Question

Find an approximate value of `int_1^1.5 (2 - x)` dx by applying the mid-point rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}

Answer 1

Here a = 1

b = 1.5

n = 5

f(x) = 2 – x

So, the width of each subinterval is

h = Δx

= `("b" - "a")/"n"`

= `(1.5- 1)/5`

= 0.1

The partition of the interval is given by

The partition of the interval is given b.y

x₀ = 1

x₁ = 1.1

x₂ = 1.2

x₃ = 1.3

x₄ = 1.4

x₅ = 1.5

The mid-point rule for Riemann sum with equal width Δx is

S = `["f"((x_0 + x_1)/2) + "f"((x_1 + x_2)/2) + ... + "f"((x_("n" - 1) + x_"n")/2]Deltax`

`"f"((x_0 + x_1)/2) = "f"((1 + 1.1)/2)`

= f(1.05)

= `2 – 1.05

= 0.95

`"f"((x_1 + x_2)/2) = "f"((1.1 + 1.2)/2)`

= f(1.15)

= `2 – 1.15

= 0.85

`"f"((x_2 + x_3)/2) = "f"((1.2 + 1.3)/2)`

= f(1.25)

= `2 – 1.25

= 0.75

`"f"((x_3 + x_4)/2) = "f"((1.3 + 1.4)/2)`

= f(1.35)

= `2 – 1.35

= 0.65

`"f"((x_4 + x_5)/2) = "f"((1.4 + 1.5)/2)`

= f(1.45)

= `2 – 1.45

= 0.55

∴ S = (0.95 + 0.85 + 0.75 + 0.65 + 0.55) × 0.1

= 3.75 × 0.1

= 0.375