Solution 03 04 2016 at 03 00am EST 1 pt Use the Midpoint Rule to approximate Z 5 1 pt Evaluate the denite integral Z 3 t ln t dt x3 dx 1 1 1 pt Solution at am EST pt Use the Midpoint Rule to approximate Z pt Evaluate the denite integral Z t ln Solution at am EST pt Use the Midpoint Rule to approximate Z pt Evaluate the Use the Midpoint Rule to approximate Z pt Evaluate the denite integral Z t ln t dt x dx pt Solution at am EST pt Use the Midpoint Rule to approximate Z pt Evaluate the denite integral Z t ln t dt x dx pt Solution at am EST pt Use the Midpoint Rule to Solution at am EST pt (Solution) 03/04/2016 at 03:00am EST. (1 pt) Use the Midpoint Rule to approximate Z 5. (1 pt) Evaluate the denite integral. Z 3 t ln(t) dt = x3 dx 1 1. (1 pt)...
Hi, I need to solve the first 15 questions. ThanksMohammed Yaaseen Gomdah
E
WeBWorK assignment due : 03/04/2016 at 03:00am EST.
1.
(1 pt) Use the Midpoint Rule to approximate
Z
5
.
5

1
.
5
x
3
dx
with
n
=
7.
2.
(1 pt) Consider the function
f
(
x
) =

x
2
2

5.
In this problem you will calculate
Z
4
0
±

x
2
2

5
²
dx
by us
ing the de?nition
Z
b
a
f
(
x
)
dx
=
lim
n
?
?
"
n
?
i
=
1
f
(
x
i
)
?
x
#
The summation inside the brackets is
R
n
which is the Rie
mann sum where the sample points are chosen to be the right
hand endpoints of each subinterval.
Calculate
R
n
for
f
(
x
) =

x
2
2

5 on the interval
[
0
,
4
]
and
write your answer as a function of
n
without any summation
signs. You will need the summation formulas on page 383 of
your textbook (page 364 in older texts).
R
n
=
lim
n
?
?
R
n
=
3.
(1
pt)
Let
Z

4

7
f
(
x
)
dx
=
1
,
Z

6

7
f
(
x
)
dx
=
9
,
Z

4

5
f
(
x
)
dx
=
1.
Find
Z

5

6
f
(
x
)
dx
=
and
Z

6

5
(
1
f
(
x
)

9
)
dx
=
4.
(1 pt) If
f
(
x
) =
Z
x
2
3
t
4
dt
then
f
0
(
x
) =
f
0
(
5
) =
5.
(1 pt) Given
f
(
x
) =
Z
x
0
t
2

4
1
+
cos
2
(
t
)
dt
At what value of
x
does the local max of
f
(
x
)
occur?
x
=
6.
(1 pt) Evaluate the de?nite integral.
Z
1
0
x
2 5
?
e
x
dx
=
7.
(1 pt) Evaluate the de?nite integral.
Z
3
1
?
t
ln
(
t
)
dt
=
8.
(1 pt) Use the Fundamental Theorem of Calculus to carry
out the following differentiation:
d
d
x
Z
?
x
1
t
t
d
t
=
.
9.
(1 pt)
Integration by Parts:
This is the most important in
tegration technique we’ve discussed in this class. It has a wide
range of applications beyond increasing our list of integration
rules.
R
z
3
ln
z
d
z
=
.
R
e
t
cos
t
d
t
=
.
R
2
?
0
sin
(
x
)
sin
(
x
+
1
)
d
x
=
.
10.
(1 pt) Use integration by parts to evaluate the integral.
Z
xe
4
x
dx
Answer:
+
C
11.
(1 pt) Use integration by parts to evaluate the de?nite
integral.
Z
7
1
?
t
ln
tdt
Answer:
12.
(1 pt) Evaluate the de?nite integral.
Z
5
3
ln
x
42
dx
Answer:
13.
(1 pt) Evaluate the inde?nite integral.
Z
ln
(
x
2
+
9
x
+
18
)
dx
1