Given the shaded region, you need to use the Trapezoidal Rule with:
[tex]n=4[/tex]In order to approximate the area of the shaded region.
By definition, the Trapezoidal Rule is:
[tex]T_n\approx\int_a^bf(x)dx=\frac{\Delta x}{2}\lbrack f(x_0)+2f(x_1)+...f(x_{n-1})+f(x_n)[/tex]You need to find:
[tex]\Delta x[/tex]This can be found with this formula:
[tex]\Delta x=\frac{b-a}{n}[/tex]Notice in the graph that, in this case, the interval is:
[tex]\lbrack-20,20\rbrack[/tex]Therefore:
[tex]\begin{gathered} a=-20 \\ b=20 \end{gathered}[/tex]Then, by substituting values into the formula, you get:
[tex]\Delta x=\frac{20-(-20)}{4}=\frac{40}{4}=10[/tex]In this case, you can determine that the subintervals must begin at -20 con adding 10 until you get to 20:
[tex]\begin{gathered} x_0=-20 \\ \\ x_1=-10 \\ \\ x_2=0 \\ \\ x_3=10 \\ \\ x_4=20 \end{gathered}[/tex]Substitute values into the Trapezoidal Rule Formula:
[tex]T_4\approx\frac{10}{2}\lbrack f(-20)+2f(-10)+2f(0)+2f(10)+f(20)\rbrack[/tex]You need to use the graph to identify the y-values that correspond to each x-value. Identify the points on the curve with those coordinates:
[tex](-20,20),(-10,30),(0,40),(10,20),(20,16)[/tex]Therefore, by substituting the corresponding y-values into the formula and evaluating, you get:
[tex]T_4\approx\frac{10}{2}\lbrack20+2(30)+2(40)+2(20)+16\rbrack[/tex][tex]T_4\approx1080[/tex]Hence, the answer is:
[tex]1080[/tex]