Estimation?

Question

Estimate :
∫ x^(1/3) dx (integral has limits : 0.5 at bottom and 1.0 at top (of integral sign))
using
(a) Trapezium rule
(b) Composite Trapezium Rule
(c) Which estimate is better?

Answer 1

(a) The trapezium rule says that you can approximate the area under a curve by a trapezium (a four sided figure). Call the function F with integration limits of x0 and x1.
The area is then approximated by:

Area = (x1 - x0)*(F(x1) + F(x0))/2

This uses a straight line to approximate the curve between the two points x0 and x1 and the area is the base (x1 - x0) times the average of the two function values at the end points (F(x1) + F(x0))/2.

F(x1) = F(1.0) = 1
F(x0) = F(0.5) = 0.7937
x1 - x0 = 1.0 - 0.5 = 0.5

Area = (0.5)*(1 + 0.7937)/2 = 0.4484

b. This breaks the area to be integrated up into a number of trapeziums and then sums over the set of such areas. In this case we will use 5 so the area becomes:

The X points will be x0, xa, xb, xc, xd, x1. The spacing is the same between each point and eual to (x1 - x0)/5

Area of Trapezium 1 = [(x1 - x0)/5]*(F(x0) + F(xa))/2
Area of Trapezium 2 = [(x1 - x0)/5]*(F(xa) + F(xb))/2
Area of Trapezium 3 = [(x1 - x0)/5]*(F(xb) + F(xc))/2
Area of Trapezium 4 = [(x1 - x0)/5]*(F(xc) + F(xd))/2
Area of Trapezium 5 = [(x1 - x0)/5]*(F(x1) + F(xd))/2
Sum all of these and get:

Area = [(x1 - x0)/5] * [F(x0) + 2F(xa) + 2F(xb) + 2F(xc) + 2F(xd) + F(x1) ]/2

F(x1) = F(1.0) = 1
F(xa) = F(0.9) = 0.9655
F(xb) = F(0.8) = 0.9283
F(xc) = F(0.7) = 0.8879
F(xd) = F(0.6) = 0.8434
F(x0) = F(0.5) = 0.7937

Area = (0.1)*(4.522) = 0.4522

c. The composite since it provides a more accurate approximation to the curve. The actual area by integration is 0.4524

Answer 2

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