Easy medal/fan, just check my work:
The table below gives selected values for the function f(x). Use a trapezoidal estimation, with 6 trapezoids to approximate the value of .
x 1 1.1 1.2 1.5 1.7 1.9 2.0
f(x) 1 3 4 6 7 8 10

my work: area of trapezoid: h*.5*(a+b) a+b= x values' displacement from next x value, h= y values
1*(.1+.1)*.5 + 3*(.1+.1)*.5 + 4*(.3+.3)*.5+6*(.2+.2)*.5+7*.5*(.2+.2)+8*(.1+.1)*.5
= 5.000 units

Question

Easy medal/fan, just check my work:
The table below gives selected values for the function f(x). Use a trapezoidal estimation, with 6 trapezoids to approximate the value of  .
x	1	1.1	1.2	1.5	1.7	1.9	2.0
f(x)	1	3	4	6	7	8	10


my work: area of trapezoid: h*.5*(a+b)  a+b= x values' displacement from next x value, h= y values 
1*(.1+.1)*.5 + 3*(.1+.1)*.5 + 4*(.3+.3)*.5+6*(.2+.2)*.5+7*.5*(.2+.2)+8*(.1+.1)*.5
= 5.000 units

boldjon · Answer

this might get confusing but  just try to follow k?
 ∇ₓ = (b - a)/n = (2 - 1)/7 = 1/7 
∫ f(x) dx [a,b] ≈ ½ ∇ₓ [f(x₁) + 2f(x₂) + 2f(x₃) .... + f(x_n)] 
f(1) = 1, f(1.1) = 3 , f(1.3) = 5, f(1.6) = 8 
f(1.7) = 10, f(1.8) = 11 , f(2) = 14 

∫ f(x) dx [1,2] 
≈ 1/14 [1 + 2*3 + 2*5 + 2*8 + 2*10+ 2*11 + 14] 
≈ 6.357