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  1. Find an upper sum for f ( x ) = 10 x 2 on [ 1 , 2 ] ; let n = 4 .
  2. Sketch the approximation.
  1. Upper sum = 8.0313 .

  2. A graph of the function f(x) = 10 − x^2 from 0 to 2. It is set up for a right endpoint approximation over the area [1,2], which is labeled a=x0 to x4. It is an upper sum.

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Finding lower and upper sums for f ( x ) = sin x

Find a lower sum for f ( x ) = sin x over the interval [ a , b ] = [ 0 , π 2 ] ; let n = 6 .

Let’s first look at the graph in [link] to get a better idea of the area of interest.

A graph of the function y = sin(x) from 0 to pi. It is set up for a left endpoint approximation from 0 to pi/2 and n=6. It is a lower sum.
The graph of y = sin x is divided into six regions: Δ x = π / 2 6 = π 12 .

The intervals are [ 0 , π 12 ] , [ π 12 , π 6 ] , [ π 6 , π 4 ] , [ π 4 , π 3 ] , [ π 3 , 5 π 12 ] , and [ 5 π 12 , π 2 ] . Note that f ( x ) = sin x is increasing on the interval [ 0 , π 2 ] , so a left-endpoint approximation gives us the lower sum. A left-endpoint approximation is the Riemann sum i = 0 5 sin x i ( π 12 ) . We have

A sin ( 0 ) ( π 12 ) + sin ( π 12 ) ( π 12 ) + sin ( π 6 ) ( π 12 ) + sin ( π 4 ) ( π 12 ) + sin ( π 3 ) ( π 12 ) + sin ( 5 π 12 ) ( π 12 ) = 0.863.
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Using the function f ( x ) = sin x over the interval [ 0 , π 2 ] , find an upper sum; let n = 6 .

A 1.125

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Key concepts

  • The use of sigma (summation) notation of the form i = 1 n a i is useful for expressing long sums of values in compact form.
  • For a continuous function defined over an interval [ a , b ] , the process of dividing the interval into n equal parts, extending a rectangle to the graph of the function, calculating the areas of the series of rectangles, and then summing the areas yields an approximation of the area of that region.
  • The width of each rectangle is Δ x = b a n .
  • Riemann sums are expressions of the form i = 1 n f ( x i * ) Δ x , and can be used to estimate the area under the curve y = f ( x ) . Left- and right-endpoint approximations are special kinds of Riemann sums where the values of { x i * } are chosen to be the left or right endpoints of the subintervals, respectively.
  • Riemann sums allow for much flexibility in choosing the set of points { x i * } at which the function is evaluated, often with an eye to obtaining a lower sum or an upper sum.

Key equations

  • Properties of Sigma Notation
    i = 1 n c = n c
    i = 1 n c a i = c i = 1 n a i
    i = 1 n ( a i + b i ) = i = 1 n a i + i = 1 n b i
    i = 1 n ( a i b i ) = i = 1 n a i i = 1 n b i
    i = 1 n a i = i = 1 m a i + i = m + 1 n a i
  • Sums and Powers of Integers
    i = 1 n i = 1 + 2 + + n = n ( n + 1 ) 2
    i = 1 n i 2 = 1 2 + 2 2 + + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6
    i = 0 n i 3 = 1 3 + 2 3 + + n 3 = n 2 ( n + 1 ) 2 4
  • Left-Endpoint Approximation
    A L n = f ( x 0 ) Δ x + f ( x 1 ) Δ x + + f ( x n 1 ) Δ x = i = 1 n f ( x i 1 ) Δ x
  • Right-Endpoint Approximation
    A R n = f ( x 1 ) Δ x + f ( x 2 ) Δ x + + f ( x n ) Δ x = i = 1 n f ( x i ) Δ x

State whether the given sums are equal or unequal.

  1. i = 1 10 i and k = 1 10 k
  2. i = 1 10 i and i = 6 15 ( i 5 )
  3. i = 1 10 i ( i 1 ) and j = 0 9 ( j + 1 ) j
  4. i = 1 10 i ( i 1 ) and k = 1 10 ( k 2 k )

a. They are equal; both represent the sum of the first 10 whole numbers. b. They are equal; both represent the sum of the first 10 whole numbers. c. They are equal by substituting j = i 1 . d. They are equal; the first sum factors the terms of the second.

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In the following exercises, use the rules for sums of powers of integers to compute the sums.

i = 5 10 i 2

385 30 = 355

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Suppose that i = 1 100 a i = 15 and i = 1 100 b i = −12 . In the following exercises, compute the sums.

i = 1 100 ( a i + b i )

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i = 1 100 ( a i b i )

15 ( −12 ) = 27

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i = 1 100 ( 3 a i 4 b i )

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i = 1 100 ( 5 a i + 4 b i )

5 ( 15 ) + 4 ( −12 ) = 27

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In the following exercises, use summation properties and formulas to rewrite and evaluate the sums.

k = 1 20 100 ( k 2 5 k + 1 )

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j = 1 50 ( j 2 2 j )

j = 1 50 j 2 2 j = 1 50 j = ( 50 ) ( 51 ) ( 101 ) 6 2 ( 50 ) ( 51 ) 2 = 40 , 375

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j = 11 20 ( j 2 10 j )

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k = 1 25 [ ( 2 k ) 2 100 k ]

4 k = 1 25 k 2 100 k = 1 25 k = 4 ( 25 ) ( 26 ) ( 51 ) 9 50 ( 25 ) ( 26 ) = −10 , 400

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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