(adsbygoogle = window.adsbygoogle || []).push({}); Integration as antiderivative

Question:Why isn't a distinction made between anti-total-derivatives and anti-partial-derivatives in common usage of integration?

Consider the functions ##G_1(x, t)## and ##F(x, t)## such that ##F(x, t)=\frac{d}{dx}G_1(x, t)=\frac{\partial G}{\partial x}+\frac{dt}{dx}\frac{\partial G}{\partial t}=\frac{\partial G}{\partial x}+\frac{1}{v}\frac{\partial G}{\partial t}##, where ##v=\frac{dx}{dt}## is the velocity.

Then we define ##G_1(x, t)## as the anti-total-derivative (also known as the total-integration) of ##F(x, t)## with respect to ##x##. ##F(x, t)=\frac{d}{dx}G_1(x, t)\iff G_1(x, t)=\int_a^xF(x', t(x'))d_{full}x'+c##, where ##a## and ##c## are constants.

Next, consider the functions ##G_2(x, t)## and ##F(x, t)## such that ##F(x, t)=\frac{\partial}{\partial x}G_2(x, t)##.

Then we define ##G_2(x, t)## as the anti-partial-derivative (also known as the partial-integration) of ##F(x, t)## with respect to ##x##. ##F(x, t)=\frac{\partial}{\partial x}G_2(x, t)\iff G_2(x, t)=\int_{a(y)}^xF(x', t(x'))d_{partial}x'+c(y)##.

Example: Partial integration

Question:Is equation (2) below correct?

Consider a force ##F## that varies with displacement ##x## and time ##t##. In general, the acceleration ##a## and velocity ##v## of a particle subjected to such a force also vary with ##x## and ##t##.

##a=\frac{\partial v}{\partial x}\frac{dx}{dt}+\frac{\partial v}{\partial t}=v\frac{\partial v}{\partial x}+\frac{\partial v}{\partial t}##

Since ##F=ma=mv\frac{\partial v}{\partial x}+m\frac{\partial v}{\partial t}##,

##mv\frac{\partial v}{\partial x}=F-m\frac{\partial v}{\partial t}.\,\,\,\,\,## -------- (1)

Integrating (partial-integration) both sides wrt ##x##,

##\int mv\frac{\partial v}{\partial x} d_{partial}x=\int(F(x, t)-m\frac{\partial v}{\partial t})d_{partial}x##

##\frac{1}{2}mv^2=g(x, t)+h(t)\,\,\,\,\,## --------(2) Is this correct?

where ##\frac{\partial}{\partial x}g(x, t)=F(x, t)-m\frac{\partial v(x, t)}{\partial t}##.

Example: Total differential

Question: I obtain below ##\frac{\partial v}{\partial t}=0##, which is inconsistent with ##F(x, t)##. What's wrong?

Multiplying ##dx## to both sides of (1),

##mv\frac{\partial v}{\partial x}dx=(F-m\frac{\partial v}{\partial t})dx##.

Adding ##mv\frac{\partial v}{\partial t}dt## to both sides,

##mv(\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial t}dt)=(F-m\frac{\partial v}{\partial t})dx+mv\frac{\partial v}{\partial t}dt.\,\,\,\,\,## -------- (3)

Since ##dv=\frac{\partial v}{\partial x}dx+\frac{\partial v}{\partial t}dt## and ##vdt=dx##,

##mvdv=(F-m\frac{\partial v}{\partial t})dx+m\frac{\partial v}{\partial t}dx=Fdx##.

Integrating both sides,

##\frac{1}{2}mv^2=\int F(x, t)dx##.

I believe the integration on the RHS should be a partial-integration. (But why?) Then

##\frac{1}{2}mv^2=G(x, t)+H(t)## (This is consistent with equation (2).)

where ##\frac{\partial}{\partial x}G(x, t)=F(x, t)##.

But simplifying the RHS of (3), we have

##mv\frac{\partial v}{\partial x}dx+mv\frac{\partial v}{\partial t}dt=Fdx+0dt##

By comparing terms with ##dt##, we deduce

##\frac{\partial v}{\partial t}=0##.

But this is inconsistent with a time-dependent force ##F(x, t)##. What's wrong?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# I Partial integration vs total integration and time-dependent force

Have something to add?

Draft saved
Draft deleted

Loading...

Similar Threads - Partial integration total | Date |
---|---|

I Verifying derivative of multivariable integral equation | Sep 7, 2016 |

B Complex Integration By Partial Fractions | Apr 8, 2016 |

Partial Derivative of a Definite Integral | Feb 25, 2016 |

Integral total and partial of a function? | Mar 31, 2014 |

Total vs partial integration | Oct 31, 2012 |

**Physics Forums - The Fusion of Science and Community**