Derivation Of Composite Trapezoidal Rule
Derivation Of Composite Trapezoidal Rule :
b
Let us find the value of ∫ f(x) dx
a
numerically by composite trapezoidal rule.
a
numerically by composite trapezoidal rule.
Solution :
Let [a,b] be divided into n equal sub intervals of length h each . Let a= x₀,x₁,x₂,.......xₙ=b be the points of sub division .
i.e x₁ = x₀+ih for i= 0,1,2,......n
By Newton's Forward Difference Interpolation
f(x) = f(x₀)+uΔf(x₀) +u(u-1)/2! Δ²f(x₀)+........
where x = x₀+uh
b
∫ f(x) dx =
a xₙ
∫ [f(x₀)+uΔf(x₀)+u(u-1)/2! Δ²f(x₀)+...]dx x₀
n
= h∫ [f(x₀)+uΔf(x₀)+u(u-1)/2 Δ²f(x₀)+...]dx
0
=h[uf(x₀) +u²/2 Δf(x₀) +
n
(u³/6 -u²/4)Δ²f(x₀)+.....]
0
=h[nf(x₀) +n²/2 Δf(x₀) +(n³/6 - n²/4)Δ²f(x₀)+...]
For n=1
x₁
∫ f(x) dx = h[1.f(x₀) + 1/2 Δf(x₀)]
x₀
= h[f(x₀) + 1/2 {f(x₁) - f(x₀)}
= h/2 [f(x₀) + f(x₁)]
x₁
⇒ ∫ f(x) dx = h/2 [f(x₀) + f(x₁)]
x₀
Similarly,
x₂
∫ f(x) dx = h/2 [f(x₁) + f(x₂)]
x₁
x₃
∫ f(x) dx = h/2 [f(x₂) + f(x₃)]
x₂
.........................................
.........................................
xₙ
∫ f(x) dx = h/2 [f(xₙ₋₁) + f(xₙ)]
xₙ₋₁
Adding respective sides , we get
x₁ x₂ xₙ
∫ f(x) dx + ∫ f(x) dx +.........+∫ f(x) dx
x₀ x₁ xₙ₋₁
= h/2 [f(x₀)+f(x₁)+f(x₁)+f(x₂)+f(x₂)+...
+ f(xₙ₋₁)+f(xₙ)]
xₙ
⇒∫ f(x) dx = h/2 {f(x₀)+2[f(x₁)+f(x₂)+....f(xₙ₋₁)]
x₀
+ f(xₙ)}
For Example :
Approximate the integral of f(x)= x³ on the interval [1,2] with four sub intervals using Composite Trapezoidal Rule .
Solution :
Here h= (2-1) /4 = 0.25
and so we have
2
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