Integral of [1-(e^(t/a))]^2 | LR Circuit Power

  • Thread starter harshasunder
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In summary, the conversation is about an integral problem involving the function [1-(e^(t/a))]^2. The problem is related to finding the total power through a LR circuit. The solution involves using substitution to manipulate the integrand and then integrating each part separately. The final answer is given as \int(1-{e}^{t/a})^{2}dt=\frac{1}{2}ae^{2t/a}-2e^{t/a}+t+C. The conversation also includes a discussion about using LaTeX to properly format mathematical expressions.
  • #1
harshasunder
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[SOLVED] integral problem

Homework Statement



hi. my problems an integral which i can't solve. the integral is this-

integral of [1-(e^(t/a))]^2 dt

a is a constant. it came when i tried to find the total power through a LR circuit


Homework Equations





The Attempt at a Solution



not sure at all.

[1-(e^(t/a))]^3
----------------
3 [ then? ]
 
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  • #2
This can be done with substitution:
[tex]\int(1-{e}^{t/a})^{2}dt[/tex]
Let [tex]u=t/a[/tex] and thus [tex]du=dt/a, dt={a}{du}[/tex].
After expanding the integrand and putting [tex]a[/tex] (the constant) on the outside, it becomes:
[tex]{a}\int{e}^{2u}-2e^u+1 du[/tex].
Again, if you don't see it already, use substitution on the first part and manually do the other two easy parts.
Let [tex]w=2u, dw=2du, \frac{1}{2}dw=du[/tex]
[tex]\frac{1}{2}a\int{e^w}dw=\frac{1}{2}ae^w=\frac{1}{2}ae^{2u}=\frac{1}{2}ae^{2t/a}[/tex]
The second part is easy:
[tex]-a \int{2e^u}du=-2e^u=-2e^{t/a}[/tex]
The third part is the easiest (I added the constant 'C' here):
[tex]a \int du=au+C=a\frac{t}{a}+C=t+C[/tex] don't forget the substitution!
Summing all of these gives these final answer:
[tex]\int(1-{e}^{t/a})^{2}dt=\frac{1}{2}ae^{2t/a}-2e^{t/a}+t+C[/tex]
 
  • #3
hey thanks! a lot. how did you get the integral signs? (the big s) and the proper format?
 
  • #4
hey thanks! a lot. how did you get the integral signs? (the big s) and the proper format?
 
  • #5
harshasunder said:
hey thanks! a lot. how did you get the integral signs? (the big s) and the proper format?

[itex]\LaTeX[/itex] is what you're looking for go to

https://www.physicsforums.com/showthread.php?t=8997

[tex]\int_0^{\infty} e^{x^2}\;dx\rightarrow \int \mid e^{x^2}\mid \lim_{x\rightarrow\infty}=[/tex]

[tex]\sum_{n=-\infty}^\infty \frac{x^n}{n!} = \lim_{n\rightarrow\infty} (1+x/n)^n[/tex]

You can click on any bit of code and cut and paste it as well. Try these ones.
 
Last edited:

1. What is the integral of [1-(e^(t/a))]^2?

The integral of [1-(e^(t/a))]^2 is a mathematical calculation that represents the area under the curve of the function [1-(e^(t/a))]^2. It is used to find the total value or quantity of a function over a given interval.

2. How do you solve the integral of [1-(e^(t/a))]^2?

To solve the integral of [1-(e^(t/a))]^2, you can use various integration techniques such as substitution, integration by parts, or partial fractions. It is important to first simplify the function and then choose the appropriate integration method to solve it.

3. What does the LR circuit power represent?

The LR circuit power is a measurement of the electrical power in a circuit that contains a resistor (R) and an inductor (L). It represents the rate at which energy is dissipated or stored in the circuit and is measured in watts.

4. How is the power in an LR circuit related to the integral of [1-(e^(t/a))]^2?

In an LR circuit, the power is directly proportional to the square of the current, as well as the integral of [1-(e^(t/a))]^2. This means that as the current or the integral value increases, the power in the circuit also increases.

5. Can the integral of [1-(e^(t/a))]^2 be used to calculate the power in an LR circuit?

Yes, the integral of [1-(e^(t/a))]^2 can be used to calculate the power in an LR circuit. This is because the integral represents the total value of the function, which in this case is the power in the circuit. By solving the integral, you can determine the power at any given time in the circuit.

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