Reduction formula question (int by parts)

In summary: So to summarize, the conversation discussed using integration by parts to find the value of I_n, which is defined as the integral of x^n * e^x from 0 to 2. The formula for integration by parts was mentioned and the correct steps were followed, but the limits of integration were not applied correctly. The final goal is to find the values of I_2 and I_3.
  • #1
thomas49th
655
0

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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  • #2
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.
 
  • #3
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
 

1. What is a reduction formula?

A reduction formula is a mathematical tool used to simplify the integration of a function that contains a repetitive pattern. It involves using integration by parts multiple times to reduce the complexity of the original integral.

2. When is a reduction formula used?

A reduction formula is typically used when the integral of a function cannot be easily evaluated, such as when the function contains a product of two or more functions or when it involves trigonometric functions.

3. How do you derive a reduction formula?

To derive a reduction formula, the following steps are typically followed:1. Perform integration by parts on the original integral.2. Substitute the resulting integral back into the original integral.3. Simplify the integral using algebraic or trigonometric identities.4. Repeat the process until a pattern emerges.5. Write the general form of the reduction formula.

4. What is the purpose of using a reduction formula?

The main purpose of using a reduction formula is to reduce the complexity of an integral and make it easier to evaluate. This is especially useful when the integral cannot be solved using other integration techniques, such as substitution or partial fractions.

5. Are there any limitations to using a reduction formula?

Yes, there are some limitations to using a reduction formula. It may not always be possible to find a reduction formula for a given integral, and even if one is found, it may not always lead to a closed form solution. Additionally, reduction formulas can be time-consuming and tedious to derive for more complex integrals.

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