Definite Integral Calculator - Symbolab 4d1i4c
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31/03/2015
Definite Integral Calculator Symbolab
Solutions
My Notebook
Practice
Blog
English
New!
Help
Deצnite Integral Calculator Solve deﺡnite integrals step by step Enter a topic
Algebra Matrices & Vectors Functions & Graphing
Like
full pad »
x
2
x
□
d dx
(☐)'
x□
log□
√☐
√☐
□
≤
≥
□ □
∂ ∂x
∫
∫□
□
lim
∑
∞
( f ◦ g)
x
◦
θ
Follow
π
Trigonometry Calculus Limits
d ( x ) − x − t2 + 3 = 0 dt t
Go
Integrals Examples »
Indefinite Integrals Definite Integrals Antiderivatives
Solution
Double Integrals Triple Integrals Multiple Integrals Derivatives Derivative Applications Series ODE (new) Laplace Transform (new) Statistics
d( ) x(t ) 2 dt x ( t ) − t − t + 3 = 0:
(
2
x ( t ) = t t − 3ln ( t ) + c1 2
« Hide Steps
)
Steps First order linear Ordinary Differential Equation
Rs.1,169
Rs.999
Rs.902
Rs.1,149
Rs.1,149
Rs.733
Hide Definition
A first order linear ODE has the form of y' ( x ) + p ( x ) y ( x ) = q ( x )
y (x ) =
The general solution is:
∫ e∫ p
( x ) dx
∫ p ( x ) dx e
q ( x ) dx + C
Related Symbolab blog posts
d( ) x(t ) 2 x ( t) − t − t + 3 = 0 dt
Rewrite the equation in the form of a first order linear ODE:
Advanced Math Solution Integral Calculator, integra parts, Part II
y' ( x ) + p ( x ) · y ( x ) = q ( x )
2
p ( x ) = − 1 , t
q(x) = t − 3
In the previous post we covered in
by parts. Quick review: Integration
− 1 x ( t ) + d ( x ( t ) ) = t − 3 2
t
essentially the reverse of the prod
dt
Find the integration factor:
is used to transform the integral ... Hide Steps
μ ( t ) = 1 t
Find the integrating factor μ ( x ) , so that:
μ ( x ) · p ( x ) = μ' ( x )
p ( x ) = − 1 t
Advanced Math Solution Integral Calculator, integra parts
( )
d( ) () 1 dt μ ( t ) = μ t − t
\int \:uv'=uv-\int \:u'v In practice w
Divide both sides by μ ( t )
μ (t)
d (μ (t) ) dt
μ (t)
μ (t)
=
choose u such that its derivative u’
v’ such that its antiderivative v is si
( ) −1 t
we want the multiplication of u’...
μ (t)
d (μ (t) ) dt
= − 1 t
d (μ (t) ) dt d ( ) ( ) ln = μ t dt μ (t)
(
)
(
)
d ( ) = − 1t dt ln μ ( t ) Solve
(
(
)
d ( ) = − 1t : ln μ ( t ) dt
c1
μ ( t ) = e
Hide Steps
t
)
d ( ) = − 1t ln μ ( t ) dt
If
f ( x ) = g ( x ) then
(
∫f ( x ) dx = ∫g ( x ) dx, up to a constant
)
∫ d ln ( μ ( t ) ) dt = ∫ − 1 dt dt t Integrate each side of the equation
https://www.symbolab.com/solver/definiteintegralcalculator/%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft(x%5Cright)%5Cfrac%7Bx%7D%7Bt%7Dt%5E%…
1/3
31/03/2015
Definite Integral Calculator Symbolab Show Steps
∫ − 1 dt = −ln ( t ) + c1 t
∫
( dtd ( ln ( μ ( t ) ) ) ) dt = ln ( μ ( t ) ) + c2
ln ( μ ( t ) ) + c2 = −ln ( t ) + c1 Combine the constants
ln ( μ ( t ) ) = −ln ( t ) + c1 Isolate
μ( t) :
Show Steps
c1
μ ( t ) = e
t
c1
μ ( t ) = e
t
c1
μ ( t ) = e
t
The entire differential equation will be multiplied by
c
e 1 t .
c Therefore the constant part, e 1, is redundant and can be ignored
μ ( t ) = 1 t
Put the equation in the form
(μ(x) · y(x) )' = μ(x) · q(x):
(
)
3 d 1 dt t x ( t ) = t − t
Hide Steps
μ ( x ) and rewrtie the equation as
Multiply by the integration factor,
(μ(x) · y(x) )' = μ(x) · q(x) − 1 x ( t ) + d ( x ( t ) ) = t − 3 2
t
dt
Multiply both sides by the integrating factor,
1 t
− 1 1 x ( t ) + 1 d ( x ( t ) ) = 1 t − 1 3 t t t dt t t 2
Refine d (x(t ) ) dt
t
− x ( t ) = t − 3 t
t
2
( f · g ) ′ = f′ · g + f · g′
Apply the Product Rule:
f = 1 , g = x ( t ) : t
(
d (x(t ) ) dt
t
(
− x ( t ) = d 1 x ( t ) t
dt t
2
)
)
3 d 1 t x ( t ) = t − t dt
Solve
(
)
3 d 1 t x ( t ) = t − t : dt
(
(
2
x ( t ) = t t − 3ln ( t ) + c1 2
)
Hide Steps
)
3 d 1 t x ( t ) = t − t dt
If
f ( x ) = g ( x ) then
∫ d
(
∫f ( x ) dx = ∫g ( x ) dx, up to a constant
)
1 x ( ) dt = ∫ t − 3 dt t t dt t
Integrate each side of the equation 2
∫ t − 3 dt = t2 − 3ln ( t ) + c1
Show Steps
t
∫
( (
d 1 ( ) t x t dt
) ) dt = 1t x ( t ) + c2
2 1 ( ) t () t x t + c2 = 2 − 3ln t + c1
https://www.symbolab.com/solver/definiteintegralcalculator/%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft(x%5Cright)%5Cfrac%7Bx%7D%7Bt%7Dt%5E%…
2/3
31/03/2015
Definite Integral Calculator Symbolab Combine the constants 2 1 ( ) t () t x t = 2 − 3ln t + c1
Isolate
x ( t) :
(
(
2
x ( t ) = t t − 3ln ( t ) + c1
2
2
x ( t ) = t t − 3ln ( t ) + c1
(
2
2
x ( t ) = t t − 3ln ( t ) + c1 2
Show Steps
)
)
)
Save
Rs.1,149
Rs.999
Rs.902
Rs.1,149
Rs.965
Rs.901
Rs.999
Rs.1,050
Rs.277
Rs.1,133
Examples π
∫0 sin ( x ) dx b
∫a x2dx 2π
∫0
cos
2
5
∫1 2x 5
∫3
2
( θ ) dθ
−x+3 x
2
4x
2 − 8x
10
2
∫−10 xe−x
© EqsQuest 2012
8
dx
home / blog / about / team / / help
https://www.symbolab.com/solver/definiteintegralcalculator/%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft(x%5Cright)%5Cfrac%7Bx%7D%7Bt%7Dt%5E%…
3/3
Definite Integral Calculator Symbolab
Solutions
My Notebook
Practice
Blog
English
New!
Help
Deצnite Integral Calculator Solve deﺡnite integrals step by step Enter a topic
Algebra Matrices & Vectors Functions & Graphing
Like
full pad »
x
2
x
□
d dx
(☐)'
x□
log□
√☐
√☐
□
≤
≥
□ □
∂ ∂x
∫
∫□
□
lim
∑
∞
( f ◦ g)
x
◦
θ
Follow
π
Trigonometry Calculus Limits
d ( x ) − x − t2 + 3 = 0 dt t
Go
Integrals Examples »
Indefinite Integrals Definite Integrals Antiderivatives
Solution
Double Integrals Triple Integrals Multiple Integrals Derivatives Derivative Applications Series ODE (new) Laplace Transform (new) Statistics
d( ) x(t ) 2 dt x ( t ) − t − t + 3 = 0:
(
2
x ( t ) = t t − 3ln ( t ) + c1 2
« Hide Steps
)
Steps First order linear Ordinary Differential Equation
Rs.1,169
Rs.999
Rs.902
Rs.1,149
Rs.1,149
Rs.733
Hide Definition
A first order linear ODE has the form of y' ( x ) + p ( x ) y ( x ) = q ( x )
y (x ) =
The general solution is:
∫ e∫ p
( x ) dx
∫ p ( x ) dx e
q ( x ) dx + C
Related Symbolab blog posts
d( ) x(t ) 2 x ( t) − t − t + 3 = 0 dt
Rewrite the equation in the form of a first order linear ODE:
Advanced Math Solution Integral Calculator, integra parts, Part II
y' ( x ) + p ( x ) · y ( x ) = q ( x )
2
p ( x ) = − 1 , t
q(x) = t − 3
In the previous post we covered in
by parts. Quick review: Integration
− 1 x ( t ) + d ( x ( t ) ) = t − 3 2
t
essentially the reverse of the prod
dt
Find the integration factor:
is used to transform the integral ... Hide Steps
μ ( t ) = 1 t
Find the integrating factor μ ( x ) , so that:
μ ( x ) · p ( x ) = μ' ( x )
p ( x ) = − 1 t
Advanced Math Solution Integral Calculator, integra parts
( )
d( ) () 1 dt μ ( t ) = μ t − t
\int \:uv'=uv-\int \:u'v In practice w
Divide both sides by μ ( t )
μ (t)
d (μ (t) ) dt
μ (t)
μ (t)
=
choose u such that its derivative u’
v’ such that its antiderivative v is si
( ) −1 t
we want the multiplication of u’...
μ (t)
d (μ (t) ) dt
= − 1 t
d (μ (t) ) dt d ( ) ( ) ln = μ t dt μ (t)
(
)
(
)
d ( ) = − 1t dt ln μ ( t ) Solve
(
(
)
d ( ) = − 1t : ln μ ( t ) dt
c1
μ ( t ) = e
Hide Steps
t
)
d ( ) = − 1t ln μ ( t ) dt
If
f ( x ) = g ( x ) then
(
∫f ( x ) dx = ∫g ( x ) dx, up to a constant
)
∫ d ln ( μ ( t ) ) dt = ∫ − 1 dt dt t Integrate each side of the equation
https://www.symbolab.com/solver/definiteintegralcalculator/%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft(x%5Cright)%5Cfrac%7Bx%7D%7Bt%7Dt%5E%…
1/3
31/03/2015
Definite Integral Calculator Symbolab Show Steps
∫ − 1 dt = −ln ( t ) + c1 t
∫
( dtd ( ln ( μ ( t ) ) ) ) dt = ln ( μ ( t ) ) + c2
ln ( μ ( t ) ) + c2 = −ln ( t ) + c1 Combine the constants
ln ( μ ( t ) ) = −ln ( t ) + c1 Isolate
μ( t) :
Show Steps
c1
μ ( t ) = e
t
c1
μ ( t ) = e
t
c1
μ ( t ) = e
t
The entire differential equation will be multiplied by
c
e 1 t .
c Therefore the constant part, e 1, is redundant and can be ignored
μ ( t ) = 1 t
Put the equation in the form
(μ(x) · y(x) )' = μ(x) · q(x):
(
)
3 d 1 dt t x ( t ) = t − t
Hide Steps
μ ( x ) and rewrtie the equation as
Multiply by the integration factor,
(μ(x) · y(x) )' = μ(x) · q(x) − 1 x ( t ) + d ( x ( t ) ) = t − 3 2
t
dt
Multiply both sides by the integrating factor,
1 t
− 1 1 x ( t ) + 1 d ( x ( t ) ) = 1 t − 1 3 t t t dt t t 2
Refine d (x(t ) ) dt
t
− x ( t ) = t − 3 t
t
2
( f · g ) ′ = f′ · g + f · g′
Apply the Product Rule:
f = 1 , g = x ( t ) : t
(
d (x(t ) ) dt
t
(
− x ( t ) = d 1 x ( t ) t
dt t
2
)
)
3 d 1 t x ( t ) = t − t dt
Solve
(
)
3 d 1 t x ( t ) = t − t : dt
(
(
2
x ( t ) = t t − 3ln ( t ) + c1 2
)
Hide Steps
)
3 d 1 t x ( t ) = t − t dt
If
f ( x ) = g ( x ) then
∫ d
(
∫f ( x ) dx = ∫g ( x ) dx, up to a constant
)
1 x ( ) dt = ∫ t − 3 dt t t dt t
Integrate each side of the equation 2
∫ t − 3 dt = t2 − 3ln ( t ) + c1
Show Steps
t
∫
( (
d 1 ( ) t x t dt
) ) dt = 1t x ( t ) + c2
2 1 ( ) t () t x t + c2 = 2 − 3ln t + c1
https://www.symbolab.com/solver/definiteintegralcalculator/%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft(x%5Cright)%5Cfrac%7Bx%7D%7Bt%7Dt%5E%…
2/3
31/03/2015
Definite Integral Calculator Symbolab Combine the constants 2 1 ( ) t () t x t = 2 − 3ln t + c1
Isolate
x ( t) :
(
(
2
x ( t ) = t t − 3ln ( t ) + c1
2
2
x ( t ) = t t − 3ln ( t ) + c1
(
2
2
x ( t ) = t t − 3ln ( t ) + c1 2
Show Steps
)
)
)
Save
Rs.1,149
Rs.999
Rs.902
Rs.1,149
Rs.965
Rs.901
Rs.999
Rs.1,050
Rs.277
Rs.1,133
Examples π
∫0 sin ( x ) dx b
∫a x2dx 2π
∫0
cos
2
5
∫1 2x 5
∫3
2
( θ ) dθ
−x+3 x
2
4x
2 − 8x
10
2
∫−10 xe−x
© EqsQuest 2012
8
dx
home / blog / about / team / / help
https://www.symbolab.com/solver/definiteintegralcalculator/%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft(x%5Cright)%5Cfrac%7Bx%7D%7Bt%7Dt%5E%…
3/3
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