Question

E5 (6607623)
Due:
Wed Dec 3 2014 11:59 PM AKST
Question
1.
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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.)
R4 = Sketch the graph and the rectangles.

Is your estimate an underestimate or an overestimate?
underestimate
overestimate (b) Repeat part (a) using left endpoints.
L4 = Sketch the graph and the rectangles.
Is your estimate an underestimate or an overestimate? underestimate
overestimate Solution or Explanation
Click to View Solution
2.
Question Details
SCalc7 4.1.005. [2524211]
(a) Estimate the area under the graph of f(x) = 8 + 2x2 from x = −1 to x = 2 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.

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?
M6
R6 L6
Solution or Explanation
Click to View Solution
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 of
the 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.
(a) and (b) are neither lower nor upper estimates since v is neither an increasing nor decreasing
function 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.
Solution or Explanation
Click to View Solution
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.
f(x) = x2 + 1 + 2x , 3 ≤ x ≤ 5
n
lim n → ∞ i = 1
Solution or Explanation
f(x) = x2 + 1 + 2x , 3 ≤ x ≤ 5. Δx = (5 − 3)/n = 2/n and xi = 3 + iΔx = 3 + 2i/n.
n
n → ∞ n → ∞ n
f(xi)Δx = lim A = lim Rn = lim i = 1
n → ∞ i = 1
(3 + 2i/n)2 + 1 + 2(3 + 2i/n) · 2
.
n

5.
Question Details
SCalc7 4.1.022. [2850758]

SCalc7 4.1.023. [1679273]

Determine a region whose area is equal to the given limit. Do not evaluate the limit.
n
2
2i 12
9 + n
n n → ∞ i = 1
lim 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]
Solution or Explanation
Click to View Solution
6.
Question Details
Determine a region whose area is equal to the given limit. Do not evaluate the limit.
n
lim n → ∞ i = 1
π tan iπ 8n
8n
x tan(x) on 0, π
8
x tan(x) on − π , π 8 8
tan(x) on [0, 8π]
tan(x) on 0, π
8
tan(x) on − π , π
8 8
Solution or Explanation
Click to View Solution
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 2π/n, show that the following is true.
An = 1 2
nr sin 2π
2
n
(b) Show that lim An = πr2. [Hint: Use the equation.] n → ∞ Solution or Explanation
(a) The diagram shows one of the n congruent triangles, Δ AOB, with central angle 2π/n. O is the center of the circle and
AB is one of the sides of the polygon. Radius OC is drawn so as to bisect ∠AOB. It follows that OC intersects AB at right
angles and bisects AB. Thus, ΔAOB is divided into two right triangles with legs of length
1
(AB) = r sin(π/n) and r cos(π/n). ΔAOB has area 2
1
1
1
2 · [r sin(π/n)][r cos(π/n)] = r2 sin(π/n) cos π/n) = r2 sin(2π/n), so An = n · area(ΔAOB) = nr2 sin(2π/n).
2
2
2
(b) To use the equation, lim θ → 0 sin θ
θ
= 1, 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.
lim An = lim n → ∞ 1
n → ∞ 2
nr2 sin(2π/n)
sin θ
πr2 = lim πr2 = (1) πr2 = πr2.
2π/n
θ → 0 θ

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 of
approximating rectangles.
A = lim Rn = lim [f(x1)Δx + f(x2)Δx + . . . + f(xn)Δx]
n → ∞ n → ∞ Use this definition to find an expression for the area under the graph of f as a limit. Do not evaluate the limit.
f(x) = 5
x , 1 ≤ x ≤ 19
n
A = lim n → ∞ i = 1
Solution or Explanation
Click to View Solution
9.
Question Details
SCalc7 4.1.JIT.003. [1805435]
Find the sum.
5
2k − 1 k = 1

10.
Question Details
SCalc7 4.2.008. [1678756]
9
The table gives the values of a function obtained from an experiment. Use them to estimate f(x) dx using three equal
3
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
9
(a) Estimate 3
f(x) dx 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 greater
than the exact value of the integral?
less than
greater than one cannot say
9
(b) Estimate 3
f(x) dx 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 greater
than the exact value of the integral?
less than
greater than one cannot say
9
(c) Estimate 3
f(x) dx 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 greater
than the exact value of the integral?
less than
greater than one cannot say
Solution or Explanation
Click to View Solution

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.
96
sin 0
x dx, n = 4 Solution or Explanation
Δx = (96 − 0)/4 = 24, so the endpoints are 0, 24, 48, 72, and 96, and the midpoints are 12, 36, 60, and 84. The Midpoint
96
Rule gives 4
sin 0
12.
x dx ≈ f(xi)Δx = 24(sin
12 + sin
36 + sin
60 + sin
84 ) ≈ 24(0.6545) = 15.7087. i = 1
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.
10
(a) g(x) dx
0
30
(b) 10
g(x) dx
35
(c) 0
g(x) dx
Solution or Explanation
Click to View Solution

13.
Question Details
SCalc7 4.2.037. [1875190]

SCalc7 4.2.040. [1875077]

SCalc7 4.2.041. [1874795]

SCalc7 4.2.042. [1875154]

Evaluate the integral by interpreting it in terms of areas.
0
2 + −9
81 − x2 dx
Solution or Explanation
Click to View Solution
14.
Question Details
Evaluate the integral by interpreting it in terms of areas.
6
|x − 3| dx
0
Solution or Explanation
Click to View Solution
15.
Question Details
π
Evaluate π
sin5 x cos5 x dx. Solution or Explanation
Click to View Solution
16.
Question Details
1
Given that 9x
0
x2 + 4 dx = 15
Solution or Explanation
Click to View Solution
0
5 − 24, what is 1
9u
u2 + 4 du ?
17.
Question Details
SCalc7 4.2.052. [1805606]
x
If F(x) = f(t) dt, where f is the function whose graph is given, which of the following values is largest?
8
F(0)
F(4) F(8)
F(12)
F(16)
Solution or Explanation
0
F(0) = 8
8
f(t) dt = − f(t) dt, so F(0) is negative, and similarly, so is F(4). F(12) and F(16) are negative since they
0
8
represent negatives of areas below the xaxis. Since F(8) = 8
largest.
f(t) dt = 0 is the only nonnegative value, F(8) is the

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
2
−4
[f(x) + 2x + 3] dx. Solution or Explanation
I = 2
−4
[f(x) + 2x + 3] dx = 2
−4
f(x) dx + 2
2
−4
x dx + 2
−4
3 dx = I1 + 2I2 + I3
I1 = −3 [area below xaxis] + 3 − 3 = −3
1
1
I2 = − (4)(4) [area of triangle, see figure] + (2)(2) = −8 + 2 = −6
2
2
I3 = 3 [2 − (−4)] = 3(6) = 18
Thus, I = −3 + 2(−6) + 18 = 3.

19.
Question Details
SCalc7 4.2.054. [1875362]
4
Suppose f has absolute minimum value m and absolute maximum value M. Between what two values must
f(x) dx lie?
2
(smaller value)
(larger value)
Which property of integrals allows you to make your conclusion?
b
If f(x) ≥ 0 for a ≤ x ≤ b, then f(x) dx ≥ 0.
a
b
If m ≤ f(x) ≤ M for a ≤ x ≤ b, then m(b − a) ≤ a
a
a
c
f(x) dx = 0
b
f(x) dx + a
a
f(x) dx = −
b
f(x) dx = c
b
f(x) dx ≤ M(b − a). f(x) dx a
a
b
f(x) dx Solution or Explanation
Click to View Solution
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
b
m(b − a) ≤ a
f(x) dx ≤ M(b − a).
Use this property to estimate the value of the integral.
16
3
x dx
1
(smaller value)
(larger value)
Solution or Explanation
Click to View Solution

21.
Question Details
SCalc7 4.2.072. [1874330]

SCalc7 4.3.009. [1928030]

SCalc7 4.3.010. [1875076]

SCalc7 4.3.012.MI. [1874750]

Express the limit as a definite integral.
1
n → ∞ n
n
8
lim i = 1
1
1 + (i/n)2
dx
0
Solution or Explanation
Click to View Solution
22.
Question Details
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
s
g(s) = 9
(t − t8)3dt g'(s) = Solution or Explanation
s
f(t) = (t − t8)3 and g(s) = 9
23.
8 3
(t − t8)3dt, so by FTC1, g'(s) = f(s) = (s − s ) .
Question Details
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
r
g(r) = 5
x2 + 3 dx
g'(r) = Solution or Explanation
Click to View Solution
24.
Question Details
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
6
G(x) = cos 7t dt
x
G'(x) = Solution or Explanation
Click to View Solution
25.
Question Details
SCalc7 4.3.015. [1874340]

SCalc7 4.3.017. [1875013]

SCalc7 4.3.023. [1874469]

SCalc7 4.3.027. [1874620]

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
tan x
y = t dt
3t + 1
y' = Solution or Explanation
Click to View Solution
26.
Question Details
Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
y = 2
2 − 3x
y' = u3
1 + u2
du
Solution or Explanation
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27.
Question Details
Evaluate the integral.
9
4
x dx
Solution or Explanation
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28.
Question Details
Evaluate the integral.
1
0
(u + 3)(u − 4) du Solution or Explanation
1
0
(u + 3)(u − 4) du = 1
0
(u2 − u − 12) du = 1
1 3
1
1
1
73
u − u2 − 12u = − − 12 − 0 = − 3
2
3
2
6
0
29.
Question Details
SCalc7 4.3.029.MI. [1679161]

SCalc7 4.3.031. [1680994]

SCalc7 4.3.037. [1679131]

SCalc7 4.3.039. [1805244]

Evaluate the integral.
16
x − 3
x
1
dx
Solution or Explanation
Click to View Solution
30.
Question Details
Evaluate the integral.
π/4
3 sec2 t dt
0
Solution or Explanation
Click to View Solution
31.
Question Details
Evaluate the integral.
π
f(x) dx where f(x) = 0
5 sin x if 0 ≤ x < π/2
5 cos x if π/2 ≤ x ≤ π
Solution or Explanation
Click to View Solution
32.
Question Details
What is wrong with the equation?
4
4
−2
7 x−3 dx = x
= 288
−2 −3
−3
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.
The lower limit is less than 0, so FTC2 cannot be applied.
There is nothing wrong with the equation.
Solution or Explanation
f(x) = x−3 is not continuous on the interval [−3, 4], so FTC2 cannot be applied. In fact, f has an infinite discontinuity at x = 0, so 4
−3
x−3 dx does not exist.
33.
Question Details
SCalc7 4.3.052. [1927571]

SCalc7 4.4.002. [1805398]

SCalc7 4.4.005. [2560330]

SCalc7 4.4.011. [2560793]

Find the derivative of the function.
1
3x2
g(x) = 5 + t3
tan x
dt
g'(x) = Solution or Explanation
Click to View Solution
34.
Question Details
State whether the following is true or false by differentiation.
cos2x dx = 1
1
x + sin 2x + C
2
4
True
False Solution or Explanation
d
dx
35.
1
1
1
1
1
1
x + sin 2x + C = + cos 2x · 2 + 0 = + cos 2x
2
4
2
2
2
4
1
1
1
1
2
2 cos x − 1 = + cos2 x − = cos2 x
= + 2
2
2
2
Question Details
Find the general indefinite integral. (Use C for the constant of integration.)
(5x2 + 6x−2) dx
Solution or Explanation
Click to View Solution
36.
Question Details
Find the general indefinite integral. (Use C for the constant of integration.)
7x3 − 8
x
x
Solution or Explanation
Click to View Solution
dx
37.
Question Details
SCalc7 4.4.013. [2560194]

SCalc7 4.4.014. [2560399]

SCalc7 4.4.015. [2560110]

SCalc7 4.4.025. [1927572]

Find the general indefinite integral. (Use C for the constant of integration.)
(7θ − 4 csc θ cot θ) dθ
Solution or Explanation
Click to View Solution
38.
Question Details
Find the general indefinite integral. (Use C for the constant of integration.)
sec t (5 sec t + 9 tan t) dt
Solution or Explanation
Click to View Solution
39.
Question Details
Find the general indefinite integral. (Use C for the constant of integration.)
8(1 + tan2 α) dα
Solution or Explanation
Click to View Solution
40.
Question Details
Evaluate the integral.
π
0
(7 sin θ − 17 cos θ) dθ
Solution or Explanation
Click to View Solution
41.
Question Details
SCalc7 4.4.029. [1927699]

SCalc7 4.4.045. [1874633]

Evaluate the integral.
16
1
13
dx
x
Solution or Explanation
Click to View Solution
42.
Question Details
The area of the region that lies to the right of the yaxis and to the left of the parabola x = 8y − y2 (the shaded region in
8
the figure) is given by the integral (8y − y2) dy. (Turn your head clockwise and think of the region as lying below the
0
curve x = 8y − y2 from y = 0 to y = 8.) Find the area of the region.
Solution or Explanation
Click to View Solution
43.
Question Details
SCalc7 4.4.046. [1927586]
The boundaries of the shaded region are the yaxis, the line y = 8, and the curve y = 8
4

x . 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
44.
Question Details
SCalc7 4.4.055. [1678684]
The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = 3t − 8, 0 ≤ t ≤ 3
(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

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 receive
any 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 receive
any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise
Evaluate the indefinite integral.
cos x
sin7 x
dx

49.
Question Details
SCalc7 4.5.024. [2560374]

SCalc7 4.5.040. [1874363]

SCalc7 4.5.060. [1875385]

Evaluate the indefinite integral. (Use C for the constant of integration.)
dt
cos2 t 3
6 + tan t
Solution or Explanation
Click to View Solution
50.
Question Details
Evaluate the definite integral.
1/14
1/42
csc 7πt cot 7πt dt
Solution or Explanation
Click to View Solution
51.
Question Details
36
If f is continuous and 0
Solution or Explanation
Click to View Solution
f(x) dx = 8, find 6
0
xf(x2) dx. 52.
Question Details
SCalc7 4.5.063. [1874607]
If a and b are positive numbers, show that
1
0
xa(1 − x)bdx = 1
0
xb(1 − x)adx.
Let u = 1 − x. Then x = and du = ?
.
Use this substitution to rewrite the integral in terms of u.
1
0
ub(−du) =
xa(1 − x)b dx = ub(1 − u)adu
1
Then replacing u with x results in the integral
1
(1 − x)adx.
0
Solution or Explanation
Let u = 1 − x. Then x = 1 − u and dx = −du, so
1
0
xa(1 − x)bdx = 0
1
(1 − u)aub(−du) = 1
0
ub(1 − u)a du = 1
0
xb(1 − x)adx.
Assignment Details
Name (AID): E5 (6607623)
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