Maybe add in chad’s idea of the area of polygons limiting to the area of a circle. Nice animation to link to online At

The computation of the area under the curve is a very old idea in Mathematics The Phythagoreans tried in 225 BC Since their time, improvements have been made to this concept f(x) = x 2

Exact calculation The use of Calculus allows the user to calculate the exact value of the area in question. Now we know that the area under the curve f(x) = x 2 from x =0 to x =20 is A = units 2 The concepts needed for this calculation did not develop overnight.

Beginnings… We begin our investigation of the available methods for computing the area under the curve using rectangles….. Right-hand endpoints

Rh midpoints n =2 A = A 1 + A 2 A 1 = b*h = 10*100 = 1000 A 2 = b*h = 10*400 = 4000 A = 5000 units 2 A2A2 A1A1 n = 2 x = 10

Or we could use the following… A = f(x 1 ) x + f(x 2 ) x with x = 10 and x 1 = 10 x 2 = 20 Then A = (10) 2 (10) + (20) 2 (10)= 5000 units 2

overestimate The Area = 5000 units 2, is a terrible approximation. It’s clearly an overestimate. How can this approximation be improved?

Divide the length of the base in half… now A = A 1 + A 2 +A 3 +A 4 A 1 = b*h = 5*25 = 125 A 2 = b*h = 5*100 = 500 A 3 = b*h = 5*225 = 1125 A 4 = b*h = 5*400 = 2000 A = 3750 units 2 n = 4 x = 5

If we repeat this process of dividing the base in half, a pattern develops….. n x~ AreaError = = = = =

It seems that this pattern is limiting toward something….. We can see that as n gets larger, x gets smaller, the error is smaller and the overestimation is reduced overestimate But it is still an overestimate. What can we do to improve our method of approximating the area under the curve, f(x) = x 2 ?

Would we be better off using the left-hand endpoints? Left-hand endpoints A = A 0 + A 1 A 0 = b*h = 5*0 A 1 = b*h = 10*100 = 1000 A = 1000 units 2 n = 2 x = 10

A = A 0 + A 1 + A 2 +A 3 A 0 = b*h = 5*0 A 1 = b*h = 5*25 = 250 A 2 = b*h = 5*100 = 500 A 3 = b*h = 5*225 = 1125 A = 1875 units 2 Left-hand endpoints n = 4 x = 5

Again, we see a similar pattern… That is… as n gets larger, x gets smaller underestimate But now each approximation is an underestimate. n x~ AreaError = = = = = Any other ideas??????

How about using the midpoints… mid-points

1 st Midpoint A = A 1 + A 2 A 1 = b*h = 10*25 = 250 A 2 = b*h = 10*225 = 2250 A = 2500 units 2

And again… A = A 1 + A 2 + A 3 + A 4 A 1 = b*h = 5*6.25 = A 2 = b*h = 5*56.25 = A 3 = b*h = 5* = A 2 = b*h = 5* = A = 2635 units 2 n = 4 x = 5

n x~ AreaError = = = = = Results using the midpoint

Results: 1. The accuracy is improved as more rectangles are used ( n   ) 2.Midpoint is the best method for calculating the Area.

Mathematically speaking... Def: The area A of the region that lies under the graph of a continuous function is the limit of the sum of the areas of the approximating rectangles: A = lim (f(x 0 ) x + f(x 1 ) x + …. f(x n ) x) n   pg 320

Other ideas?????? What about more complex geometric shapes?? They need to be shapes that we are able easily compute the area of.

The trapezoid is one possibility: The area can be determined by averaging the heights of the two sides. A1A1 A2A2 A3A3

h 1 = 225 A3A3 h 2 = 400 A 3 = (h 1 + h 2 ) x 2 A 3 = ( ) * 5 2 A 3 = 312.5* 5 = But if we want to do this for n trapezoids, we need to have a way to generalize what we are doing…. Notice: h n = f(x n )

Trapezoidal Rule In general: T n =. x ( f(x 1 ) + f(x 2 ) + f(x 3 ) + f(x 4 ) + …. + f(x n-1 )+f(x n ) ) 2 2 T 3 = x( f(x 1 )+f(x 2 ) + f(x 2 )+f(x 3 ) + f(x 3 )+f(x 4 ) ) T 3 =. x( f(x 1 )+f(x 2 ) + f(x 2 )+f(x 3 ) + f(x 3 )+f(x 4 ) ) 2 T 3 =. x( f(x 1 )+2f(x 2 ) + 2f(x 3 ) +f(x 4 ) ) 2 T 3 = x( f(x 1 )+ f(x 2 ) + f(x 3 ) +f(x 4 ) ) 2 Animation of Trap Rule

Clearly, you can all see the need for adding many areas together To help with this, Section 5.2 will introduce Reimann Sums Many of you have already seen summation notation before. Only difference here is that we are summing up a number of areas rather than a list of numbers

Area expressed with this notation Suppose the interval is [a,b] (in our example [0,20]) Then x = b-a n So far we have been introduced to the notation n and x How are they related?

A = lim (f(x 0 ) x + f(x 1 ) x + …. + f(x n ) x) n   = lim  x(f(x 0 ) + f(x 1 ) + …. + f(x n )) n   = lim  x f (x i ) n   i=1 n

Example: Use left hand endpoints with n=4 to approximate the area under the curve y = sin(x) from x = 0 to x = .  x =  -0 =   /4  /2 3  /4  x i = 0,  /4,  /2, 3  /4 A =  /4 (f(0) + f(  /4 ) + f(  /2 ) + f( 3  /4 )) A =  /4 (sin(0) + sin(  /4 ) + sin(  /2 ) + sin( 3  /4 )) A =  /4 (0 +  2/  2/2) =  /4 *(1+  2)  1.9 More generally, x i = 0+i*  /4

Repeat with midpoints Still…  x =  -0 =   /4  /2 3  /2  x 0 =  /4-0 =  2 8 x 1 =  /8 + 1*  /4 =3  /8 x 2 =  /8 + 2(  /4) =5  /8 x 3 =  /8 + 3(  /4) =7  /8 A =  /4 (f(  /8 ) + f( 3  /8 ) + f( 5  /8 ) + f( 7  /8 )) A =  /4 (sin(  /8 ) + sin( 3  /8 ) + sin( 5  /8 ) + sin( 7  /8 )) In general: x i =  /8 + i*  /4 A =  /4 ( ) A =  /4 ( )=

Introducing the Integral a = lower bound b = upper bound f(x) is the integrand

= lim f (x i * ) x n   Definition: pg. 326 If f(x) is a continuous function defined for a <x <b, we divide the interval [a,b] into n subintervals of equal width  x = (b-a)/n We let x 0,x 1,x 2 ….x n be the endpoints of these intervals and we choose sample points x 1 *, x 2 *,…. x n * in these subintervals. Then the Definite integral of f(x) from a to b is: i=1 n

Rules for Sigma: i=1 n n n i = n(n+1) 2 i 2 = n(n+1)(2n+1) 6 c = nc and more on pg 329

To be continued….

Other ideas?????? What about more complex geometric shapes?? They need to be shapes that we are able easily compute the area of.

The trapezoid is one possibility: The area can be determined by averaging the heights of the two sides. A1A1 A2A2 A3A3

h 1 = 225 A3A3 h 2 = 400 A 3 = (h 1 + h 2 ) x 2 A 3 = ( ) * 5 2 A 3 = 312.5* 5 = But if we want to do this for n trapezoids, we need to have a way to generalize what we are doing…. Notice: h n = f(x n )

Trapezoidal Rule In general: T n =. x ( f(x 1 ) + f(x 2 ) + f(x 3 ) + f(x 4 ) + …. + f(x n-1 )+f(x n ) ) 2 2 T 3 = x( f(x 1 )+f(x 2 ) + f(x 2 )+f(x 3 ) + f(x 3 )+f(x 4 ) ) T 3 =. x( f(x 1 )+f(x 2 ) + f(x 2 )+f(x 3 ) + f(x 3 )+f(x 4 ) ) 2 T 3 =. x( f(x 1 )+2f(x 2 ) + 2f(x 3 ) +f(x 4 ) ) 2 T 3 = x( f(x 1 )+ f(x 2 ) + f(x 3 ) +f(x 4 ) ) 2

Trapezoidal Rule A =. x ( f(x 1 ) + 2f(x 2 ) + 2f(x 3 ) + 2f(x 4 ) + …. + 2f(x n-1 )+f(x n ) ) 2 id.2/index.html