Integral of a vector valued function

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SUMMARY

The discussion centers on the integral of a vector-valued function, specifically the equation \(\int \mathbf{A}(t) \vec{w}(t) dt = \int \mathbf{B}(t) dt\). The user attempts to isolate \(\vec{w}(t)\) by multiplying both sides by \(\left(\int \mathbf{A}(t) dt\right)^{-1}\). The correct formulation is established as \(\vec{w}(t) = \left(\int \mathbf{A}(t) dt\right)^{-1} \int \mathbf{B}(t) dt\), which is validated through integration by parts. The discussion highlights the importance of correctly applying integration techniques to derive accurate results.

PREREQUISITES
  • Understanding of vector-valued functions
  • Familiarity with integral calculus
  • Knowledge of integration by parts
  • Proficiency in manipulating integrals and functions
NEXT STEPS
  • Study the properties of vector-valued functions in calculus
  • Learn advanced integration techniques, including integration by parts
  • Explore the implications of isolating variables in integral equations
  • Investigate the application of the inverse of integrals in solving equations
USEFUL FOR

Mathematicians, physics students, and anyone involved in advanced calculus or vector analysis will benefit from this discussion, particularly those working with integrals of vector-valued functions.

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I came across this integral of a vector valued function.
\int \mathbf A(t) \vec{w(t)} dt = \int \mathbf B(t).
I want to isolate \vec{w(t)} and so I multiply by \left (\int \mathbf A(t) dt \right)^{-1} on both sides.
\left (\int \mathbf A(t) dt \right)^{-1} \int \mathbf A(t) \vec{w(t)} dt = \left (\int \mathbf A(t) dt\right)^{-1} \int \mathbf B(t) dt

I thought the correct form would be
\int \vec{w(t)} dt = \left (\int \mathbf A(t) dt\right)^{-1} \int \mathbf B(t) dt.

But it turns out I get the right answer if I take
\vec{w(t)} = \left (\int \mathbf A(t) dt \right)^{-1} \int \mathbf B(t) dt.

Can anyone show why the second form is correct?
 
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Both formulas are wrong. Check it with ##\omega(t)=t^2## and ##A(t)=e^t##. All we have is the integration by parts: ##\displaystyle{\int } A(t)\omega(t)dt = \left(\int A(t)dt\right) \omega(t) - \int \left(\int A(t)dt\right)\omega'(t)dt##
 

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