Finding Approximate Areas Under Curves

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 split the area up into a number of trapeziums each of width h and find the area of each. The y coordinates are given by y 0, y 1, y 2, y 3 etc

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

Simpsons Rule Who was William Simpson?

Simpson was born in Market Bosworth, Leicestershire. The son of a weaver, Simpson taught himself mathematics, then turned to astrology after seeing a solar eclipse. He also dabbled in witchcraft and caused fits in a girl after 'raising a devil' from her. After this incident, he and his wife had to flee to London. From 1743, he taught mathematics at the Royal Military Academy, Woolwich. Apparently, the method that became known as Simpson's rule was well known and used earlier by Bonaventura Cavalieri (a student of Galileo) in It was later rediscovered by James Gregory (who Simpson succeeded as Professor of Mathematics at the University of St Andrews) but was only attributed to Simpson. In 1758, Simpson was elected a foreign member of the Royal Swedish Academy of Sciences. Simpson's rule is a staple of scientific data analysis and engineering. It is widely used, for example, by Naval architects to calculate the capacity of a ship or lifeboat.

Simpsons Rule This approximates the area under the curve using a quadratic curve to join the tops of each y value rather than a straight line as in the trapezium rule y 0 y 1 y 2 y 3 y 4 y 5

e.g. To estimate we’ll take 4 strips. The rule fits a quadratic curve to the 1 st 3 points at the top edge of the strips. x x x

x x x e.g. To estimate we’ll take 4 strips. The rule fits a quadratic curve to the 1 st 3 points at the top edge of the strips. Another quadratic curve is fitted to the 3 rd, 4 th and 5 th points.

x e.g. To estimate we’ll take 4 strips. The rule fits a quadratic curve to the 1 st 3 points at the top edge of the strips. Another quadratic curve is fitted to the 3 rd, 4 th and 5 th points. x x

x x x e.g. To estimate we’ll take 4 strips. The rule fits a quadratic curve to the 1 st 3 points at the top edge of the strips. Another quadratic curve is fitted to the 3 rd, 4 th and 5 th points.

Consider three points which have coordinates (-h, y 0 ), (0, y 1 ), (h, y 2 ). The tops of each red bar are joined by a quadratic curve(parabola) rather than straight lines as in the trapezium rule The parabola has the equation : -h h 0 y0y0 y1y1 y2y2 (-h, y 0 ), (0, y 1 ) (h, y 2 ) y = ax 2 + bx + c

The object is to find the area of the 2 strips shown using the y coordinates and then extend the idea to the next 2 strips. -h h 0 y0y0 y1y1 y2y2 y = ax 2 + bx + c.

-h h 0 y0y0 y1y1 y2y2 A= Now substitute in the boundaries i.e x=h and x=-h y = ax 2 + bx + c.

-h h 0 y0y0 y1y1 y2y2 y = ax 2 + bx + c.

-h h 0 y0y0 y1y1 y2y2 A= y = ax 2 + bx + c.

When we substitute the coordinates of the three points (-h, y 0 ), (0, y 1 ), (h, y 2 ) into the equation for the parabola, we obtain the three equations y 0 = ah 2 - bh + c, y 1 = c, y 2 = ah 2 + bh + c. Remember the equation of the quadratic curve is y = ax 2 + bx + c If we add the 1 st and 3 rd equation then we obtain y 0 + y 2 = 2ah 2 + 2c Now make a the subject

y 0 = ah 2 - bh + c, y 1 = c, y 2 = ah 2 + bh + c. Remember the equation of the quadratic curve is y = ax 2 + bx + c y 0 + y 2 = 2ah 2 + 2c Now make a the subject

Finally, substitute the above expression for a and y 1 for c in the equation for the area : Cancel the h 2 terms and the 2 y 1 = c h is a common factor

h h h h h h

The formula works providing The number of y values is…………. And the number of strips is ……….. ODD EVEN h h

e.g. (a) Use Simpson’s rule with 4 strips to estimate giving your answer to 4 d.p. Solution: (a) y0y0 y2y2 y1y1 y4y4 y3y3

Solution: X yy 0 =1y 1 = y 2 =0.8y 3 =0.64y 4 =0.5

(a) (b) Use your calculator to check the answer using the integration button

Exercise using Simpson’s rule with 2 strips, giving your answer to 4 d.p. 1. (a) Estimate (b) Improve your answer using 4 strips

Solution: using Simpson’s rule with 2 strips, giving your answer to 4 d.p. 1. (a) Estimate y0y0 y1y1 y2y2

Solution: using Simpson’s rule with 2 strips, giving your answer to 4 d.p. 1. (a) Estimate X123 yy 0 =1y 1 =0.5y 2 =

Solution: using Simpson’s rule with 4 strips, giving your answer to 4 d.p. 1. (a) Estimate