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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 9 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 9 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 9 months ago
Question
(
1
×
(
(
1
2
)
n
−
1
)
)
/
(
(
1
2
)
−
1
)
\left(1\times\left(\left(\frac{1}{2}\right)^n-1\right)\right)\bigg/\left(\left(\frac{1}{2}\right)-1\right)
(
1
×
(
(
2
1
)
n
−
1
)
)
/
(
(
2
1
)
−
1
)
Get tutor help
Posted 8 months ago
Question
Solve
∫
1
(
x
−
2
)
(
x
2
+
x
+
1
)
d
x
\int \frac{1}{(x-2)(x^{2}+x+1)}\,dx
∫
(
x
−
2
)
(
x
2
+
x
+
1
)
1
d
x
.
Get tutor help
Posted 8 months ago
Question
What is the integral of the function
f
(
x
)
=
sin
(
2
x
)
f(x) = \sin(2x)
f
(
x
)
=
sin
(
2
x
)
?
\newline
−
1
2
cos
(
x
)
+
C
\frac{-1}{2}\cos(x) + C
2
−
1
cos
(
x
)
+
C
\newline
1
2
sin
(
x
)
+
C
\frac{1}{2}\sin(x) + C
2
1
sin
(
x
)
+
C
\newline
−
1
2
cos
(
2
x
)
+
C
\frac{-1}{2}\cos(2x) + C
2
−
1
cos
(
2
x
)
+
C
\newline
1
2
sin
(
2
x
)
+
C
\frac{1}{2}\sin(2x) + C
2
1
sin
(
2
x
)
+
C
Get tutor help
Posted 8 months ago
Question
What is the integral of the function
f
(
x
)
=
s
i
n
(
2
x
)
\ f(x) = sin(2x)
f
(
x
)
=
s
in
(
2
x
)
?
\newline
−
1
2
cos
(
x
)
+
C
\frac{-1}{2}\cos(x) + C
2
−
1
cos
(
x
)
+
C
\newline
1
2
sin
(
x
)
+
C
\frac{1}{2} \sin(x) + C
2
1
sin
(
x
)
+
C
\newline
−
1
2
cos
(
2
x
)
+
C
\frac{-1}{2}\cos(2x) + C
2
−
1
cos
(
2
x
)
+
C
\newline
1
2
sin
(
2
x
)
+
c
\frac{1}{2} \sin(2x) + c
2
1
sin
(
2
x
)
+
c
Get tutor help
Posted 8 months ago
Question
Suppose
∫
2
4
f
(
2
x
)
d
x
=
10
\int_{2}^{4} f(2x) \, dx = 10
∫
2
4
f
(
2
x
)
d
x
=
10
. Then
∫
4
8
f
(
u
)
d
u
=
\int_{4}^{8} f(u) \, du =
∫
4
8
f
(
u
)
d
u
=
Get tutor help
Posted 8 months ago
Question
Find
d
d
t
∫
2
t
4
e
x
2
d
x
\frac{d}{dt}\int_{2}^{t^{4}}e^{x^{2}}dx
d
t
d
∫
2
t
4
e
x
2
d
x
Get tutor help
Posted 8 months ago
Question
Evaluate and write the answer in scientific notation:
\newline
(
1.48
×
1
0
3
)
(
7
×
1
0
4
)
÷
8.2
×
1
0
12
(1.48\times10^{3})(7\times10^{4})\div8.2\times10^{12}
(
1.48
×
1
0
3
)
(
7
×
1
0
4
)
÷
8.2
×
1
0
12
Get tutor help
Posted 8 months ago
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