Wed Dec 3 2014 11:59 PM AKST - University of Alaska...
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Question 123456789101112131415161718192021222324252627282930313233343536373 E5 (6607623) Due: Wed Dec 3 2014 11:59 PM AKST 1. Question Details SCalc7 4.1.003.MI. [1874320] (a) Estimate the area under the graph of f(x)= 5 cos(x) from x = 0 to x = π/2 using four approximating rectangles and right endpoints. (Round your answers to four decimal places.) R 4 = Sketch the graph and the rectangles.
Transcript of Wed Dec 3 2014 11:59 PM AKST - University of Alaska...
Question 12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152
E5 (6607623)
Due: Wed Dec 3 2014 11:59 PM AKST
1. Question Details SCalc7 4.1.003.MI. [1874320]
(a) Estimate the area under the graph of f(x) = 5 cos(x) from x = 0 to x = π/2 using four approximating rectangles andright endpoints. (Round your answers to four decimal places.)R4 =
Sketch the graph and the rectangles.
Is your estimate an underestimate or an overestimate?
(b) Repeat part (a) using left endpoints.L4 =
Sketch the graph and the rectangles.
underestimate
overestimate
Is your estimate an underestimate or an overestimate?
Solution or Explanation
Click to View Solution
underestimate
overestimate
2. Question Details SCalc7 4.1.005. [2524211]
(a) Estimate the area under the graph of using three rectangles and right endpoints.R3 =
Then improve your estimate by using six rectangles.R6 =
Sketch the curve and the approximating rectangles for R3.
f(x) = 8 + 2x2 from x = −1 to x = 2
Sketch the curve and the approximating rectangles for R6.
(b) Repeat part (a) using left endpoints.L3=L6=
Sketch the curve and the approximating rectangles for L3.
Sketch the curve and the approximating rectangles for L6.
(c) Repeat part (a) using midpoints.M3=M6=
Sketch the curve and the approximating rectangles for M3.
Sketch the curve and the approximating rectangles for M6.
(d) From your sketches in parts (a)(c), which appears to be the best estimate?
Solution or Explanation
Click to View Solution
M6
R6
L6
3. Question Details SCalc7 4.1.014. [1679076]
Speedometer readings for a motorcycle at 12second intervals are given in the table.
t (s) 0 12 24 36 48 60
v (ft/s) 30 27 25 22 24 28
(a) Estimate the distance traveled by the motorcycle during this time period using the velocities at the beginning ofthe time intervals.
ft
(b) Give another estimate using the velocities at the end of the time periods. ft
(c) Are your estimates in parts (a) and (b) upper and lower estimates? Explain.
Solution or Explanation
Click to View Solution
(a) and (b) are neither lower nor upper estimates since v is neither an increasing nor decreasingfunction of t.
(b) is a lower estimate and (a) is an upper estimate since v is a decreasing function of t.
(a) is a lower estimate and (b) is an upper estimate since v is an increasing function of t.
4. Question Details SCalc7 4.1.020. [1874313]
Use the Definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.
Solution or Explanation
f(x) = x2 + , 3 ≤ x ≤ 51 + 2x
lim n → ∞
n
i = 1
f(x) = x2 + , 3 ≤ x ≤ 5.1 + 2x Δx = (5 − 3)/n = 2/n and xi = 3 + iΔx = 3 + 2i/n.
A = Rn = = lim n → ∞
lim n → ∞
n
f(xi)Δxi = 1
lim n → ∞
n(3 + 2i/n)2 + · .
i = 11 + 2(3 + 2i/n) 2
n
5. Question Details SCalc7 4.1.022. [2850758]
Determine a region whose area is equal to the given limit. Do not evaluate the limit.
Solution or Explanation
Click to View Solution
lim n → ∞
n9 +
i = 1
2n
2in
12
y = (9 + x)16 on [9, 11]
y = (9 + x)11 on [0, 2]
y = (9 + x)11 on [9, 11]
y = x12 on [0, 2]
y = (9 + x)12 on [0, 2]
6. Question Details SCalc7 4.1.023. [1679273]
Determine a region whose area is equal to the given limit. Do not evaluate the limit.
Solution or Explanation
Click to View Solution
lim n → ∞
n tan
i = 1
π8n
iπ8n
x tan(x) on 0, π8
x tan(x) on − , π8
π8
tan(x) on [0, 8π]
tan(x) on 0, π8
tan(x) on − , π8
π8
7. Question Details SCalc7 4.1.030. [1805695]
(a) Let An be the area of a polygon with n equal sides inscribed in a circle with radius r.
By dividing the polygon into n congruent triangles with central angle show that the following is true.
(b) Show that [Hint: Use the equation.]
Solution or Explanation
(a) The diagram shows one of the n congruent triangles, Δ with central angle O is the center of the circle andAB is one of the sides of the polygon. Radius OC is drawn so as to bisect ∠AOB. It follows that OC intersects AB at rightangles and bisects AB. Thus, ΔAOB is divided into two right triangles with legs of length
ΔAOB has area
so
(b) To use the equation, we need to have the same expression in the denominator as we have in the
argument of the sine function—in this case, 2π/n.
2π/n,
An = nr2 sin12
2πn
An = πr2.lim n → ∞
AOB, 2π/n.
(AB) = r sin(π/n) and r cos(π/n).12
2 · [r sin(π/n)][r cos(π/n)] = r2 sin(π/n) cos π/n) = r2 sin(2π/n),12
12
An = n · area(ΔAOB) = nr2 sin(2π/n).12
= 1,lim θ → 0
sin θθ
An = nr2 πr2 = πr2 = (1) πr2 = πr2.lim n → ∞
lim n → ∞
12
sin(2π/n)2π/n
lim θ → 0
sin θθ
8. Question Details SCalc7 4.1.501.XP. [1874565]
The area A of the region S that lies under the graph of the continuous function is the limit of the sum of the areas ofapproximating rectangles.
Use this definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.
Solution or Explanation
Click to View Solution
A = Rn = [f(x1)Δx + f(x2)Δx + . . . + f(xn)Δx]lim n → ∞
lim n → ∞
f(x) = , 1 ≤ x ≤ 19x5
A = lim n → ∞
n
i = 1
9. Question Details SCalc7 4.1.JIT.003. [1805435]
Find the sum.
5
2k − 1k = 1
10. Question Details SCalc7 4.2.008. [1678756]
The table gives the values of a function obtained from an experiment. Use them to estimate using three equal
subintervals with right endpoints, left endpoints, and midpoints.
x 3 4 5 6 7 8 9
f(x) −3.5 −2.3 −0.6 0.3 0.7 1.5 1.8
(a) Estimate using three equal subintervals with right endpoints.
R3 =
If the function is known to be an increasing function, can you say whether your estimate is less than or greaterthan the exact value of the integral?
(b) Estimate using three equal subintervals with left endpoints.
L3 =
If the function is known to be an increasing function, can you say whether your estimate is less than or greaterthan the exact value of the integral?
(c) Estimate using three equal subintervals with midpoints.
M3 =
If the function is known to be an increasing function, can you say whether your estimate is less than or greaterthan the exact value of the integral?
Solution or Explanation
Click to View Solution
f(x) dx9
3
f(x) dx9
3
less than
greater than
one cannot say
f(x) dx9
3
less than
greater than
one cannot say
f(x) dx9
3
less than
greater than
one cannot say
11. Question Details SCalc7 4.2.009. [1805242]
Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places.
Solution or Explanation
so the endpoints are 0, 24, 48, 72, and 96, and the midpoints are 12, 36, 60, and 84. The Midpoint
Rule gives
sin dx, n = 496
x0
Δx = (96 − 0)/4 = 24,
sin dx ≈ 96
x4f(xi)Δx = 24(sin + sin + sin + sin ) ≈ 24(0.6545) = 15.7087.
i = 112 36 60 84
0
12. Question Details SCalc7 4.2.034.MI. [3164742]
The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral.
(a)
(b)
(c)
Solution or Explanation
Click to View Solution
g(x) dx10
0
g(x) dx30
10
g(x) dx35
0
13. Question Details SCalc7 4.2.037. [1875190]
Evaluate the integral by interpreting it in terms of areas.
Solution or Explanation
Click to View Solution
2 + dx0
81 − x2−9
14. Question Details SCalc7 4.2.040. [1875077]
Evaluate the integral by interpreting it in terms of areas.
Solution or Explanation
Click to View Solution
|x − 3| dx6
0
15. Question Details SCalc7 4.2.041. [1874795]
Evaluate
Solution or Explanation
Click to View Solution
sin5 x cos5 x dx.π
π
16. Question Details SCalc7 4.2.042. [1875154]
Given that what is
Solution or Explanation
Click to View Solution
= 15 − 24,9x dx1
x2 + 40
5 ?9u du0
u2 + 41
17. Question Details SCalc7 4.2.052. [1805606]
If where f is the function whose graph is given, which of the following values is largest?
Solution or Explanation
so F(0) is negative, and similarly, so is F(4). F(12) and F(16) are negative since they
represent negatives of areas below the xaxis. Since is the only nonnegative value, F(8) is the
largest.
F(x) = ,f(t) dtx
8
F(0)
F(4)
F(8)
F(12)
F(16)
F(0) = = − f(t) dt0
8f(t) dt,8
0
F(8) = f(t) dt = 08
8
18. Question Details SCalc7 4.2.053. [1804975]
Each of the regions A, B, and C bounded by the graph of f and the xaxis has area 3. Find the value of
Solution or Explanation
= −8 + 2 = −6
Thus,
[f(x) + 2x + 3] dx.2
−4
I = = + 2 + = I1 + 2I2 + I3[f(x) + 2x + 3] dx2
−4f(x) dx2
−4x dx2
−43 dx2
−4I1 = −3 [area below xaxis] + 3 − 3 = −3
I2 = − (4)(4) [area of triangle, see figure] + (2)(2)12
12
I3 = 3 [2 − (−4)] = 3(6) = 18I = −3 + 2(−6) + 18 = 3.
19. Question Details SCalc7 4.2.054. [1875362]
Suppose f has absolute minimum value m and absolute maximum value M. Between what two values must lie?
(smaller value)
(larger value)
Which property of integrals allows you to make your conclusion?
Solution or Explanation
Click to View Solution
f(x) dx4
2
If f(x) ≥ 0 for a ≤ x ≤ b, then ≥ 0.f(x) dxb
a
If m ≤ f(x) ≤ M for a ≤ x ≤ b, then m(b − a) ≤ ≤ M(b − a).f(x) dxb
a
= 0f(x) dxa
a
+ = f(x) dxc
af(x) dxb
cf(x) dxb
a
= − f(x) dxb
af(x) dxa
b
20. Question Details SCalc7 4.2.059. [1875255]
If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is the absolute minimum and M is the absolute maximum of f on the interval[a, b], then
Use this property to estimate the value of the integral.
(smaller value)
(larger value)
Solution or Explanation
Click to View Solution
m(b − a) ≤ ≤ M(b − a).f(x) dxb
a
3 dx16
x1
21. Question Details SCalc7 4.2.072. [1874330]
Express the limit as a definite integral.
Solution or Explanation
Click to View Solution
lim n→∞
1n
n
i = 1
81 + (i/n)2
dx1
0
22. Question Details SCalc7 4.3.009. [1928030]
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
=
Solution or Explanation
g(s) = (t − t8)3dts
9
g'(s)
f(t) = (t − t8)3 and g(s) = , so by FTC1, g'(s) = f(s) = (s − s8)3.(t − t8)3dts
9
23. Question Details SCalc7 4.3.010. [1875076]
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
g'(r) =
Solution or Explanation
Click to View Solution
g(r) = dxr
x2 + 35
24. Question Details SCalc7 4.3.012.MI. [1874750]
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
G'(x) =
Solution or Explanation
Click to View Solution
G(x) = cos dt6
7tx
25. Question Details SCalc7 4.3.015. [1874340]
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
y' =
Solution or Explanation
Click to View Solution
y = dttan x
3t + t1
26. Question Details SCalc7 4.3.017. [1875013]
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
y' =
Solution or Explanation
Click to View Solution
y = du2 u3
1 + u22 − 3x
27. Question Details SCalc7 4.3.023. [1874469]
Evaluate the integral.
Solution or Explanation
Click to View Solution
dx9
x4
28. Question Details SCalc7 4.3.027. [1874620]
Evaluate the integral.
Solution or Explanation
(u + 3)(u − 4) du1
0
= = u3 − u2 − 12u = − − 12 − 0 = − (u + 3)(u − 4) du1
0(u2 − u − 12) du1
0
13
12
1
0
13
12
736
29. Question Details SCalc7 4.3.029.MI. [1679161]
Evaluate the integral.
Solution or Explanation
Click to View Solution
dx16 x − 3
x1
30. Question Details SCalc7 4.3.031. [1680994]
Evaluate the integral.
Solution or Explanation
Click to View Solution
3 sec2 t dtπ/4
0
31. Question Details SCalc7 4.3.037. [1679131]
Evaluate the integral.
Solution or Explanation
Click to View Solution
where f(x) = f(x) dxπ
0
5 sin x if 0 ≤ x < π/25 cos x if π/2 ≤ x ≤ π
32. Question Details SCalc7 4.3.039. [1805244]
What is wrong with the equation?
Solution or Explanation
is not continuous on the interval [−3, 4], so FTC2 cannot be applied. In fact, f has an infinite discontinuity at
does not exist.
x−3 dx = = 4 x−2
−2
4
−3
7288−3
The lower limit is less than 0, so FTC2 cannot be applied.
There is nothing wrong with the equation.
f(x) = x−3 is not continuous on the interval [−3, 4] so FTC2 cannot be applied.
f(x) = x−3 is continuous on the interval [−3, 4] so FTC2 cannot be applied.
f(x) = x−3 is not continuous at x = −3, so FTC2 cannot be applied.
f(x) = x−3
x = 0, so x−3 dx4
−3
33. Question Details SCalc7 4.3.052. [1927571]
Find the derivative of the function.
Solution or Explanation
Click to View Solution
g(x) = dt3x2 1
5 + t3tan x
g'(x) =
34. Question Details SCalc7 4.4.002. [1805398]
State whether the following is true or false by differentiation.
Solution or Explanation
cos2x dx = x + sin 2x + C12
14
True
False
x + sin 2x + C = + cos 2x · 2 + 0 = + cos 2x
= + 2 cos2 x − 1 = + cos2 x − = cos2 x
ddx
12
14
12
14
12
12
12
12
12
12
35. Question Details SCalc7 4.4.005. [2560330]
Find the general indefinite integral. (Use C for the constant of integration.)
Solution or Explanation
Click to View Solution
(5x2 + 6x−2) dx
36. Question Details SCalc7 4.4.011. [2560793]
Find the general indefinite integral. (Use C for the constant of integration.)
Solution or Explanation
Click to View Solution
dx7x3 − 8x
x
37. Question Details SCalc7 4.4.013. [2560194]
Find the general indefinite integral. (Use C for the constant of integration.)
Solution or Explanation
Click to View Solution
(7θ − 4 csc θ cot θ) dθ
38. Question Details SCalc7 4.4.014. [2560399]
Find the general indefinite integral. (Use C for the constant of integration.)
Solution or Explanation
Click to View Solution
sec t (5 sec t + 9 tan t) dt
39. Question Details SCalc7 4.4.015. [2560110]
Find the general indefinite integral. (Use C for the constant of integration.)
Solution or Explanation
Click to View Solution
8(1 + tan2 α) dα
40. Question Details SCalc7 4.4.025. [1927572]
Evaluate the integral.
Solution or Explanation
Click to View Solution
(7 sin θ − 17 cos θ) dθπ
0
41. Question Details SCalc7 4.4.029. [1927699]
Evaluate the integral.
Solution or Explanation
Click to View Solution
dx16 13
x1
42. Question Details SCalc7 4.4.045. [1874633]
The area of the region that lies to the right of the yaxis and to the left of the parabola (the shaded region in
the figure) is given by the integral (Turn your head clockwise and think of the region as lying below the
curve from y = 0 to y = 8.) Find the area of the region.
Solution or Explanation
Click to View Solution
x = 8y − y2
(8y − y2) dy.8
0x = 8y − y2
43. Question Details SCalc7 4.4.046. [1927586]
The boundaries of the shaded region are the yaxis, the line and the curve Find the area of this region
by writing x as a function of y and integrating with respect to y.
Solution or Explanation
Click to View Solution
y = 8, y = 8 .x4
44. Question Details SCalc7 4.4.055. [1678684]
The velocity function (in meters per second) is given for a particle moving along a line.
(a) Find the displacement. m
(b) Find the distance traveled by the particle during the given time interval. m
Solution or Explanation
Click to View Solution
v(t) = 3t − 8, 0 ≤ t ≤ 3
45. Question Details SCalc7 4.5.003.MI.SA. [2631984]
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receiveany points for the skipped part, and you will not be able to come back to the skipped part.
Tutorial Exercise
Evaluate the integral by making the given substitution.
x2 dx, u = x3 + 4x3 + 4
46. Question Details SCalc7 4.5.006.MI.SA. [2947488]
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receiveany points for the skipped part, and you will not be able to come back to the skipped part.
Tutorial Exercise
Evaluate the integral by making the given substitution.
dx, u = 1/x2sec2(1/x2)x3
47. Question Details SCalc7 4.5.007.MI.SA. [2947529]
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receiveany points for the skipped part, and you will not be able to come back to the skipped part.
Tutorial Exercise
Evaluate the indefinite integral.
x4 sin(x5) dx
48. Question Details SCalc7 4.5.021.MI.SA. [2947508]
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receiveany points for the skipped part, and you will not be able to come back to the skipped part.
Tutorial Exercise
Evaluate the indefinite integral.
dxcos xsin7 x
49. Question Details SCalc7 4.5.024. [2560374]
Evaluate the indefinite integral. (Use C for the constant of integration.)
Solution or Explanation
Click to View Solution
dtcos2 t 6 + tan t3
50. Question Details SCalc7 4.5.040. [1874363]
Evaluate the definite integral.
Solution or Explanation
Click to View Solution
csc 7πt cot 7πt dt1/14
1/42
51. Question Details SCalc7 4.5.060. [1875385]
If f is continuous and find
Solution or Explanation
Click to View Solution
f(x) dx = 8,36
0xf(x2) dx.6
0
Name (AID): E5 (6607623)Submissions Allowed: 2Category: ExamCode:Locked: YesAuthor: Frith, Russell ( [email protected] )Last Saved: Nov 21, 2014 01:51 PM AKSTPermission: ProtectedRandomization: PersonWhich graded: Last
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52. Question Details SCalc7 4.5.063. [1874607]
If a and b are positive numbers, show that
Let and du = ? .
Use this substitution to rewrite the integral in terms of u.
Then replacing u with x results in the integral
Solution or Explanation
Let
= .xa(1 − x)bdx1
0xb(1 − x)adx1
0
u = 1 − x. Then x =
= xa(1 − x)b dx1
0 1 ub(−du) = ub(1 − u)adu
(1 − x)adx.1
0
u = 1 − x. Then x = 1 − u and dx = −du, so
= = = xa(1 − x)bdx1
0(1 − u)aub(−du)0
1ub(1 − u)a du1
0xb(1 − x)adx.1
0
