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Find the antiderivative F f(x)=(√x+x^4)/(x^2), F(1)=2
Related Topics
Wize University Calculus 1 Textbook > Integrals
Initial Conditions
3 Activities
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
−
2
x
−
x
3
3
−
11
3
\dfrac{-2}{\sqrt{x}}-\dfrac{x^3}{3}-\dfrac{11}{3}
x
−
2
−
3
x
3
−
3
11
−
2
x
−
x
3
3
+
11
3
\dfrac{-2}{\sqrt{x}}-\dfrac{x^3}{3}+\dfrac{11}{3}
x
−
2
−
3
x
3
+
3
11
−
2
x
+
x
3
3
+
11
3
\dfrac{-2}{\sqrt{x}}+\dfrac{x^3}{3}+\dfrac{11}{3}
x
−
2
+
3
x
3
+
3
11
2
x
+
x
3
3
+
11
3
\dfrac{2}{\sqrt{x}}+\dfrac{x^3}{3}+\dfrac{11}{3}
x
2
+
3
x
3
+
3
11
I don't know
Check Submission
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
.
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)=..."
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
)
=
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