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Let’s check out your problem:
g
(
x
)
=
∫
−
7
x
(
t
−
t
2
)
d
t
g(x)=\int_{-7}^{x}\left(t-t^{2}\right) d t
g
(
x
)
=
∫
−
7
x
(
t
−
t
2
)
d
t
\newline
g
′
(
10
)
=
g^{\prime}(10)=
g
′
(
10
)
=
View step-by-step help
Home
Math Problems
Calculus
Evaluate definite integrals using the power rule
Full solution
Q.
g
(
x
)
=
∫
−
7
x
(
t
−
t
2
)
d
t
g(x)=\int_{-7}^{x}\left(t-t^{2}\right) d t
g
(
x
)
=
∫
−
7
x
(
t
−
t
2
)
d
t
\newline
g
′
(
10
)
=
g^{\prime}(10)=
g
′
(
10
)
=
Find Derivative of
g
(
x
)
g(x)
g
(
x
)
:
First, let's find the derivative of
g
(
x
)
g(x)
g
(
x
)
using the Fundamental Theorem of Calculus.
g
′
(
x
)
=
d
d
x
[
∫
−
7
x
(
t
−
t
2
)
d
t
]
=
x
−
x
2
g'(x) = \frac{d}{dx} \left[\int_{-7}^{x}(t - t^2) dt\right] = x - x^2
g
′
(
x
)
=
d
x
d
[
∫
−
7
x
(
t
−
t
2
)
d
t
]
=
x
−
x
2
More problems from Evaluate definite integrals using the power rule
Question
The number of subscribers to a magazine is changing at a rate of
r
(
t
)
r(t)
r
(
t
)
subscribers per month (where
t
t
t
is time in months).
\newline
What does
∫
8
10
r
′
(
t
)
d
t
=
7
\int_{8}^{10} r^{\prime}(t) d t=7
∫
8
10
r
′
(
t
)
d
t
=
7
mean?
\newline
Choose
1
1
1
answer:
\newline
(A) The rate of change of number of subscribers increased by
7
7
7
subscribers per month between
t
=
8
t=8
t
=
8
and
t
=
10
t=10
t
=
10
months.
\newline
(B) As of month
10
10
10
, the magazine had
7
7
7
subscribers.
\newline
(C) The number of subscribers increased by
7
7
7
between
t
=
8
t=8
t
=
8
and
t
=
10
t=10
t
=
10
months.
\newline
(D) The average rate of change in subscribers between month
8
8
8
and month
10
10
10
was
7
7
7
subscribers per month.
Get tutor help
Posted 3 months ago
Question
Consider the following problem:
\newline
The water level under a bridge is changing at a rate of
r
(
t
)
=
40
sin
(
π
t
6
)
r(t)=40 \sin \left(\frac{\pi t}{6}\right)
r
(
t
)
=
40
sin
(
6
π
t
)
centimeters per hour (where
t
t
t
is the time in hours). At time
t
=
3
t=3
t
=
3
, the water level is
91
91
91
centimeters. By how much does the water level change during the
4
th
4^{\text {th }}
4
th
hour?
\newline
Which expression can we use to solve the problem?
\newline
Choose
1
1
1
answer:
\newline
(A)
∫
0
4
r
(
t
)
d
t
\int_{0}^{4} r(t) d t
∫
0
4
r
(
t
)
d
t
\newline
(B)
∫
4
5
r
(
t
)
d
t
\int_{4}^{5} r(t) d t
∫
4
5
r
(
t
)
d
t
\newline
(C)
∫
3
4
r
(
t
)
d
t
\int_{3}^{4} r(t) d t
∫
3
4
r
(
t
)
d
t
\newline
(D)
∫
4
4
r
(
t
)
d
t
\int_{4}^{4} r(t) d t
∫
4
4
r
(
t
)
d
t
Get tutor help
Posted 3 months ago
Question
The base of a solid is the region enclosed by the graphs of
y
=
sin
(
x
)
y=\sin (x)
y
=
sin
(
x
)
and
y
=
4
−
x
y=4-\sqrt{x}
y
=
4
−
x
, between
x
=
2
x=2
x
=
2
and
x
=
7
x=7
x
=
7
.
\newline
Cross sections of the solid perpendicular to the
x
x
x
-axis are rectangles whose height is
2
x
2 x
2
x
.
\newline
Which one of the definite integrals gives the volume of the solid?
\newline
Choose
1
1
1
answer:
\newline
(A)
∫
2
7
[
sin
(
x
)
+
x
−
4
]
⋅
2
x
d
x
\int_{2}^{7}[\sin (x)+\sqrt{x}-4] \cdot 2 x d x
∫
2
7
[
sin
(
x
)
+
x
−
4
]
⋅
2
x
d
x
\newline
(B)
∫
2
7
[
sin
2
(
x
)
−
(
4
−
x
)
2
]
d
x
\int_{2}^{7}\left[\sin ^{2}(x)-(4-\sqrt{x})^{2}\right] d x
∫
2
7
[
sin
2
(
x
)
−
(
4
−
x
)
2
]
d
x
\newline
(C)
∫
2
7
[
4
−
x
−
sin
(
x
)
]
⋅
2
x
d
x
\int_{2}^{7}[4-\sqrt{x}-\sin (x)] \cdot 2 x d x
∫
2
7
[
4
−
x
−
sin
(
x
)]
⋅
2
x
d
x
\newline
(D)
∫
2
7
[
(
4
−
x
)
2
−
sin
2
(
x
)
]
d
x
\int_{2}^{7}\left[(4-\sqrt{x})^{2}-\sin ^{2}(x)\right] d x
∫
2
7
[
(
4
−
x
)
2
−
sin
2
(
x
)
]
d
x
Get tutor help
Posted 3 months ago
Question
Nork/en/geometry/unit/
9
9
9
/lesson/
1
1
1
\newline
Coursework
\newline
Edgext
\newline
Wall
\newline
Menu
\newline
Triangle
\newline
△
O
G
K
\triangle OGK
△
OG
K
is shown where
\newline
O
G
=
22.3
OG=22.3
OG
=
22.3
centimeters (cond
\newline
G
K
=
32.5
cm
GK=32.5\text{cm}
G
K
=
32.5
cm
,
O
K
=
21.9
cm
OK=21.9\text{cm}
O
K
=
21.9
cm
. Height of the triangle is
\newline
15
cm
15\text{cm}
15
cm
.
\newline
What is the area of
\newline
△
O
G
K
\triangle OGK
△
OG
K
?
\newline
(A)
\newline
244.185
cm
2
244.185\text{cm}^2
244.185
cm
2
\newline
(B)
\newline
243.75
cm
2
243.75\text{cm}^2
243.75
cm
2
\newline
(C)
\newline
167.25
cm
2
167.25\text{cm}^2
167.25
cm
2
\newline
(D)
\newline
487.5
cm
2
487.5\text{cm}^2
487.5
cm
2
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Posted 2 months ago
Question
∫
(
e
cot
−
1
(
2
z
)
)
1
4
+
z
2
d
z
\int \left(e^{\cot^{-1}\left(\frac{2}{z}\right)}\right)\frac{1}{4+z^{2}}dz
∫
(
e
c
o
t
−
1
(
z
2
)
)
4
+
z
2
1
d
z
Get tutor help
Posted 2 months ago
Question
∫
0
π
(
1
−
cos
(
x
)
)
d
x
=
\int_{0}^{\pi}(1-\cos(x))\,dx=
∫
0
π
(
1
−
cos
(
x
))
d
x
=
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Posted 2 months ago
Question
∫
(
1
+
3
x
)
x
2
d
x
\int (1+3x)x^{2}\,dx
∫
(
1
+
3
x
)
x
2
d
x
Get tutor help
Posted 2 months ago
Question
∫
sin
6
(
x
)
cos
3
(
x
)
d
x
\int \frac{\sin^6(x)}{\cos^3(x)} \, dx
∫
c
o
s
3
(
x
)
s
i
n
6
(
x
)
d
x
Get tutor help
Posted 2 months ago
Question
Find
∫
(
2
+
5
x
−
1
(
x
−
2
)
2
)
d
x
\int(2+5x-\frac{1}{(x-2)^{2}})dx
∫
(
2
+
5
x
−
(
x
−
2
)
2
1
)
d
x
Get tutor help
Posted 2 months ago
Question
Consider the reaction between
Na
2
CO
3
(
aq
)
\text{Na}_2\text{CO}_3(\text{aq})
Na
2
CO
3
(
aq
)
and
H
3
PO
4
(
aq
)
\text{H}_3\text{PO}_4(\text{aq})
H
3
PO
4
(
aq
)
as shown in the following balanced equation:
\newline
3
Na
2
CO
3
(
aq
)
+
2
H
3
PO
4
(
aq
)
→
2
Na
3
PO
4
(
aq
)
+
3
H
2
O
(
l
)
+
3
CO
2
(
g
)
3\text{Na}_2\text{CO}_3(\text{aq})+2\text{H}_3\text{PO}_4(\text{aq})\rightarrow 2\text{Na}_3\text{PO}_4(\text{aq})+3\text{H}_2\text{O}(\text{l})+3\text{CO}_2(\text{g})
3
Na
2
CO
3
(
aq
)
+
2
H
3
PO
4
(
aq
)
→
2
Na
3
PO
4
(
aq
)
+
3
H
2
O
(
l
)
+
3
CO
2
(
g
)
\newline
\newline
Chemical
\newline
MW
\newline
\newline
Na
2
CO
3
\text{Na}_2\text{CO}_3
Na
2
CO
3
\newline
105.991
105.991
105.991
\newline
\newline
H
3
PO
4
\text{H}_3\text{PO}_4
H
3
PO
4
\newline
97.9977
97.9977
97.9977
\newline
\newline
Na
3
PO
4
\text{Na}_3\text{PO}_4
Na
3
PO
4
\newline
163.944
163.944
163.944
\newline
\newline
H
2
O
\text{H}_2\text{O}
H
2
O
\newline
18.0158
18.0158
18.0158
\newline
\newline
H
3
PO
4
(
aq
)
\text{H}_3\text{PO}_4(\text{aq})
H
3
PO
4
(
aq
)
0
0
0
\newline
H
3
PO
4
(
aq
)
\text{H}_3\text{PO}_4(\text{aq})
H
3
PO
4
(
aq
)
1
1
1
\newline
\newline
How many gram of
Na
2
CO
3
\text{Na}_2\text{CO}_3
Na
2
CO
3
is used if
H
3
PO
4
(
aq
)
\text{H}_3\text{PO}_4(\text{aq})
H
3
PO
4
(
aq
)
3
3
3
is made assuming all reactant is converted into products?
Get tutor help
Posted 2 months ago
