c7hw-dun42rkq.pdf
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Student: Sam Henry
Date: 06/14/21
Instructor:
Course: Assignment:
Evaluate the integral.
∫
− 11
− 10
dx
x
∫
− 11
− 10
dx
x
= ln
10
11
Evaluate the integral.
∫ 4y3dy
y4 − 3
∫ 4y3dy
y4 − 3
= ln + cy4 − 3
Evaluate .∫ dx
14 x + 14x
∫ dx
14 x + 14x
= + c
ln x + 1
7
Evaluate the integral .21 e dx∫ 7x + 3
21 e dx∫ 7x + 3 = 3 e + c7x + 3
Evaluate the integral .e dx∫
ln 8
ln 125 x
3
e dx∫
ln 8
ln 125 x
3 = 9
(Simplify your answer.)
Evaluate the integral .15t e dt∫ 4 − t5
15t e dt∫ 4 − t5 = − 3 e + c− t5
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Evaluate the integral .e ( − 6 sin (6x))dx∫ 6 cos (6x)
e ( − 6 sin (6x))dx∫ 6 cos (6x) = + ce
6 cos (6x)
6
Evaluate the integral.
dy∫ 7 e
7y
4 + e 7y
dy∫ 7 e 7y
4 + e 7y
= ln + Ce 7y + 4
Evaluate the integral.
x2 dx∫
1
3
x2
(Type an exact answer.)x2 dx∫
1
3
x2 =
3
ln 2
Evaluate the following integral.
(1 + ln x)x dx∫
1
3
x
(Type an exact answer.)(1 + ln x)x dx∫
1
3
x = 26
Evaluate the integral .x dx∫
0
9
2 + 1 2
x dx∫
0
9
2 + 1 2 = XXXXXXXXXX
(Type an exact answer, using radicals as needed.)
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Evaluate the integral.
dx∫
0
8
log 64(x + 8)
x + 8
dx∫
0
8
log 64(x + 8)
x + 8
=
(Type an exact answer.)
Evaluate the integral .∫ dx8x log 5x
∫ dx8x log 5x =
Solve the initial value problem.
, y( )
dy
dt = 5 e sec 5π e
− t 2 − t ln 4 = 6 / π
y = − tan 5π e +
1
π
− t 7
π
(Type an exact answer, using as needed.) π
Solve the initial value problem.
, y(1) , (1)
d2y
dt2
= 1 − e 2t = − 1 y′ = 2
y = − e − 1 −
t2 + e 2 + 2 t
2
1
4
2t
e
2 + 6
4
Solve the initial value problem.
; ; = 5 sec x
d2y
dx2
2 y′
π
4 = 0 y(0) = 0
The solution is y .= − 5 ln − 5xcos (x)
(Type an exact answer, using as needed.)π
Find the length of the following curve.
, y = − 4 ln x
x2
32 5 ≤ x ≤ 30
The length of the curve is . (Type an exact answer.)
18. Instead of approximating near x 1, approximate near x 0 to obtain a simpler formula this way.ln x = ln XXXXXXXXXXx =
(a) Find the linearization of at x 0.ln XXXXXXXXXXx =
(b) Estimate the five decimal place e
or involved in replacing with x on the interval [0,0.1].ln XXXXXXXXXXx
(c) Graph and x together for 0 x 0.5. Use different colors, if available. At what points does the approximation of
seem the best? Least good? By reading coordinates from the graph, find as good an upper bound for the e
or as
the graphing utility will allow.
ln XXXXXXXXXXx ≤ ≤
ln XXXXXXXXXXx
(a) The linearization of at x 0 is L(x) .ln XXXXXXXXXXx = ≈
(b) The approximation e
or is . (Round to five decimal places as needed.)
(c) Choose the co
ect graph of y and y x shown both for 0 x 0.5.= ln XXXXXXXXXXx = ≤ ≤
A.
XXXXXXXXXX
-0.5
0.5
x
y
B.
XXXXXXXXXX
-0.5
0.5
x
y
C.
XXXXXXXXXX
-0.5
0.7
x
y
D.
XXXXXXXXXX
-0.5
0.5
x
y
Find the x-value in the interval 0 x 0.5 at which the approximation of seems the best. Choose the co
ect
answer below.
≤ ≤ ln XXXXXXXXXXx
A. 0.5
B. 0
C. 0.3
D. 0.2
Find the x-value in the interval 0 x 0.5 at which the approximation of seems the least good. Choose the co
ect
answer below.
≤ ≤ ln XXXXXXXXXXx
A. 0.2
B. 0.1
C. 0.5
D. 0.4
By reading the coordinates from the graph, the upper bound for the e
or is .
(Round to four decimal places as needed.)
y = x
y = ln XXXXXXXXXXx y = ln XXXXXXXXXXx
y = x
y = x
y = ln XXXXXXXXXXx
y = ln XXXXXXXXXXx
y = x
19. Which is larger, or ? Answer this question by completing parts (a) through (e) below.πe e π
a. Find an equation for the line through the origin that is tangent to the graph of y .= ln x
[ 3,6] by [ 3,3]− −
The equation for the line through the origin tangent to the graph of y is y .= ln x =
x
e
(Type an exact answer.)
. Give an argument based on the graphs of y and the tangent line to explain why for all positive x .= ln x ln x
x
e
≠ e
To determine why for all positive x , first, notice that when , .ln x
x
e
≠ e x = e ln x =
x
e
=
Consider how each function changes over time by finding the derivative of each.
(Simplify your answer.)ln x
d
dx
=
(Simplify your answer.)
d
dx
x
e
=
For , the derivative of the function is (1) than the derivative of the function . Therefore, as x
decreases from , decreases (2) than .
0 < x < e
x
e
ln x
e ln x
x
e
Similarly, for , the derivative of the function is (3) than the derivative of the function .
Therefore, as x increases from , increases (4) than .
e < x
x
e
ln x
e ln x
x
e
Based on the previous results, what can be concluded?
A. The graph of y lies below the tangent line for all x .= ln x > e
B. The graph of y lies above the tangent line for all x .= ln x > e
C. The graph of y lies above the tangent line for all positive x .= ln x ≠ e
D. The graph of y lies below the tangent line for all positive x .= ln x ≠ e
c. Show that for all positive x .ln < xxe ≠ e
ln (x) < x
e
x Multiply both sides by .e
ln < x Use the Power Rule
(1) less
greate
(2) faste
more slowly
(3) greate
less
(4) faste
more slowly
d. Conclude that for all positive x . Using the result of the previous step how can this conclusion be drawn?
Select the co
ect choice below and, if necessary, fill in the answer box to complete your choice.
x < ee x ≠ e
A. Raise to the power of each side of the inequality , and simplify.ln < xxe
B. Raise each side of the inequality to the power of , and simplify.ln < xxe
C. This cannot be concluded as the statement is false.
e. So, which is larger, or ?πe e π
e
π
π
e
YOU ANSWERED: 2x
20. Check whether each of the following functions is a solution of the differential equation .3 + 7yy′ = 4 e − x
(a) y = e − x (b) y = e − x + e − (7 / 3)x
(c) y C= e − x + e − (7 / 3)x
(a) Find , y, and for y .3y′ 7 3 + 7yy′ = e − x
3y′ = − 3 e − x
y7 = 7 e − x
3 + 7yy′ = 4 e − x
Is the function y a solution of ? Choose the co
ect answer below.= e − x 3 + 7yy′ = 4 e − x
No
Yes
(b) Find , y, and for y .3y′ 7 3 + 7yy′ = e + e− x − (7 / 3)x
3y′ = − 3 e − 7 e− x − (7 / 3)x
y7 = 7 e + 7 e− x − (7 / 3)x
3 + 7yy′ = 4 e − x
Is the function y a solution of ? Choose the co
ect answer below.= e + e− x − (7 / 3)x 3 + 7yy′ = 4 e − x
No
Yes
(c) Find , y, and for y C .3y′ 7 3 + 7yy′ = e − x + e − (7 / 3)x
3y′ = − 3 e − 7C e− x − (7 / 3)x
y7 = 7 e + 7C e− x − (7 / 3)x
3 + 7yy′ = 4 e − x
Is the function y C a solution of ? Choose the co
ect answer below.= e − x + e − (7 / 3)x 3 + 7yy′ = 4 e − x
No
Yes
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Check whether the function y is a solution of the differential equation y with the
initial condition .
= e tan 6 e−