Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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O edcc.instructure. X
O Acing Google Co x
O Kevawn «Redund x
M Whole Course Re X
M Invitation: CS&
e Papago
C Get Homework E X
Get Homework
A variables2 - Solve X
A WAMAP - Posts x
5.2 Calc.pdf
+
A https://wamaps3.s3.amazonaws.com/ufiles/222249/5.2_Calc.pdf
These are equal by the distributive property. The
-1
other formulas are discussed in Appendix F.
11
(а, — b) —
E b.
a, -
EXAMPLE 2 Evaluating an integral as a limit of Riemann sums
(a) Evaluate the Riemann sum for f(x) = x' – 6.x, taking the sample points to be right
endpoints and a = 0, b = 3, and n = 6.
(b) Evaluate (:
x' - 6x) dx.
SOLUTION
(a) With n = 6 the interval width is
b - a
3 - 0
Ar =
%3D
6
2
and the right endpoints are x, = 0.5, r2 = 1.0, x; = 1.5, x, = 2.0, xs = 2.5, and
X, = 3.0. So the Riemann sum is
R, =
i-
= f(0.5) Ar + f(1.0) Ar + f(1.5) Ar + f(2.0) Ax + f(2.5) Ax + f(3.0) Ax
5
y =x'- 6x
= }(-2,875 – 5 – 5.625 – 4 + 0.625 + 9)
= -3.9375
Notice that f is not a positive function and so the Riemann sum does not represent a sum
of areas of rectangles. But it does represent the sum of the areas of the blue rectangles
(above the x-axis) minus the sum of the areas of the gold rectangles (below the x-axis) in
FIGURE 5
Figure 5.
347
SECTION 5.2 THE DEFINITE INTEGRAL
(b) With n subintervals we have
b - a
3
Thus Xp = 0, x, = 3/n, x, = 6/n, X= 9/n, and, in general, x, = 3i/n. Since we are
using right endpoints, we can use Theorem 4:
3i 3
[(r' - 6x) dx = lim E S(x,) Ax = lim
te sum, N is a constant (unlike
nve 3n in front of the E sign
so we can
= lim
Equation 9 withe
오후 12:57
P Type here to search
^O G ) O
2021-01-07
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Transcribed Image Text:O edcc.instructure. X O Acing Google Co x O Kevawn «Redund x M Whole Course Re X M Invitation: CS& e Papago C Get Homework E X Get Homework A variables2 - Solve X A WAMAP - Posts x 5.2 Calc.pdf + A https://wamaps3.s3.amazonaws.com/ufiles/222249/5.2_Calc.pdf These are equal by the distributive property. The -1 other formulas are discussed in Appendix F. 11 (а, — b) — E b. a, - EXAMPLE 2 Evaluating an integral as a limit of Riemann sums (a) Evaluate the Riemann sum for f(x) = x' – 6.x, taking the sample points to be right endpoints and a = 0, b = 3, and n = 6. (b) Evaluate (: x' - 6x) dx. SOLUTION (a) With n = 6 the interval width is b - a 3 - 0 Ar = %3D 6 2 and the right endpoints are x, = 0.5, r2 = 1.0, x; = 1.5, x, = 2.0, xs = 2.5, and X, = 3.0. So the Riemann sum is R, = i- = f(0.5) Ar + f(1.0) Ar + f(1.5) Ar + f(2.0) Ax + f(2.5) Ax + f(3.0) Ax 5 y =x'- 6x = }(-2,875 – 5 – 5.625 – 4 + 0.625 + 9) = -3.9375 Notice that f is not a positive function and so the Riemann sum does not represent a sum of areas of rectangles. But it does represent the sum of the areas of the blue rectangles (above the x-axis) minus the sum of the areas of the gold rectangles (below the x-axis) in FIGURE 5 Figure 5. 347 SECTION 5.2 THE DEFINITE INTEGRAL (b) With n subintervals we have b - a 3 Thus Xp = 0, x, = 3/n, x, = 6/n, X= 9/n, and, in general, x, = 3i/n. Since we are using right endpoints, we can use Theorem 4: 3i 3 [(r' - 6x) dx = lim E S(x,) Ax = lim te sum, N is a constant (unlike nve 3n in front of the E sign so we can = lim Equation 9 withe 오후 12:57 P Type here to search ^O G ) O 2021-01-07
See Example 2 in Section 5.2. Suppose that in part (b) we change the upper limit of integration to 2,
resulting in
| (2 – 6z)dr. To find the value of this integral, work through the following step-by-step
process.
a) We know that a =
and b =
b) Using n subintervals, Ar =
c) Assume that the sample points in each interval are right endpoints. Find the following:
21 = a + Az =
22 = a + 2Az =
23 = a + 3Az =
In general, the ith sample point is r; = a + iAr =
Note: simplify your answer for this
question!
d) Now find the sum of the areas of n approximating rectangles. Note: your answer will be an expression
in terms of n only. You really need to use the information in example 2 to help you complete this question!
f(r:)Az =
i=1
e) Finally, find the value of the integral by letting the number of rectangles approach infinity.
[ (2" – 6z)dz = lim E f(#:)Az =
1=1
Question Help: O Message instructor D Post to forum
Submit Question
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Transcribed Image Text:See Example 2 in Section 5.2. Suppose that in part (b) we change the upper limit of integration to 2, resulting in | (2 – 6z)dr. To find the value of this integral, work through the following step-by-step process. a) We know that a = and b = b) Using n subintervals, Ar = c) Assume that the sample points in each interval are right endpoints. Find the following: 21 = a + Az = 22 = a + 2Az = 23 = a + 3Az = In general, the ith sample point is r; = a + iAr = Note: simplify your answer for this question! d) Now find the sum of the areas of n approximating rectangles. Note: your answer will be an expression in terms of n only. You really need to use the information in example 2 to help you complete this question! f(r:)Az = i=1 e) Finally, find the value of the integral by letting the number of rectangles approach infinity. [ (2" – 6z)dz = lim E f(#:)Az = 1=1 Question Help: O Message instructor D Post to forum Submit Question
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