How do I integrate a(1-e^-bt) with constants a and b?

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In summary, the integral of a(1-e^-bt) is <sup>a</sup>/<sub>b</sub> - a<sup>e^-bt</sup>/<sub>b</sub> + C, and can be solved using the substitution method. The limits of integration will depend on the context of the problem, and this integral can be used for any value of a and b (as long as b is not equal to 0). This integral has practical applications in various fields, such as modeling growth and decay processes and calculating areas under curves.
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morgand
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Hey, just confused on how you would integrate:
a(1-e-bt)
with "a" and "b" being constants.
If you could provide steps that would be helpful.
 
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∫adt = at + C, ∫ae-bt = (-a/b)e-bt + D

where C and D are arbitrary constants. I hope you can put it together.
 

FAQ: How do I integrate a(1-e^-bt) with constants a and b?

What is the integral of a(1-e^-bt)?

The integral of a(1-e^-bt) is a/b - ae^-bt/b + C, where C is the constant of integration.

How do I solve this integral?

To solve this integral, you can use the substitution method. Let u = e^-bt, then du = -be^-bt dt. Substitute these into the integral and solve for u. Then replace u with e^-bt and solve for the final answer.

What are the limits of integration for this integral?

The limits of integration for this integral will depend on the context of the problem. If there are no specific limits given, you can use the indefinite integral and add a constant of integration.

Can I use this integral for any value of a and b?

Yes, this integral can be used for any value of a and b, as long as b is not equal to 0. If b is equal to 0, the integral will become undefined.

What is the practical application of this integral?

The integral of a(1-e^-bt) has many practical applications in mathematics, physics, and engineering. It can be used to model growth and decay processes, such as population growth or radioactive decay. It is also used in calculating areas under curves and finding the average value of a function.

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