Step-by-step explanation:
The width of each interval is Δx = (4−1)/6 = 1/2.
a) Evaluate the function at the beginning and end of each interval.
f(x) = 7√(ln x)
f(1) = 7√(ln 1) = 0
f(1.5) = 7√(ln 1.5) ≈ 4.4573300
f(2) = 7√(ln 2) ≈ 5.8278823
f(2.5) = 7√(ln 2.5) ≈ 6.7006153
f(3) = 7√(ln 3) ≈ 7.3370295
f(3.5) = 7√(ln 3.5) ≈ 7.8348826
f(4) = 7√(ln 4) ≈ 8.2418702
Calculate the area of each trapezoid.
T₁ = ½ (0 + 4.4573300) (1/2) = 1.1143325
T₂ = ½ (4.4573300 + 5.8278823) (1/2) = 2.5713031
T₃ = ½ (5.8278823 + 6.7006153) (1/2) = 3.1321244
T₄ = ½ (6.7006153 + 7.3370295) (1/2) = 3.5094112
T₅ = ½ (7.3370295 + 7.8348826) (1/2) = 3.7929780
T₆ = ½ (7.8348826 + 8.2418702) (1/2) = 4.0191882
The total area is therefore:
T = 1.1143325 + 2.5713031 + 3.1321244 + 3.5094112 + 3.7929780 + 4.0191882
T = 18.139337
b) Evaluate the function at the midpoint of each interval.
f(1.25) = 7√(ln 1.25) ≈ 3.3066651
f(1.75) = 7√(ln 1.75) ≈ 5.2365230
f(2.25) = 7√(ln 2.25) ≈ 6.3036165
f(2.75) = 7√(ln 2.75) ≈ 7.0404861
f(3.25) = 7√(ln 3.25) ≈ 7.5996115
f(3.75) = 7√(ln 3.75) ≈ 8.0477348
Calculate the area of each rectangle.
M₁ = (3.3066651) (1/2) = 1.6533325
M₂ = (5.2365230) (1/2) = 2.6182615
M₃ = (6.3036165) (1/2) = 3.1518082
M₄ = (7.0404861) (1/2) = 3.5202431
M₅ = (7.5996115) (1/2) = 3.7998057
M₆ = (8.0477348) (1/2) = 4.0238674
The total area is therefore:
M = 1.6533325 + 2.6182615 + 3.1518082 + 3.5202431 + 3.7998057 + 4.0238674
M = 18.767318
c) Simpson's rule can be calculated as:
S₆ = Δx/3 [f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + 2f(x₄) + 4f(x₅) + f(x₆)]
S₆ = (1/2)/3 [0 + 4(4.4573300) + 2(5.8278823) + 4(6.7006153) + 2(7.3370295) + 4(7.8348826) + 8.2418702]
S₆ = 18.423834
