Can an integral that is a variable of itself be solved?

In summary, the conversation discusses determining the relationship between current I and time t under the assumption of a given voltage equation. It also mentions the use of integration and differentiation to solve the problem and the use of initial conditions. The final post refers to a related question.
  • #1
kmarinas86
979
1
Under the assumption that the voltage is [itex]V_f\left(1-e^{-\frac{t}{RC}}\right)[/itex], where [itex]V_f[/itex] is the final voltage, how would I determine the relationship between current [itex]I[/itex] and time [itex]t[/itex]?

[tex]I = \int_0^T \frac{V_f\left(1-e^{-\frac{t}{RC}}\right) - RI}{L} \,dt \,[/tex]

[itex]L[/itex] the magnetic inductance, [itex]R[/itex] the resistance, and [itex]C[/itex] the capacitance, are constants.

How would I plot current [itex]I[/itex] as a function of time [itex]t[/itex]? (The only variables here are [itex]I[/itex] and [itex]t[/itex].) Let's assume initial conditions of [itex]I=0[/itex] and [itex]t=0[/itex]. My problem here is that the variable I am trying to calculate is a variable inside the integral that is used in deriving the variable itself! How are such problems handled? Any help is appreciated! :smile:
 
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  • #3
So, I take it that you mean that [itex]\displaystyle I(T\,) = \int_0^T \frac{V_f\left(1-e^{-\frac{t}{RC}}\right) - RI(t)}{L} \,dt\,.[/itex]

Differentiate both sides w.r.t T and solve for I'(T). See if you can integrate the result to get I(τ).
 
  • #4
OP has a related question in this other post.

note:
[itex]I=\int RI.dt[/itex] - differentiate both sides gives

[itex]dI/I = R.dt[/itex]

... what was the problem?
 

1. Can an integral be a variable of itself?

Yes, it is possible for an integral to be a variable of itself. This is known as a "nested integral" or a "double integral".

2. How do you solve an integral that is a variable of itself?

To solve a nested integral, you can use the method of substitution where you replace the inner integral with a new variable and then integrate with respect to that variable. This will result in a single integral that can be solved using traditional methods.

3. What are the applications of nested integrals?

Nested integrals are commonly used in physics and engineering to calculate the volume, mass, and other properties of three-dimensional objects. They are also used in probability and statistics to calculate the probability of events with multiple variables.

4. Can nested integrals have more than two variables?

Yes, it is possible for nested integrals to have more than two variables. These are called "triple integrals" and can be solved using a similar method of substitution as double integrals. They are often used in advanced calculus and physics problems.

5. Are there any techniques for simplifying nested integrals?

Yes, there are several techniques for simplifying nested integrals, such as using symmetry properties, changing the order of integration, and using trigonometric identities. It is important to carefully analyze the integral and choose the most efficient method for solving it.

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