Reduction formula question (int by parts)

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SUMMARY

The discussion centers on the integration by parts method applied to the integral \( I_{n} = \int^{2}_{0} x^{n}e^{x} dx \) for positive integers \( n \). The user, Thomas, attempts to derive the formula \( 2^{n}e^{2} - nI_{n-1} \) but encounters confusion regarding the application of limits in the integration by parts process. Key insights include the correct application of the integration by parts formula and the importance of applying limits to both terms in the equation.

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thomas49th
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Homework Statement


Let [tex]I_{n} = \int^{2}_{0} x^{n}e^{x} dx[/tex] where n is a positive integer. Use integration by parts to show
that

[tex] 2^{n}e^{2} - nI_{n-1}[/tex]

By first finding

[tex]I_{1} = \int ^{2}_{0} xe^{x} dx[/tex]
find I2 and I3.

Homework Equations



I'm sure your all aware of the formula for Int by parts. We'll take the [tex]e^{x}[/tex] function as the one to integrate and the [tex]x^{n}[/tex] as the one to differentiate.

The Attempt at a Solution

So use integration by parts to find In:

[tex]x^{n}e^{x} - n\int^{2}_{0}x^{n-1}[/tex]

Well that's
[tex]x^{n}e^{x} - nI_{n-1}[/tex]
is it not?

So now put the limits in

[tex][x^{n}e^{x} - nI_{n-1}]^{2}_{0}[/tex]

[tex][2^{n}e^{2} - nI_{n-1}] - [ - nI_{n-1}][/tex]
but that's isn't right because I'm getting no [tex]nI_{n-1}[/tex] because they cancel!

Where have I gone wrong?

Thanks
Thomas
 
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thomas49th said:
So use integration by parts to find In:

[tex]x^{n}e^{x} - n\int^{2}_{0}x^{n-1}[/tex]

Well that's
[tex]x^{n}e^{x} - nI_{n-1}[/tex]
is it not?

The step is correct but neither of them represent [itex]I_n[/itex]. The Integration by parts formula for definite integrals is [tex]\int^b_a u(x) \frac{dv(x)}{dx} dx = u(b)v(b)-u(a)v(a) - \int^b_a v(x) \frac{du(x)}{dx}[/tex].

You have put in the limits of integration for the second term, but not the first.
 
thomas49th said:

Homework Statement


Let [tex]I_{n} = \int^{2}_{0} x^{n}e^{x} dx[/tex] where n is a positive integer. Use integration by parts to show
that

[tex] 2^{n}e^{2} - nI_{n-1}[/tex]
You omitted part of the statement above. Show that 2ne2 - nIn - 1 equals what or does what?
thomas49th said:
By first finding

[tex]I_{1} = \int ^{2}_{0} xe^{x} dx[/tex]
find I2 and I3.


Homework Equations



I'm sure your all aware of the formula for Int by parts. We'll take the [tex]e^{x}[/tex] function as the one to integrate and the [tex]x^{n}[/tex] as the one to differentiate.

The Attempt at a Solution




So use integration by parts to find In:

[tex]x^{n}e^{x} - n\int^{2}_{0}x^{n-1}[/tex]

Well that's
[tex]x^{n}e^{x} - nI_{n-1}[/tex]
is it not?

So now put the limits in

[tex][x^{n}e^{x} - nI_{n-1}]^{2}_{0}[/tex]

[tex][2^{n}e^{2} - nI_{n-1}] - [ - nI_{n-1}][/tex]
but that's isn't right because I'm getting no [tex]nI_{n-1}[/tex] because they cancel!

Where have I gone wrong?

Thanks
Thomas
 

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