MATH 12002 - CALCULUS I 4.2: Riemann Sum Examples Professor Donald - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §4.2: Riemann Sum Examples

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Riemann Sums

Definition

Let y = f (x) be a function defined on an interval [a, b]. Let n be a positive integer. Break the interval [a, b] into n equal subintervals with endpoints a = x0 < x1 < x2 < · · · < xn−1 < xn = b, so that each subinterval has length ∆x = b−a

n .

For each i = 1, . . . , n, choose a point x∗

i in the ith subinterval [xi−1, xi].

The nth Riemann Sum for f on [a, b] with this choice of sample points is Sn = f (x∗

1)∆x + f (x∗ 2)∆x + · · · + f (x∗ n)∆x.

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Riemann Sums

The nth right Riemann sum Rn is obtained by letting x∗

i = xi,

the right endpoint of the ith subinterval [xi−1, xi]: Rn = f (x1)∆x + f (x2)∆x + · · · + f (xn)∆x. The nth left Riemann sum Ln is obtained by letting x∗

i = xi−1,

the left endpoint of the ith subinterval [xi−1, xi]: Ln = f (x0)∆x + f (x1)∆x + · · · + f (xn−1)∆x.

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Example

Example

Compute the sixth left Riemann sum L6 and the sixth right Riemann sum R6 for the function f (x) = x3 − 5 on the interval [0, 3].

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Example

In the notation of the definition, n = 6, a = 0, b = 3, so ∆x = 3−0

6

= 1

2.

The partition of our interval [0, 3] is

1 2 3

1 2 3 2 5 2

The left endpoints of the intervals are 0, 1

2, 1, 3 2, 2, 5 2,

the right endpoints of the intervals are 1

2, 1, 3 2, 2, 5 2, 3, and we have

f (0) = 03 − 5 = 0 − 5 = −5 f (1

2)

= ( 1

2)3 − 5 = 1 8 − 5 = −39 8

f (1) = 13 − 5 = 1 − 5 = −4 f (3

2)

= ( 3

2)3 − 5 = 27 8 − 5 = −13 8

f (2) = 23 − 5 = 8 − 5 = 3 f (5

2)

= ( 5

2)3 − 5 = 125 8 − 5 = 85 8

f (3) = 33 − 5 = 27 − 5 = 22.

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Example

Therefore, we have L6 = f (0) · 1

2 + f (1 2) · 1 2 + f (1) · 1 2 + f (3 2) · 1 2 + f (2) · 1 2 + f (5 2) · 1 2

= (−5) · 1

2 + (−39 8 ) · 1 2 + (−4) · 1 2 + (−13 8 ) · 1 2 + (3) · 1 2 + (85 8 ) · 1 2

=

−5 2 − 39 16 − 2 − 13 16 + 3 2 + 85 16 = −15 16 = −0.9375

and R6 = f (1

2) · 1 2 + f (1) · 1 2 + f (3 2) · 1 2 + f (2) · 1 2 + f (5 2) · 1 2 + f (3) · 1 2

= (−39

8 ) · 1 2 + (−4) · 1 2 + (−13 8 ) · 1 2 + (3) · 1 2 + (85 8 ) · 1 2 + (22) · 1 2

= −39

16 − 2 − 13 16 + 3 2 + 85 16 + 11 = 201 16 = 12.5625.

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Example

Both L6 = −0.9375 and R6 = 12.5625 are (poor) approximations for 3 x3 − 5 dx = lim

n→∞ Ln = lim n→∞ Rn.

Since f (x) is increasing, for any n we have Ln 3 x3 − 5 dx Rn, so we know −0.9375 3 x3 − 5 dx 12.5625. The larger n we use, the closer Ln and Rn will be to the integral. We will see later that the exact value is 3 x3 − 5 dx = 5.25.

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