Riemann Sums
description
Transcript of Riemann Sums
Riemann SumsLesson 14.2
Riemann Sums are used to approximatethe area between a curve and the x-axis over an interval. Riemann sums divide the areas into rectangles. By adding the areas of the rectangles, one gets an approximation for the area under the curve on the given interval.
Usually Riemann sums will use equally sized partitions of the interval to make calculations easier. By having bases of equal length, the base can be factored out when calculating the sum.
Here’s an example of how a Riemann Sum works:
).partitions of # n where4,(n partitions 4
using curve under the area theeapproximat sLet'
[1,3]. interval on the x
1yfunction heConsider t
y=1/x
interval. theof endpoints theare and where,x use
will we,partitions uniform want weSince partition.ith the
is i where,x ,partitionsour oflength thefind weFirst,
i
i
ban
ab
1x 2x 3x 4x
)1,1(
)0,1(
y=1/x
.2
1
4
1-3 be willx the, [1,3] interval,our For i
1x 2x 3x 4x
)1,1(
)0,1(
y=1/x
3
1f(3)
5
2)
2
5f(
2
1f(2)
3
2)
2
3f(
.3,2
5,2,
2
3at are endpointsright The
.rectanglesour of heights thebe will valuesThese .partitionsour of
endpointsright at thefunction theof value thecalculate weNext,
1x 2x 3x 4x
)1,1(
)0,1(
y=1/x
2
1
2
11R : wayfollowing then thearea, thefind
could We4. to1ifor ,2
1 1at are endpoints theendpoints,
right using calculated wereheights intervals' thesince Notice,
20
19)
3
1
5
2
2
1
3
2(
2
1 R be willarea The
4
1right
right
i
if
i
1x 2x 3x 4x
)1,1(
)0,1(
y=1/x
n
abi
n
abaf
n
i 1
R be willarea theand intervalsour of
endpointsright usecan we,partitionsn with b][a, intervalan
over curve aunder area theeapproximat togeneral,In
Base ofeach rectangle
Height ofeach rectangleusing the rightendpoint.
Example 1
•Calculate Riemann Sums for the function f(x) = x3 on the interval from 1 to 5 using 4 subintervals and choosing zi to be the left endpoint, right endpoint, and midpoint.
Left endpoint…
1(1) + 1(8) + 1(27) + 1(64) 100
Right endpoint…
1(8) + 1(27) + 1(64) + 1(125) 224
Midpoint…
1(3.375) + 1(15.625) + 1(42.875) + 1(91.125)
153
What if the question was…
•Calculate Riemann Sums for the function f(x) = x3 on the interval from 1 to 5 using 4, 40, 400, and 4000 subintervals and choosing zi to be right endpoint.
•We are going to need a little help from our calculator
•The idea is to create a short (7 lines) program to compute the sum automatically (in a loop) after you provide a formula for Y1 and values for a, b, n, and v (the type of the sum: v = 0 for Left, v = 1 for Right, and v = 0.5 for Midpoint). Press Prgm New 1: Create New.
•Name your program RIEM. (If you already have a program with that name, use a different name.)
•Enter the following lines in the program, pressing ENTER after each line. (Press PRGM 4 to get "For"; press PRGM 7 to get "End".)
• : 0 sto S • : (B - A) / N sto H •: For(K,1,N) • : A + (K-1+V) * H sto X • : S + Y1 sto S • : End •: H * S
•Test your program to compute the Left Riemann Sum for the function f(x)=x2 on the interval [2, 4] with 10 subintervals: ▫press QUIT to exit Prgm mode; ▫define Y1=x2 in the Equation Editor; ▫assign the appropriate values to A (left
end), B (right end), N (number of subintervals), and V (type of sum): 2 sto A 4 sto B 10 sto N 0 sto V
▫Correct answer: 17.48
Back to example…4 intervals 40 intervals 400 intervals 4000 intervals
1 sto A 1 sto A 1 sto A 1 sto A
5 sto B 5 sto B 5 sto B 5 sto B
4 sto N 4 sto N 4 sto N 4 sto N
1 sto V 1 sto V 1 sto V 1 sto V
224 162.26 156.62 156.06
Homework
Pages 825 – 826
1 – 6, 11, 12