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Exponential Functions: $f$ from $f''$
Related Topics
Wize University Calculus 1 Textbook > Integrals
Antiderivatives of Exponential Functions
4 Activities
Wize University Calculus 1 Textbook > Integrals
Initial Conditions
3 Activities
Given
f
′
′
(
x
)
=
5
e
x
+
2
sin
x
f''(x)=5e^x+2\sin x
f
′′
(
x
)
=
5
e
x
+
2
sin
x
and given
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
,
f
(
π
)
=
0
f(\pi)=0
f
(
π
)
=
0
, find
f
(
x
)
f(x)
f
(
x
)
. Enter "f(x)=..."
Answer
I don't know
Check Submission
More Antiderivatives of Exponential Functions Questions:
Evaluate the indefinite integral
∫
6
x
+
e
−
6
x
−
3
x
2
+
1
d
x
\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx
∫
6
x
+
e
−
6
x
−
x
2
+
1
3
d
x
.
Evaluate the indefinite integral
∫
6
x
+
e
−
6
x
−
3
x
2
+
1
d
x
\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx
∫
6
x
+
e
−
6
x
−
x
2
+
1
3
d
x
.
Practice: Definite Integral (~F2017 Final Q26)
Practice: Definite Integral
Evaluate
∫
0
3
1
9
+
x
2
+
2
x
ln
2
d
x
\int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx
∫
0
3
9
+
x
2
1
+
2
x
ln
2
d
x
.
Antiderivatives: Exponential Functions
Find the indefinite integral
∫
(
6
x
+
e
−
6
x
)
d
x
\displaystyle \int(6^x+e^{-6x}) \ dx
∫
(
6
x
+
e
−
6
x
)
d
x
Exponential Functions: $f$ from $f''$
Given
f
′
′
(
x
)
=
5
e
x
+
2
sin
x
f''(x)=5e^x+2\sin x
f
′′
(
x
)
=
5
e
x
+
2
sin
x
and given
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
,
f
(
π
)
=
0
f(\pi)=0
f
(
π
)
=
0
, find
f
(
x
)
f(x)
f
(
x
)
. Enter "f(x)=..."
Indefinite Integrals with Exponential
Compute the following integral:
∫
3
−
x
d
x
\displaystyle \int 3^{-x}\,\text{d}x
∫
3
−
x
d
x
.
Use upper case
C
C
C
to denote any constants.
Antiderivatives: Exponential Functions
Which of the following is the most general antiderivative of the function
e
3
x
+
7
e^{3x+7}
e
3
x
+
7
? In the functions below,
c
c
c
is an arbitrary constant.
Find
∫
(
c
o
s
7
)
e
3
x
d
x
\displaystyle \int (cos7)e^{3x}dx
∫
(
cos
7
)
e
3
x
d
x
Antiderivatives: Exponential Functions
Find an antiderivative of
f
(
x
)
=
2
e
3
x
−
2
f(x) = 2e^{3x-2}
f
(
x
)
=
2
e
3
x
−
2
Practice: Definite Integral
Evaluate
∫
0
3
1
9
+
x
2
+
2
x
ln
2
d
x
\displaystyle \int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx
∫
0
3
9
+
x
2
1
+
2
x
ln
2
d
x
.
Find the equation of the tangent line of the graph
y
=
e
x
cos
x
y=\frac{e^x}{\cos\ x}
y
=
c
o
s
x
e
x
at the point where 𝑥 = 0.
Indefinite Integrals with Exponential
Compute the following integral:
∫
3
−
x
d
x
\displaystyle \int 3^{-x}\,\text{d}x
∫
3
−
x
d
x
.
Use upper case
C
C
C
to denote any constants.
Antiderivatives: Exponential Functions
Find the indefinite integral
∫
(
6
x
+
e
−
6
x
)
d
x
\displaystyle \int(6^x+e^{-6x}) \ dx
∫
(
6
x
+
e
−
6
x
)
d
x
Find the equation of the tangent line to the curve
y
=
2
3
e
3
−
3
x
y=2^3e^{3-3x}
y
=
2
3
e
3
−
3
x
at
x
=
1
x=1
x
=
1
.
Enter your answer as y=...
Evaluate the indefinite integral
∫
6
x
+
e
−
6
x
−
3
x
2
+
1
d
x
\displaystyle \int6^x+e^{-6x}-\frac{3}{x^2+1}dx
∫
6
x
+
e
−
6
x
−
x
2
+
1
3
d
x
.
Higher Order Derivatives
Find a general formula for the
n
th-derivative of
f
(
x
)
=
e
3
x
f(x)=e^{3x}
f
(
x
)
=
e
3
x
.
Antiderivatives: Trigonometric and Exponential Functions
If
f
′
′
(
x
)
=
cos
(
2
x
)
−
x
4
+
e
x
f''\left(x\right)=\cos\left(2x\right)-x^4+e^x
f
′′
(
x
)
=
cos
(
2
x
)
−
x
4
+
e
x
,
f
(
0
)
=
0
=
f
′
(
0
)
f\left(0\right)=0=f'\left(0\right)
f
(
0
)
=
0
=
f
′
(
0
)
Practice: Definite Integral
Practice: Definite Integral
Evaluate
∫
0
3
1
9
+
x
2
+
2
x
ln
2
d
x
\int_0^3\frac{1}{9+x^2}+2^x\ln2\ dx
∫
0
3
9
+
x
2
1
+
2
x
ln
2
d
x
.
More Initial Conditions Questions:
Indefinite Integral
Find
f
(
x
)
f\left(x\right)
f
(
x
)
, given that
f
′
(
x
)
=
x
2
−
2
+
sin
x
f'\left(x\right)=x^2-2+\sin x
f
′
(
x
)
=
x
2
−
2
+
sin
x
and
f
(
0
)
=
3
f\left(0\right)=3
f
(
0
)
=
3
.
Applications of Integration: Position, Velocity and Acceleration
If the accelertion of a particle is given by
a
(
x
)
=
2
x
+
x
a\left(x\right)=2^x+x
a
(
x
)
=
2
x
+
x
, and
v
(
0
)
=
s
(
0
)
=
0
v\left(0\right)=s\left(0\right)=0
v
(
0
)
=
s
(
0
)
=
0
, then the displacement of the particle
s
(
x
)
s\left(x\right)
s
(
x
)
is
Practice: Indefinite Integral
Find
f
(
x
)
f\left(x\right)
f
(
x
)
, given that
f
′
(
x
)
=
x
2
−
2
+
sin
x
f'\left(x\right)=x^2-2+\sin x
f
′
(
x
)
=
x
2
−
2
+
sin
x
and
f
(
0
)
=
3
f\left(0\right)=3
f
(
0
)
=
3
.
Practice: Indefinite Integral
Practice: Indefinite Integral
Find
f
(
x
)
f\left(x\right)
f
(
x
)
, given that
f
′
(
x
)
=
x
2
−
2
+
sin
x
f'\left(x\right)=x^2-2+\sin x
f
′
(
x
)
=
x
2
−
2
+
sin
x
and
f
(
0
)
=
3
f\left(0\right)=3
f
(
0
)
=
3
.
Indefinite Integral
Find
f
(
x
)
f\left(x\right)
f
(
x
)
, given that
f
′
(
x
)
=
x
2
−
2
+
sin
x
f'\left(x\right)=x^2-2+\sin x
f
′
(
x
)
=
x
2
−
2
+
sin
x
and
f
(
0
)
=
3
f\left(0\right)=3
f
(
0
)
=
3
.
Practice: Indefinite Integral (~F2018 Final Q26)
Practice: Indefinite Integral
Find
f
(
x
)
f\left(x\right)
f
(
x
)
, given that
f
′
(
x
)
=
x
2
−
2
+
sin
x
f'\left(x\right)=x^2-2+\sin x
f
′
(
x
)
=
x
2
−
2
+
sin
x
and
f
(
0
)
=
3
f\left(0\right)=3
f
(
0
)
=
3
.
Exponential Functions: $f$ from $f''$
Given
f
′
′
(
x
)
=
5
e
x
+
2
sin
x
f''(x)=5e^x+2\sin x
f
′′
(
x
)
=
5
e
x
+
2
sin
x
and given
f
(
0
)
=
0
f(0)=0
f
(
0
)
=
0
,
f
(
π
)
=
0
f(\pi)=0
f
(
π
)
=
0
, find
f
(
x
)
f(x)
f
(
x
)
. Enter "f(x)=..."
Practice: Indefinite Integral
Find
f
(
x
)
f\left(x\right)
f
(
x
)
, given that
f
′
′
(
x
)
=
x
2
−
2
+
sin
x
f''\left(x\right)=x^2-2+\sin x
f
′′
(
x
)
=
x
2
−
2
+
sin
x
,
f
(
0
)
=
3
f\left(0\right)=3
f
(
0
)
=
3
and
f
′
(
3
)
=
1
f'(3)=1
f
′
(
3
)
=
1
Practice: Indefinite Integral
Find
f
(
x
)
f\left(x\right)
f
(
x
)
, given that
f
′
(
x
)
=
x
2
−
2
+
sin
x
f'\left(x\right)=x^2-2+\sin x
f
′
(
x
)
=
x
2
−
2
+
sin
x
and
f
(
0
)
=
3
f\left(0\right)=3
f
(
0
)
=
3
.
Applications of Integration: Position, Velocity and Acceleration
If the acceleartion of a particle is given by
a
(
x
)
=
2
x
+
x
a\left(x\right)=2^x+x
a
(
x
)
=
2
x
+
x
, and
v
(
0
)
=
s
(
0
)
=
0
v\left(0\right)=s\left(0\right)=0
v
(
0
)
=
s
(
0
)
=
0
, then the displacement of the particle
s
(
x
)
s\left(x\right)
s
(
x
)
is
Initial Conditions
Find the function
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
if
y
′
′
=
3
x
−
5
/
2
y
′
(
1
)
=
1
y
(
4
)
=
6
y''=3x^{-5/2} \quad y'(1)=1 \quad y(4)=6
y
′′
=
3
x
−
5/2
y
′
(
1
)
=
1
y
(
4
)
=
6
Initial Conditions
Solve the initial value problem
y
′
=
x
2
+
1
x
y'=\frac{x^2+1}{\sqrt{x}}
y
′
=
x
x
2
+
1
y
(
1
)
=
3
y(1)=3
y
(
1
)
=
3
Initial Conditions
Solve the initial value problem
y
′
′
=
e
x
/
2
y
′
(
0
)
=
3
y
(
0
)
=
5
y''=e^{x/2} \quad y'(0)=3 \quad y(0)=5
y
′′
=
e
x
/2
y
′
(
0
)
=
3
y
(
0
)
=
5
Initial Conditions
Find the function
f
(
x
)
f(x)
f
(
x
)
whose derivative is
4
x
−
1
/
x
2
4x-1/x^2
4
x
−
1/
x
2
for
x
>
0
x>0
x
>
0
and for which
f
(
1
)
=
−
1
f(1)=-1
f
(
1
)
=
−
1
.
Solve the initial value problem
f
(
x
)
=
4
x
+
x
2
2
+
5
,
F
(
1
)
=
1
/
3
\displaystyle f(x)=4x+\frac{x^2}{2}+5, F(1)=1/3
f
(
x
)
=
4
x
+
2
x
2
+
5
,
F
(
1
)
=
1/3
Solve the initial value problem
f
(
x
)
=
x
+
x
4
x
2
,
F
(
0
)
=
2
\displaystyle f(x)=\frac{\sqrt{x}+x^4}{x^2}, F(0)=2
f
(
x
)
=
x
2
x
+
x
4
,
F
(
0
)
=
2
Solve the initial value problem
y
′
=
sec
2
x
−
e
−
x
,
y
(
0
)
=
2
y'=\sec^2x-e^{-x},\ \ \ \ y\left(0\right)=2
y
′
=
sec
2
x
−
e
−
x
,
y
(
0
)
=
2
Consider Newton's law of cooling
d
T
d
t
=
2
−
T
10
\displaystyle\frac{\text{d}T}{\text{d}t}=2-\frac{T}{10}
d
t
d
T
=
2
−
10
T
. Determine
T
(
t
)
T(t)
T
(
t
)
if
T
(
0
)
=
100
T(0)=100
T
(
0
)
=
100
Indefinite Integral
Find
f
(
x
)
f\left(x\right)
f
(
x
)
, given that
f
′
(
x
)
=
x
2
−
2
+
sin
x
f'\left(x\right)=x^2-2+\sin x
f
′
(
x
)
=
x
2
−
2
+
sin
x
and
f
(
0
)
=
3
f\left(0\right)=3
f
(
0
)
=
3
.
Practice: Rearrange First
Q
:
\bf{Q:}
Q
:
Solve the initial value problem
t
2
d
y
d
t
+
1
=
t
3
t^2\dfrac{dy}{dt} + 1 = t^3
t
2
d
t
d
y
+
1
=
t
3
with
y
(
1
)
=
1
y(1) = 1
y
(
1
)
=
1
Q
:
\bf{Q:}
Q
:
Solve the initial value problem
y
′
=
x
3
+
sin
2
x
y'=x^3+\sin2x
y
′
=
x
3
+
sin
2
x
given
y
(
π
2
)
=
0
y\left(\dfrac{\pi}{2}\right)=0
y
(
2
π
)
=
0
Q
:
\bf{Q:}
Q
:
Solve the initial value problem
y
′
=
sec
2
x
−
e
−
x
y'=\sec^2x-e^{-x}
y
′
=
sec
2
x
−
e
−
x
with
y
(
0
)
=
2
y\left(0\right)=2
y
(
0
)
=
2
Solve the initial-value problem
y
′
=
x
2
+
1
x
y
(
1
)
=
3
\begin{aligned} y' &=\dfrac{x^2 + 1}{\sqrt{x}}\\ y(1) & = 3 \end{aligned}
y
′
y
(
1
)
=
x
x
2
+
1
=
3
Practice: From Second Derivative to Function
Q
:
\bf{Q:}
Q
:
Solve the initial-value problem
y
′
′
=
sin
2
x
y'' = \sin 2x
y
′′
=
sin
2
x
,
y
′
(
0
)
=
3
y'(0) = 3
y
′
(
0
)
=
3
,
y
(
0
)
=
5
y(0) = 5
y
(
0
)
=
5
.
Solve the initial value problem
y
′
′
=
e
x
/
2
y
′
(
0
)
=
3
y
(
0
)
=
5
\begin{aligned} y''& = e^{x/2}\\ y'(0)& = 3\\ y(0) &= 5 \end{aligned}
y
′′
y
′
(
0
)
y
(
0
)
=
e
x
/2
=
3
=
5
Solve the initial value problem
y
′
=
sec
2
x
−
e
−
x
,
y
(
0
)
=
2
y'=\sec^2x-e^{-x},\ \ y\left(0\right)=2
y
′
=
sec
2
x
−
e
−
x
,
y
(
0
)
=
2
.
Practice: Indefinite Integral
Practice: Indefinite Integral
Find
f
(
x
)
f\left(x\right)
f
(
x
)
, given that
f
′
(
x
)
=
x
2
−
2
+
sin
x
f'\left(x\right)=x^2-2+\sin x
f
′
(
x
)
=
x
2
−
2
+
sin
x
and
f
(
0
)
=
3
f\left(0\right)=3
f
(
0
)
=
3
.
Find the antiderivative
F
F
F
f
(
x
)
=
x
+
x
4
x
2
,
F
(
1
)
=
2
\displaystyle f(x)=\frac{\sqrt{x}+x^4}{x^2}, F(1)=2
f
(
x
)
=
x
2
x
+
x
4
,
F
(
1
)
=
2
Find the antiderivative
F
F
F
f
(
x
)
=
4
x
+
x
2
2
+
5
,
F
(
1
)
=
1
/
3
\displaystyle f(x)=4x+\frac{x^2}{2}+5, F(1)=1/3
f
(
x
)
=
4
x
+
2
x
2
+
5
,
F
(
1
)
=
1/3