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Math Problems
Calculus
Evaluate definite integrals using the power rule
∫
2
x
d
x
=
\int 2^{x} d x=
∫
2
x
d
x
=
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∫
1
e
2
x
3
+
1
x
d
x
=
\int_{1}^{e^{2}} \frac{x^{3}+1}{x} d x=
∫
1
e
2
x
x
3
+
1
d
x
=
Get tutor help
The population of a town grows at a rate of
r
(
t
)
r(t)
r
(
t
)
people per year (where
t
t
t
is time in years).
\newline
What does
∫
2
4
r
(
t
)
d
t
\int_{2}^{4}r(t)\,dt
∫
2
4
r
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The average rate at which the population grew between the second and the fourth year.
\newline
(B) The change in number of people between the second and the fourth year.
\newline
(C) The number of people in the town on the fourth year.
\newline
(D) The time it took for the town to grow from a population of
2
2
2
people to a population of
4
4
4
people.
Get tutor help
g
(
x
)
=
∫
−
10
x
(
10
−
3
t
)
d
t
g(x)=\int_{-10}^{x}(10-3 t) d t
g
(
x
)
=
∫
−
10
x
(
10
−
3
t
)
d
t
\newline
g
′
(
−
4
)
=
g^{\prime}(-4)=
g
′
(
−
4
)
=
Get tutor help
g
(
x
)
=
∫
−
12
x
(
4
t
−
7
)
d
t
g(x)=\int_{-12}^{x}(4 t-7) d t
g
(
x
)
=
∫
−
12
x
(
4
t
−
7
)
d
t
\newline
g
′
(
5
)
=
g^{\prime}(5)=
g
′
(
5
)
=
Get tutor help
g
(
x
)
=
∫
−
10
x
t
2
+
11
d
t
g(x)=\int_{-10}^{x} \sqrt{t^{2}+11} d t
g
(
x
)
=
∫
−
10
x
t
2
+
11
d
t
\newline
g
′
(
−
5
)
=
g^{\prime}(-5)=
g
′
(
−
5
)
=
Get tutor help
g
(
x
)
=
∫
0
x
(
5
t
2
−
t
)
d
t
g(x)=\int_{0}^{x}\left(5 t^{2}-t\right) d t
g
(
x
)
=
∫
0
x
(
5
t
2
−
t
)
d
t
\newline
g
′
(
2
)
=
g^{\prime}(2)=
g
′
(
2
)
=
Get tutor help
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
)
=
Get tutor help
F
(
x
)
=
∫
1
2
x
1
+
t
3
d
t
F(x)=\int_{1}^{2 x} \sqrt{1+t^{3}} d t
F
(
x
)
=
∫
1
2
x
1
+
t
3
d
t
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
g
(
x
)
=
∫
0
x
t
2
−
t
t
d
t
g(x)=\int_{0}^{x} \frac{t^{2}-t}{\sqrt{t}} d t
g
(
x
)
=
∫
0
x
t
t
2
−
t
d
t
\newline
g
′
(
4
)
=
g^{\prime}(4)=
g
′
(
4
)
=
Get tutor help
g
(
x
)
=
∫
0
x
5
+
4
tan
t
d
t
g(x)=\int_{0}^{x} \sqrt{5+4 \tan t} d t
g
(
x
)
=
∫
0
x
5
+
4
tan
t
d
t
\newline
g
′
(
π
4
)
=
g^{\prime}\left(\frac{\pi}{4}\right)=
g
′
(
4
π
)
=
Get tutor help
F
(
x
)
=
∫
2
2
x
15
−
t
d
t
F(x)=\int_{2}^{2 x} \sqrt{15-t} d t
F
(
x
)
=
∫
2
2
x
15
−
t
d
t
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
F
(
x
)
=
∫
−
2
2
x
(
3
t
2
+
2
t
)
d
t
F(x)=\int_{-2}^{2 x}\left(3 t^{2}+2 t\right) d t
F
(
x
)
=
∫
−
2
2
x
(
3
t
2
+
2
t
)
d
t
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
F
(
x
)
=
∫
1
2
x
t
2
t
2
+
1
d
t
F(x)=\int_{1}^{2 x} \frac{t^{2}}{t^{2}+1} d t
F
(
x
)
=
∫
1
2
x
t
2
+
1
t
2
d
t
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
g
(
x
)
=
∫
2
x
(
10
t
+
2
)
d
t
g(x)=\int_{2}^{x}(10 t+2) d t
g
(
x
)
=
∫
2
x
(
10
t
+
2
)
d
t
\newline
g
′
(
3
)
=
g^{\prime}(3)=
g
′
(
3
)
=
Get tutor help
g
(
x
)
=
∫
0
x
5
+
4
cos
t
d
t
g(x)=\int_{0}^{x} \sqrt{5+4 \cos t} d t
g
(
x
)
=
∫
0
x
5
+
4
cos
t
d
t
\newline
g
′
(
π
)
=
g^{\prime}(\pi)=
g
′
(
π
)
=
Get tutor help
g
(
x
)
=
∫
−
π
x
sin
(
t
)
d
t
g(x)=\int_{-\pi}^{x} \sin (t) d t
g
(
x
)
=
∫
−
π
x
sin
(
t
)
d
t
\newline
g
′
(
π
)
=
g^{\prime}(\pi)=
g
′
(
π
)
=
Get tutor help
F
(
x
)
=
∫
0
x
2
t
d
t
F(x)=\int_{0}^{\sqrt{x}} 2 t d t
F
(
x
)
=
∫
0
x
2
t
d
t
\newline
where
x
>
0
x>0
x
>
0
.
\newline
F
′
(
x
)
=
F^{\prime}(x)=
F
′
(
x
)
=
Get tutor help
H
(
x
)
=
−
x
−
5
H(x)=-x-5
H
(
x
)
=
−
x
−
5
\newline
h
(
x
)
=
H
′
(
x
)
h(x)=H^{\prime}(x)
h
(
x
)
=
H
′
(
x
)
\newline
∫
−
3
6
h
(
x
)
d
x
=
\int_{-3}^{6} h(x) d x=
∫
−
3
6
h
(
x
)
d
x
=
Get tutor help
F
(
x
)
=
x
+
7
F(x)=\sqrt{x+7}
F
(
x
)
=
x
+
7
\newline
f
(
x
)
=
F
′
(
x
)
f(x)=F^{\prime}(x)
f
(
x
)
=
F
′
(
x
)
\newline
∫
2
9
f
(
x
)
d
x
=
\int_{2}^{9} f(x) d x=
∫
2
9
f
(
x
)
d
x
=
Get tutor help
G
(
x
)
=
3
x
G(x)=\sqrt{3 x}
G
(
x
)
=
3
x
\newline
g
(
x
)
=
G
′
(
x
)
g(x)=G^{\prime}(x)
g
(
x
)
=
G
′
(
x
)
\newline
∫
3
12
g
(
x
)
d
x
=
\int_{3}^{12} g(x) d x=
∫
3
12
g
(
x
)
d
x
=
Get tutor help
F
(
x
)
=
5
x
F(x)=5^{x}
F
(
x
)
=
5
x
\newline
f
(
x
)
=
F
′
(
x
)
f(x)=F^{\prime}(x)
f
(
x
)
=
F
′
(
x
)
\newline
∫
0
3
f
(
x
)
d
x
=
\int_{0}^{3} f(x) d x=
∫
0
3
f
(
x
)
d
x
=
Get tutor help
G
(
x
)
=
cos
(
3
x
)
G(x)=\cos (3 x)
G
(
x
)
=
cos
(
3
x
)
\newline
g
(
x
)
=
G
′
(
x
)
g(x)=G^{\prime}(x)
g
(
x
)
=
G
′
(
x
)
\newline
∫
0
π
g
(
x
)
d
x
=
\int_{0}^{\pi} g(x) d x=
∫
0
π
g
(
x
)
d
x
=
Get tutor help
G
(
x
)
=
2
x
3
−
4
G(x)=2 x^{3}-4
G
(
x
)
=
2
x
3
−
4
\newline
g
(
x
)
=
G
′
(
x
)
g(x)=G^{\prime}(x)
g
(
x
)
=
G
′
(
x
)
\newline
∫
1
2
g
(
x
)
d
x
=
\int_{1}^{2} g(x) d x=
∫
1
2
g
(
x
)
d
x
=
Get tutor help
H
(
x
)
=
1
0
x
H(x)=10^{x}
H
(
x
)
=
1
0
x
\newline
h
(
x
)
=
H
′
(
x
)
h(x)=H^{\prime}(x)
h
(
x
)
=
H
′
(
x
)
\newline
∫
−
3
2
h
(
x
)
d
x
=
\int_{-3}^{2} h(x) d x=
∫
−
3
2
h
(
x
)
d
x
=
Get tutor help
Evaluate.
\newline
k
λ
4
R
∫
−
π
2
π
/
2
sec
θ
2
⋅
d
θ
\frac{k \lambda}{4 R} \int_{-\frac{\pi}{2}}^{\pi / 2} \sec \frac{\theta}{2} \cdot d \theta
4
R
kλ
∫
−
2
π
π
/2
sec
2
θ
⋅
d
θ
Get tutor help
Evaluate the limit:
\newline
lim
x
→
0
(
1
x
−
cot
x
)
=
\lim _{x \rightarrow 0}\left(\frac{1}{x}-\cot x\right)=
x
→
0
lim
(
x
1
−
cot
x
)
=
Get tutor help
∫
0
3
x
2
+
4
x
+
5
x
+
3
d
x
=
\int_{0}^{3} \frac{x^{2}+4 x+5}{x+3} d x=
∫
0
3
x
+
3
x
2
+
4
x
+
5
d
x
=
Get tutor help
∫
(
4
x
2
−
6
)
(
5
x
2
+
3
)
d
x
=
_
_
_
_
_
\int\left(4 x^{2}-6\right)\left(5 x^{2}+3\right) d x=\_\_\_\_\_
∫
(
4
x
2
−
6
)
(
5
x
2
+
3
)
d
x
=
_____
Get tutor help
∫
2
x
(
x
2
+
1
)
2
d
x
=
\int 2 x\left(x^{2}+1\right)^{2} d x=
∫
2
x
(
x
2
+
1
)
2
d
x
=
Get tutor help
Integrate.
\newline
∫
x
+
5
x
2
−
2
x
−
3
d
x
\int \frac{x+5}{x^{2}-2 x-3} d x
∫
x
2
−
2
x
−
3
x
+
5
d
x
Get tutor help
Expand the logarithm fully using the properties of logs. Express the final answer in terms of
log
x
,
log
y
\log x, \log y
lo
g
x
,
lo
g
y
, and
log
z
\log z
lo
g
z
.
\newline
log
x
5
y
4
z
4
3
\log \frac{x^{5} y^{4}}{\sqrt[3]{z^{4}}}
lo
g
3
z
4
x
5
y
4
\newline
Answer:
Get tutor help
Expand the logarithm fully using the properties of logs. Express the final answer in terms of
log
x
,
log
y
\log x, \log y
lo
g
x
,
lo
g
y
, and
log
z
\log z
lo
g
z
.
\newline
log
x
3
z
5
y
\log \frac{\sqrt{x^{3}}}{z^{5} y}
lo
g
z
5
y
x
3
\newline
Answer:
Get tutor help
Expand the logarithm fully using the properties of logs. Express the final answer in terms of
log
x
,
log
y
\log x, \log y
lo
g
x
,
lo
g
y
, and
log
z
\log z
lo
g
z
.
\newline
log
x
5
y
3
z
4
\log \frac{\sqrt{x^{5}}}{y^{3} z^{4}}
lo
g
y
3
z
4
x
5
\newline
Answer:
Get tutor help
Expand the logarithm fully using the properties of logs. Express the final answer in terms of
log
x
,
log
y
\log x, \log y
lo
g
x
,
lo
g
y
, and
log
z
\log z
lo
g
z
.
\newline
log
x
5
z
3
y
2
\log \frac{\sqrt{x^{5}}}{z^{3} y^{2}}
lo
g
z
3
y
2
x
5
\newline
Answer:
Get tutor help
Expand the logarithm fully using the properties of logs. Express the final answer in terms of
log
x
,
log
y
\log x, \log y
lo
g
x
,
lo
g
y
, and
log
z
\log z
lo
g
z
.
\newline
log
y
3
z
5
3
x
2
\log \frac{y^{3}}{\sqrt[3]{z^{5}} x^{2}}
lo
g
3
z
5
x
2
y
3
\newline
Answer:
Get tutor help
Find the numerical value of the log expression.
\newline
log
a
=
−
4
log
b
=
−
5
log
c
=
−
9
log
c
9
a
3
b
8
\begin{array}{c} \log a=-4 \quad \log b=-5 \quad \log c=-9 \\ \log \frac{c^{9}}{a^{3} b^{8}} \end{array}
lo
g
a
=
−
4
lo
g
b
=
−
5
lo
g
c
=
−
9
lo
g
a
3
b
8
c
9
\newline
Answer:
Get tutor help
Find the numerical value of the log expression.
\newline
log
a
=
10
log
b
=
12
log
c
=
−
3
log
a
2
b
9
c
5
3
\begin{array}{c} \log a=10 \quad \log b=12 \quad \log c=-3 \\ \log \frac{a^{2} b^{9}}{\sqrt[3]{c^{5}}} \end{array}
lo
g
a
=
10
lo
g
b
=
12
lo
g
c
=
−
3
lo
g
3
c
5
a
2
b
9
\newline
Answer:
Get tutor help
Find the numerical value of the log expression.
\newline
log
a
=
−
11
log
b
=
−
5
log
c
=
6
log
a
7
c
5
b
7
\begin{array}{c} \log a=-11 \quad \log b=-5 \quad \log c=6 \\ \log \frac{a^{7} c^{5}}{b^{7}} \end{array}
lo
g
a
=
−
11
lo
g
b
=
−
5
lo
g
c
=
6
lo
g
b
7
a
7
c
5
\newline
Answer:
Get tutor help
Find the numerical value of the log expression.
\newline
log
a
=
−
12
log
b
=
−
8
log
c
=
−
6
log
b
3
c
3
a
9
\begin{array}{c} \log a=-12 \quad \log b=-8 \quad \log c=-6 \\ \log \frac{\sqrt{b^{3} c^{3}}}{a^{9}} \end{array}
lo
g
a
=
−
12
lo
g
b
=
−
8
lo
g
c
=
−
6
lo
g
a
9
b
3
c
3
\newline
Answer:
Get tutor help
Find the numerical value of the log expression.
\newline
log
a
=
9
log
b
=
12
log
c
=
3
log
a
3
c
3
b
2
3
\begin{array}{c} \log a=9 \quad \log b=12 \quad \log c=3 \\ \log \frac{a^{3} c^{3}}{\sqrt[3]{b^{2}}} \end{array}
lo
g
a
=
9
lo
g
b
=
12
lo
g
c
=
3
lo
g
3
b
2
a
3
c
3
\newline
Answer:
Get tutor help
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
∫
1
x
3
+
1
d
x
=
\int \frac{1}{x^{3}+1} \, dx =
∫
x
3
+
1
1
d
x
=
Get tutor help
Rewrite the expression as a product of four linear factors:
\newline
(
x
2
+
8
x
)
2
+
19
(
x
2
+
8
x
)
+
84
\left(x^{2}+8 x\right)^{2}+19\left(x^{2}+8 x\right)+84
(
x
2
+
8
x
)
2
+
19
(
x
2
+
8
x
)
+
84
\newline
Answer:
Get tutor help
Rewrite the expression as a product of four linear factors:
\newline
(
x
2
−
x
)
2
−
22
(
x
2
−
x
)
+
40
\left(x^{2}-x\right)^{2}-22\left(x^{2}-x\right)+40
(
x
2
−
x
)
2
−
22
(
x
2
−
x
)
+
40
\newline
Answer:
Get tutor help
Rewrite the expression as a product of four linear factors:
\newline
(
2
x
2
−
13
x
)
2
−
(
2
x
2
−
13
x
)
−
42
\left(2 x^{2}-13 x\right)^{2}-\left(2 x^{2}-13 x\right)-42
(
2
x
2
−
13
x
)
2
−
(
2
x
2
−
13
x
)
−
42
\newline
Answer:
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∫
sin
3
(
x
)
cos
2
(
x
)
=
\int \frac{\sin ^{3}(x)}{\cos ^{2}(x)}=
∫
c
o
s
2
(
x
)
s
i
n
3
(
x
)
=
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∫
d
x
x
e
(
40
x
)
=
\int \frac{d x}{\sqrt{x} e^{(40 \sqrt{x})}}=
∫
x
e
(
40
x
)
d
x
=
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Calculate the integral and write your answer in simplest form.
\newline
∫
5
x
5
4
6
d
x
\int \frac{5 \sqrt[4]{x^{5}}}{6} \mathrm{dx}
∫
6
5
4
x
5
dx
\newline
Answer:
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Calculate the integral and write your answer in simplest form.
\newline
∫
5
x
3
4
2
d
x
\int \frac{5 \sqrt[4]{x^{3}}}{2} \mathrm{dx}
∫
2
5
4
x
3
dx
\newline
Answer:
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1
2
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