1 Section 5.2

1 Section 5.2

2.f (x) = ln x − 1, 1 ≤ x ≤ 4, ∆x = ⁴⁻¹₆ = ¹₂ Since we are using left endpoints x^∗_i = x_i−1 L6 =^Pn_i=1f (xi−1)∆x

= (∆x)[f (x₀) + f (x₁) + f (x₂) + f (x₃) + f (x₄) + f (x₅)]

= ¹₂[f (1) + f (³₂) + f (2) + f (⁵₂) + f (3) + f (₇⁽)2)]

= ¹₂[(−1) + ln³₂ − 1 + ln 2 − 1 + ln⁵₂ − 1 + ln 3 − 1 + ln⁷₂ − 1]

= ¹₂(−6 + ln¹⁰⁵₄ )

The Reimann sum represents the sum of the areas of the rectangles above the x-axis minus the sum of the areas of the rectangles below the x-axis ; that is , the bet area of the rectangles with respect to the x-axis.

5.∆x = ^b−a_n = ⁸⁻⁰₄ = 2

(a)Using the right endpoints to approximate ^R₀⁸f (x)dx,we have

P4

i=1f (x_i)∆x = 2[f (2) + f (4) + f (6) + f (8)] ≈ 4

(b)Using the left endpoints to approximate ^R₀⁸f (x)dx,we have

P4

i=1f (x_i−1)∆x = 2[f (0) + f (2) + f (4) + f (6)] ≈ 6

(c)Using the midpoint of each subinterval to approximate ^R₀⁸f (x)dx,we have

P4

i=1f (xi)∆x = 2[f (1) + f (3) + f (5) + f (7)] ≈ 10 8. ∆x = ⁽_b − a)n = ⁹⁻³₃ = 2

(a)Using the right endpoints to approximate ^R₃⁹f (x)dx,we have

P3

i=1f (x_i)∆x = 2[f (5) + f (7) + f (9)] = 2(−0.6 + 0.9 + 1.8) = 4.2 (b)Using the left endpoints to approximate ^R₃⁹f (x)dx,we have

1

P3

i=1f (x_i−1)∆x = 2[f (3) + f (5) + f (7)] = 2(−3.4 − 0.6 + 0.9) = −6.2 (c)Using the midpoint of each subinterval to approximate ^R₃⁹f (x)dx,we have

P3

i=1f (x_i)∆x = 2[f (4) + f (6) + f (8)] = 2(−2.1 + 0.3 + 1.4) = −0.8

We can not say anything about the midpoint compared to the exact value of the integral.

9.∆x = ¹⁰⁻²₄ = 2,so the endpoints are 2, 4, 6, 8, 10 and the midpoints are 3, 5, 7, 9.

The Midpoint Rule gives ^R10₂√

x³+ 1dx ≈ ^P4_i=1f (x_i)∆x = 2(√

3³+ 1 +

√5³+ 1 +√

7³+ 1 +√

9³+ 1) ≈ 124.1644

14.See the solution to Exercise5.1.7 for a possible algorithm to calculate the sums. With ∆x = ¹⁻⁰₁₀₀ = 0.01 and subinterval endpoints 1,1.01,1.02,...,1.99,2, we calculate that the left Riemann sum is L₁₀₀ =^P100_i=1sin(x²_i−1)∆x ≈ 0.30607, and the right Riemann sum is R100=^P100_i=1sin(x²_i)∆x ≈ 0.31448.

Since f (x) = sin(x²) is an increasig function, we have L₁₀₀ ≤^R₀¹sin(x²)dx ≤ R₁₀₀, so 0.306 < L₁₀₀ ≤^R₀¹sin(x²)dx ≤ R₁₀₀ < 0.315.

Therefore, the approximate value 0.3084≈ 0.31 in Exercise must be accurate to two decimal places.

18.On [π, 2π], lim_n→∞^Pn_i=1^{cos x}_x ⁱ

i ∆x =^R_π^{2π cos x}_x dx 30. ∆x = ¹⁰⁻¹_n = _n⁹ and x_i = 1 + i∆x = 1 +⁹ⁱ_n, so

R1

01(x − 4 ln x)dx = limn→∞Rn = limn→∞Pn

i=1[(1 + ⁹ⁱ_n) − 4 ln(1 + ⁹ⁱ_n)]_n⁹ 33.

(a) Think of^R₀²f (x)dx as the area of a trapezoid with bases 1 and 3 and height 2. The area of a trapezoid is A = ¹₂(b + B)h, so ^R₀²f (x)dx = ¹₂(1 + 3)2 = 4.

2

(b)^R₀⁵f (x)dx =^R₀²f (x)dx +^R₂³f (x)dx +^R₃⁵f (x)dx =trapezoid + rectangle + triangle = ¹₂(1 + 3)2 + 3 ˙1 + ¹₂˙2˙3 = 4 + 3 + 3 = 10

(c)^R₅⁷f (x)dx is the negative of the area of the triangle with base 2 and height 3. ^R₅⁷f (x)dx = ⁻¹₂ ˙2˙3 = −3.

(d) ^R₇⁹f (x)dx is the negative of the area of a trapezoid with base 3 and 2 and height 2 , so it equals −¹₂(B + b)h = −¹₂(3 + 2)2 = −5. Thus,

R9

0 f (x)dx =^R₀⁵f (x)dx +^R₅⁷f (x)dx +^R₇⁹f (x)dx = 10 + (−3) + (−5) = 2.

36. ^R₋₂² √

4 − x²dx can be interpreted as the area under the graph of f (x) =

√4 − x² between x = −2 and x = 2 . This is equal to half the area of the

circle with radius 2 , so ^R₋₂² √

4 − x²dx = ¹₂π ˙2² = 2π.

41.^R_π^πsin²x cos⁴xdx = 0 since the limits of integration are equal.

47.^R₋₂² f (x)dx+^R₂⁵f (x)dx−^R₋₂⁻¹f (x)dx =^R₋₂⁵ f (x)dx+^R₋₁⁻²f (x)dx =^R₋₁⁵ f (x)dx

66.Since | sin 2x| ≤ 1, |^R₀^2πf (x) sin 2xdx| ≤^R₀^2π|f (x) sin 2x|dx ≤ ^R₀^2π|f (x)|| sin 2x|dx ≤

R_2π

0 |f (x)|dx

70. It’s an n-partition on [0, 1],so ¹⁻⁰_n = ¹_n, and let f (x) = _1+x¹ 2 and then lim_n→∞n^{1 Pn}_i=1 ¹

1+_nⁱ² =^R₀¹f (x)dx =^R₀¹ _1+x¹ 2dx

3