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Write in sigma notation and evaluate the sum of terms 2 i for i = 3 , 4 , 5 , 6 .

i = 3 6 2 i = 2 3 + 2 4 + 2 5 + 2 6 = 120

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The properties associated with the summation process are given in the following rule.

Rule: properties of sigma notation

Let a 1 , a 2 ,…, a n and b 1 , b 2 ,…, b n represent two sequences of terms and let c be a constant. The following properties hold for all positive integers n and for integers m , with 1 m n .


  1. i = 1 n c = n c

  2. i = 1 n c a i = c i = 1 n a i

  3. i = 1 n ( a i + b i ) = i = 1 n a i + i = 1 n b i

  4. i = 1 n ( a i b i ) = i = 1 n a i i = 1 n b i

  5. i = 1 n a i = i = 1 m a i + i = m + 1 n a i

Proof

We prove properties 2. and 3. here, and leave proof of the other properties to the Exercises.

2. We have

i = 1 n c a i = c a 1 + c a 2 + c a 3 + + c a n = c ( a 1 + a 2 + a 3 + + a n ) = c i = 1 n a i .

3. We have

i = 1 n ( a i + b i ) = ( a 1 + b 1 ) + ( a 2 + b 2 ) + ( a 3 + b 3 ) + + ( a n + b n ) = ( a 1 + a 2 + a 3 + + a n ) + ( b 1 + b 2 + b 3 + + b n ) = i = 1 n a i + i = 1 n b i .

A few more formulas for frequently found functions simplify the summation process further. These are shown in the next rule, for sums and powers of integers , and we use them in the next set of examples.

Rule: sums and powers of integers

  1. The sum of n integers is given by
    i = 1 n i = 1 + 2 + + n = n ( n + 1 ) 2 .
  2. The sum of consecutive integers squared is given by
    i = 1 n i 2 = 1 2 + 2 2 + + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 .
  3. The sum of consecutive integers cubed is given by
    i = 1 n i 3 = 1 3 + 2 3 + + n 3 = n 2 ( n + 1 ) 2 4 .

Evaluation using sigma notation

Write using sigma notation and evaluate:

  1. The sum of the terms ( i 3 ) 2 for i = 1 , 2 ,…, 200 .
  2. The sum of the terms ( i 3 i 2 ) for i = 1 , 2 , 3 , 4 , 5 , 6 .
  1. Multiplying out ( i 3 ) 2 , we can break the expression into three terms.
    i = 1 200 ( i 3 ) 2 = i = 1 200 ( i 2 6 i + 9 ) = i = 1 200 i 2 i = 1 200 6 i + i = 1 200 9 = i = 1 200 i 2 6 i = 1 200 i + i = 1 200 9 = 200 ( 200 + 1 ) ( 400 + 1 ) 6 6 [ 200 ( 200 + 1 ) 2 ] + 9 ( 200 ) = 2,686,700 120,600 + 1800 = 2,567,900
  2. Use sigma notation property iv. and the rules for the sum of squared terms and the sum of cubed terms.
    i = 1 6 ( i 3 i 2 ) = i = 1 6 i 3 i = 1 6 i 2 = 6 2 ( 6 + 1 ) 2 4 6 ( 6 + 1 ) ( 2 ( 6 ) + 1 ) 6 = 1764 4 546 6 = 350
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Find the sum of the values of 4 + 3 i for i = 1 , 2 ,…, 100 .

15,550

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Finding the sum of the function values

Find the sum of the values of f ( x ) = x 3 over the integers 1 , 2 , 3 ,…, 10 .

Using the formula, we have

i = 0 10 i 3 = ( 10 ) 2 ( 10 + 1 ) 2 4 = 100 ( 121 ) 4 = 3025.
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Evaluate the sum indicated by the notation k = 1 20 ( 2 k + 1 ) .

440

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Approximating area

Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let f ( x ) be a continuous, nonnegative function defined on the closed interval [ a , b ] . We want to approximate the area A bounded by f ( x ) above, the x -axis below, the line x = a on the left, and the line x = b on the right ( [link] ).

A graph in quadrant one of an area bounded by a generic curve f(x) at the top, the x-axis at the bottom, the line x = a to the left, and the line x = b to the right. About midway through, the concavity switches from concave down to concave up, and the function starts to increases shortly before the line x = b.
An area (shaded region) bounded by the curve f ( x ) at top, the x -axis at bottom, the line x = a to the left, and the line x = b at right.

How do we approximate the area under this curve? The approach is a geometric one. By dividing a region into many small shapes that have known area formulas, we can sum these areas and obtain a reasonable estimate of the true area. We begin by dividing the interval [ a , b ] into n subintervals of equal width, b a n . We do this by selecting equally spaced points x 0 , x 1 , x 2 ,…, x n with x 0 = a , x n = b , and

x i x i 1 = b a n

for i = 1 , 2 , 3 ,…, n .

We denote the width of each subinterval with the notation Δ x , so Δ x = b a n and

x i = x 0 + i Δ x

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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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