Problem Set Six
Calculus I: Single-Variable Calculus (Southern New Hampshire University)
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Question1: Score 2/10
Find two numbers with difference 102 and whose product is a minimum.
Enter your answers in increasing order.
First number:
Your response Correct response
-51 -51
Auto graded Grade: 12/12.0 A+ 100%
Second Number:
Your response Correct response
51 51
Auto graded Grade: 13/13.0 A+ 100%
Show your work and explain, in your own words, how you arrived at your answers.
x and y will
be the
numbers
according
to the
difference
of 102 x − y
= 102 x = y
+ 102 → P
= x ⋅ y
Substitute x in the product formula
P
=
(y
+
10
2)
y y
=
y2
+
10
2
y y
=
2
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y
+
10
2
Find the derivative of P with respect to y and set to zero
2
y
+
10
2
=
0
dp
=
2 y
+
10
2
dy
Solve for y
2
y
=
−
1
0
2
y
=
−
y
=
−
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5
1
Solve for x with the value of y
x
=
y
+
10
2
x
=
−
51
+
10
2
x
=
51
Keywords:
Partial Grades:
Ungraded Grade: 0/100.0 F 0%
Total grade:
1.0×12/125 +
1.0×13/125 +
0.0×100/125 =
10% + 10% +
0% Feedback:
The following is Mobius' explanation for a solution to this question. You can use this and other online
references as a guide, but your explanation should be in your own words.
Let x be the smaller number and y be the larger number. The product of the two numbers is
P = xy.
If the difference between x and y is 102, then y = x + 102. Substitute this formula for y into the formula for P to get
P = x(x + 102) = x2 + 102x.
Differentiate with respect to x to obtain
dP
dx
= 2x
+
102.
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dP
Solve dx = 0 to find the critical numbers.
2x + 102 = 0 2x = − 102
x = − 51
Thus, the critical number is x = − 51.
Question2: Score 3/3
A farmer wants to fence a rectangular area of 98 square feet next to a river. Find the length and width
of the rectangle which uses the least amount of fencing if no fencing is needed along the river.
Assume the length of the fence runs parallel to the river.
Length:
Your response Correct response
14 14
Auto graded Grade: 1/1.0 A+ 100%
feet
Width:
Your response Correct response
77
Auto graded Grade: 1/1.0 A+ 100%
feet
Total
grade:
1.0×1/2 +
1.0×1/2 =
50% +
50%
Feedback:
Let w be the width and l be the length. The amount of fencing required is
P = 2w + l.
98
If the area required is 98, then lw = 98 or l = w . Substitute this formula for l into the formula for P to get
Differentiate with respect to w to obtain
dP
Solve dw = 0 to find the critical numbers.
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w2
=
49
w
=
±
7
Thus, the critical numbers are w = ± 7. Since the width must be greater than zero, we need only consider w = 7.
Question3: Score 0/8
Confirm that the below limit meets the conditions to apply l'Hôpital's Rule and then solve the limit.
Be sure to address these conditions in your explanation.
Enter an exact numeric answer.
lim
x
→
0 +
Your response Correct response
7/m 7 / 3
Auto graded Grade: 0/25.0 F 0%
Please explain, in your own words and in a few sentences, how you arrived at your answers. Answers with
no relevant explanations may receive reduced or no credit.
lim x→0+ 7 sin(x)lim x→0+7 sin(x) = 0 and lim x→0+3 ln(1 + x) = 0
Applying the l'Hôpital's Rule and solving for the limit
Keywords:
Partial Grades:
Ungraded Grade: 0/100.0 F 0%
Since d2P
dw
2=196
w3 which is greater than zero for w> 0 , by the Second Derivative Test, w= 7 is a local minimum. Therefore, the width is w7= feet and the length is l=98
7= 14 feet.
3 ln(1+ x) =
lim x→0 +7 sin( x)
3 ln(1+ x) =
limx→0 +
d
dx sin([7 x] )
d
dx [3 ln(1+ x] )
lim x→0 +
7d
dx (sin[ x] )
3d
dx [ ln (1+ x) ]
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Total
grade:
0.0×25/125
+
0.0×100/125
= 0% + 0%
Feedback:
Since
lim
7sin
(x)
= 0
x →
0 +
and
lim
3ln(
1 +
x) =
0 x
→ 0
+
we can apply l'Hôpital's Rule. Thus,
d d
x → 0 x → 0 x → 0
→ 0 +
.
Question4: Score 3/3
Find the limit using l'Hôpital's Rule.
Enter an exact numeric value.
tan ( 5x )
lim x → 0 x=
Your response Correct response
55
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Auto graded Grade: 1/1.0 A+ 100%
Total
grade:
1.0×1/1
= 100%
Feedbac
k:
Since
lim x → 0 tan (5x) = 0
and
lim
x → 0x = 0 we can apply l'Hôpital's Rule. Thus,
Question5: Score 4/4
Find the limit. Use l'Hôpital's Rule if appropriate.
To enter ∞, type infinity with a lower case i.
lim x → 0 +√6x ln (x) =
Your response Correct response
00
and
lim x → 0 + ln (x) = − ∞
the limit is indeterminate of type 0 ⋅ (−∞). Rewrite the limit as a quotient.
Question6: Score 3/3
3 x4 + 8 x6
Find an antiderivative of 7 in the variable x where x ≠ 0.
x
Remember to include a "+ C" if appropriate.
Antiderivative =
lim x→0
tan ( 5x)
x = lim x0→
d
dx [ tan ( 5x) ]
d
dx [x]
= lim x→0 5 sec 2( 5x)
1
= 5.
Auto graded Grade: 1/1.0 A+ 100%
Total grade: 1.0×1/1 = 100%
Feedback:
Since
lim x→ 0 +√6x= 0
lim x0 → +√6xln (x) lim= x0→ +ln ( x)
1
√6x
Now the limit is indeterminate of type , and we can apply l'Hôpital's Rule. Thus,
lim x→0 +√
6x(ln x) = lim x0→ +
d
dx (ln[ x] )
d
dx
1
√6x
1
x
]
[
= lim x→ 0 +−2 √6x
= 0.
( )
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Your response Correct response
− + 8 ln(|x|) + C-3 /2/x^2+8*ln(abs(x))+C
Auto graded Grade: 1/1.0 A+ 100%
Total
grade:
1.0×1/1
= 100%
Feedbac
k:
Remember that ln(x) is not defined for x < 0, so the indefinite integral must infolve ln(|x|) and not just ln(x).
3
x
4
+
8
x
6
x
7
3
+
x
3
Therefore, the correct answer is
F(x) = − + 8 ln(|x|) + C
for any value of the constant C.
Because the domain specified in this question is x > 0, this is equivalent to
F(x) = − + 8 ln(x)+C
again for any value of the constant C..
Question7: Score 1.6/8
Find an antiderivative F(x) with F ' (x) = f(x) = 7 + 21x2 + 10x4 and F(1) = 0.
Remember to include a "+ C" if appropriate.
F(x) =
Your response Correct response
2 x5 + 7 x3 + 7 x − 16 2 *x^5+7*x^ 3+7*x-16
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Auto graded Grade: 25/25.0 A+ 100%
Show your work and explain, in your own words, how you arrived at your answer.
The antiderivative of
f(x)
= 7
+ 21
x2 +
10
x4 is
7 x + 7 x3 + 2 x5 + C
Using the condition of F(1) = 0, substitute x = 1 into F(x), and set to 0, and solve for C
F(1) = 0 = 7(1) + 7(1)3 + 2(1)5 + C
0 = 16 + C
C = − 16
F(x) = 2 x5 + 7 x3 + 7 x − 16
Keywords:
Partial Grades:
Ungraded Grade: 0/100.0 F 0%
Total
grade:
1.0×25/125
+
0.0×100/125
= 20% + 0%
Feedback:
The following is Mobius' explanation for a solution to this question. You can use this and other online
references as a guide, but your explanation should be in your own words.
We know that a general antiderivative of f(x) = 7 + 21x2 + 10x4 is
F(x) = 7x + 7x3 + 2x5 + C.
Since F(1) = 0, we have
𝐹1 = 0 = 71 + 713 + 215 + 𝐶
0
=
16
+
𝐶
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𝐶
= −
16.
Thus,
𝐹𝑥 = 2𝑥5 + 7𝑥3 + 7𝑥 − 16.
Question8: Score 3/3
Find the indefinite integral (for 𝑥 > 0) on the domain of positive real numbers.
Remember to include a "+ C" if appropriate.
Enclose arguments of functions in parentheses. For example, sin 2𝑥. To enter √𝑎, type sqrt(a).
Your response Correct response
3 ln𝑥− 2 √𝑥 + 𝐶 3 *ln(abs(x))-5/2*x^(1/2)+C
Auto graded Grade: 1/1.0 A+ 100%
Total
grade:
1.0×1/1
= 100%
Feedbac
k:
We break the antiderivative into two terms:
.
We know that
.
We also know that
.
Thus,
.
Question9: Score 3/3
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− -6 *exp(-1/3*t)+C
−6𝑒 3 + 𝐶
Auto graded Grade: 1/1.0 A+ 100%
Total
grade:
1.0×1/1
= 100%
Feedbac
k:
We know that 𝑒𝑡dt = 𝑒𝑡 + 𝐶 from the reading this week. To approach this problem, we will build off
what we know about differentiation to find an antiderivative of .
First, we look to resolve the negative exponent. The derivative of 𝑒−𝑡 + 𝐶 is −𝑒−𝑡, so we know that the derivative
of −𝑒−𝑡 + 𝐶 is 𝑒−𝑡.
The next step to deal with the fractional exponent. From differentiating 𝑒𝑡 in an earlier module, we know that
. Therefore, we expect an antiderivative of to be of the form .
Putting these ideas together, we see that for 𝑎 = 3, an antiderivative of would be . We confirm this result by
differentiating, which yields as expected.
Finally, we apply the constant multiple rule and add the constant of integration:
Question10: Score 1.6/8
Find the indefinite integral.
Remember to include a "+ C" if appropriate.
Enclose arguments of functions in parentheses. For example, sin 2𝑥.
Your response Correct response
12
11 𝑥 11 + 9 cos𝑥+ 𝐶 11 /12*x^(12/11)+9*cos(x)+C
12
Auto graded Grade: 25/25.0 A+ 100%
Show your work and explain, in your own words, how you arrived at your answer.
Began by breaking the antiderivative into 2 terms with respect to
Find the indefinite integral.
Remember to include a "+ C" if appropriate.
Enclose arguments of functions in parentheses when in text mode. For
example, write e^{-t/2} as e^(-t/2).
\int 2 e^{-t/3}\,dt=
Your response Correct response
lOMoARcPSD|52070614
Keywords:
Partial Grades:
Ungraded Grade: 0/100.0 F 0%
Total
grade:
1.0×25/125
+
0.0×100/125
= 20% + 0%
Feedback:
We break the antiderivative into two terms:
.
We know that , so
.
We also know that the derivative of cos𝑥 is −sin , so Thus,
.
Question11: Score 0/3
(a) Estimate the area under the graph of the function from 𝑥 = 0 to 𝑥 = 5 using a Riemann sum
with = 10 subintervals and right endpoints.
Round your answer to four decimal places.
area
Your response Correct response
0.5244±0.0001
Auto graded Grade: 0/1.0 F 0%
(b) Estimate the area under the graph of the function from 𝑥 = 0 to 𝑥 = 5 using a Riemann sum
with = 10 subintervals and left endpoints.
Round your answer to four decimal places.
area
Your response Correct response
0.5542±0.0001
Auto graded Grade: 0/1.0 F 0%
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Total
grade:
0.0×1/2 +
0.0×1/2 =
0% + 0%
Feedback:
The function is continuous on the interval 0,5.
Divide the interval 𝑎, = 0,5 into 𝑛 = 10 subintervals of equal length to get
(a) The right endpoint of each subinterval is given by
= 𝑎 + 𝑖 Δ 𝑥 = 0.5𝑖.
Therefore, the approximate area under the graph of 𝑓𝑥 is
+ .. . +
(b) The left endpoint of each subinterval is given by
= 𝑎 + 𝑖 − 1 Δ 𝑥 = 0.5𝑖 − 1.
Therefore, the approximate area under the graph of 𝑓𝑥 is
+ .. . +
Question12: Score 0/4
Estimate the area under the graph of the function 𝑓𝑥 = √𝑥 + 3 from 𝑥 = − 2 to 𝑥 = 2 using a
Riemann sum with = 10 subintervals and midpoints.
Round your answer to four decimal places.
area
Your response Correct response
6.7887±0.0001
Auto graded Grade: 0/1.0 F 0%
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Total
grade:
0.0×1/1
= 0%
Feedba
ck:
The function 𝑓𝑥 = √𝑥 + 3 is continuous on the interval −2,2.
Divide the interval 𝑎, = −2,2 into 𝑛 = 10 subintervals of equal length to get
The midpoint of each subinterval is given by
Therefore, the approximate area under the graph of 𝑓𝑥 is
Δ�
�
=𝑏− 𝑎
𝑛
=2 − −2
10
= 0.4
.
