# Finding Approximate Areas Under Curves. The Area of a Trapezium Area= 1 / 2 (a+b) x h Example a=10 b=14 h= 8 Area= 1 / 2 (10 + 14) x 8 = 96cm 2 a b h.

## Presentation on theme: "Finding Approximate Areas Under Curves. The Area of a Trapezium Area= 1 / 2 (a+b) x h Example a=10 b=14 h= 8 Area= 1 / 2 (10 + 14) x 8 = 96cm 2 a b h."— Presentation transcript:

Finding Approximate Areas Under Curves

The Area of a Trapezium Area= 1 / 2 (a+b) x h Example a=10 b=14 h= 8 Area= 1 / 2 (10 + 14) x 8 = 96cm 2 a b h

The Trapezium Rule y 0 y 1 y 2 y 3 y 4 y 5 This curve has a complicated equation so instead of integrating we will split the area up into a number of trapeziums each of width 0.2 and find the area of each. The y coordinates are given by y 0,y 1,y 2,y 3 etc

The Trapezium Rule The demonstration that follows finds the area under the curve by splitting it up into a number of trapeziums You have to determine the rule if the number of trapeziums (intervals) is given Applet

Proving the Formula Area of trapezium 1 = 1 / 2 (y 0 +y 1 ) x h Area of Trapezium 2 = 1 / 2 (y 1 +y 2 ) x h Area of Trapezium 3 = 1 / 2 (y 2 +y 3 ) x h Area of Trapezium 4 = 1 / 2 (y 3 +y 4 ) x h y 0 y 1 y 2 y 3 y 4 y 5 h= strip width = interval width h

Proving the Formula Area = 1 / 2 h(y 0 + y 1 + y 1 + y 2 + y 2 …y n ) = 1 / 2 h(y 0 + 2y 1 + 2y 2 + 2y 3 …y n ) = 1 / 2 h(y 0 + 2(y 1 + y 2 + y 3 …) + y n ) y 0 y 1 y 2 y 3 y 4 y 5 h

Try this on using 5 intervals 5 intervals from 0 to 1 means the width of each strip is 0.2. Type y1= and set up the table starting at x=0 with steps of 0.2 Using the Trapezium Rule 0 0.2 0.4 0.6 0.8 1.0

Using the Table Function on Your Calculator to Determine the y Values Enter the equation of your graph in y 1 Press Table Setup (2ndF Table) Press the down arrow to TBLStart and input the left hand boundary for the required area. Press the down arrow to TBLStep and input the width of each strip (interval) Press Table to see the y values y 0,y 1,y 2,y 3 etc

Using the Trapezium Rule Values from table  1 / 2 h(y 0 + 2(y 1 + y 2 + y 3 + y 4 ) + y 5 )  1 / 2 x 0.2(1+2(0.9615+0.8621+0.7353+0.6098)+0.5)  0.78374 xy 0.01.0000y0 0.20.9615y1 0.40.8621y2 0.60.7353y3 0.80.6098y4 1.00.5000y5

Is the Approximate Area Too Large or Small?

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