- #1

kloong

- 36

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How do I use the limits (infinity and 0) after getting the equation from integration by parts?

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In summary, when using the limits (infinity and 0) after integration by parts, you need to write the improper integral as the limit of a proper integral. This can be done by evaluating the antiderivative at the limits and taking the limit as the upper limit approaches infinity. The integral represents the first moment (mean) of an exponential distribution on x, and is equal to \lambda^{-1}.

- #1

kloong

- 36

- 0

How do I use the limits (infinity and 0) after getting the equation from integration by parts?

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- #2

Petr Mugver

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- #3

Mark44

Mentor

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[tex] \int_0^\infty \lambda x e^{-\lambda x} dx = \lim_{b \to \infty} \int_0^b \lambda x e^{-\lambda x} dx [/tex]

After you get your antiderivative, evaluate it at b and 0, and take the limit as b --> infinity.

- #4

donifan

- 12

- 0

Integrals by parts with infinity as limit is a mathematical technique used to evaluate integrals that involve products of functions. It involves using the formula ∫u dv = uv - ∫v du, where u and v are functions, and applying it repeatedly until the integral can be solved.

Integration by parts with infinity as limit is necessary when the integral involves a product of two functions, and one of the functions cannot be easily integrated. It is also used when evaluating integrals that have infinity as one of the limits.

The general rule for choosing u and dv is to choose u as the function that becomes simpler when differentiated, and dv as the function that becomes easier to integrate. However, there is no one correct way to choose u and dv, and sometimes trial and error may be needed.

Yes, integration by parts with infinity as limit can be applied to definite integrals, as long as the limits of integration are not infinity. In this case, the final answer will be a finite number.

One limitation of integration by parts with infinity as limit is that it can only be used for integrals that involve products of functions. It also may not always provide a solution for the integral, and in some cases, other techniques such as substitution may be needed.

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