, dx \)[/tex] using the Trapezoidal Rule with an approximate value of 0.335.
