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Let’s check out your problem:
Evaluate the integral:
\newline
∫
e
2
x
−
1
e
2
x
+
3
d
x
\int\frac{e^{2x}-1}{e^{2x}+3}dx
∫
e
2
x
+
3
e
2
x
−
1
d
x
View step-by-step help
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Math Problems
Calculus
Evaluate definite integrals using the chain rule
Full solution
Q.
Evaluate the integral:
\newline
∫
e
2
x
−
1
e
2
x
+
3
d
x
\int\frac{e^{2x}-1}{e^{2x}+3}dx
∫
e
2
x
+
3
e
2
x
−
1
d
x
Substitution:
Let's do a substitution: let
u
=
e
2
x
+
3
u = e^{2x} + 3
u
=
e
2
x
+
3
. Then,
d
u
=
2
e
2
x
d
x
du = 2e^{2x}dx
d
u
=
2
e
2
x
d
x
. We need to solve for
d
x
dx
d
x
, so
d
x
=
d
u
2
e
2
x
dx = \frac{du}{2e^{2x}}
d
x
=
2
e
2
x
d
u
.
Solve for dx:
Substitute
u
u
u
and
d
x
dx
d
x
into the integral:
∫
e
2
x
−
1
e
2
x
+
3
d
x
=
∫
u
−
4
2
u
⋅
d
u
2
e
2
x
\int\frac{e^{2x} - 1}{e^{2x} + 3} dx = \int\frac{u - 4}{2u} \cdot \frac{du}{2e^{2x}}
∫
e
2
x
+
3
e
2
x
−
1
d
x
=
∫
2
u
u
−
4
⋅
2
e
2
x
d
u
.
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Consider the following problem:
\newline
The water level under a bridge is changing at a rate of
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t
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r(t)=40 \sin \left(\frac{\pi t}{6}\right)
r
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Which expression can we use to solve the problem?
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\newline
(A)
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(B)
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s
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18
s
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Choose
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(A) The rate of change of enrollment is
20
20
20
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18
18
than it was in year
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18
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(C) Between years
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15
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18
18
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20
20
years.
\newline
(D) There are
20
20
20
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18
18
18
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7
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t
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1
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t
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t
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1
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t
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7
t=7
t
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7
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B) The average rate of filling between
t
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1
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t
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7
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t
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t
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1
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t
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1
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t
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7
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t
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(D) The amount of water filled between
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t
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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
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y
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x
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x
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2
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x
=
2
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x
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7
x=7
x
=
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x
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2 x
2
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Which one of the definite integrals gives the volume of the solid?
\newline
Choose
1
1
1
answer:
\newline
(A)
∫
2
7
[
(
4
−
x
)
2
−
sin
2
(
x
)
]
d
x
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∫
2
7
[
(
4
−
x
)
2
−
sin
2
(
x
)
]
d
x
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(B)
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2
7
[
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(
x
)
+
x
−
4
]
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x
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x
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2
7
[
sin
(
x
)
+
x
−
4
]
⋅
2
x
d
x
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(C)
∫
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
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(D)
∫
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
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∫
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