- #1

- 635

- 0

I am very confuzed with all the notation is involved, can someone explain and do a few examples? Thanks.

- Thread starter PrudensOptimus
- Start date

- #1

- 635

- 0

I am very confuzed with all the notation is involved, can someone explain and do a few examples? Thanks.

- #2

Tom Mattson

Staff Emeritus

Science Advisor

Gold Member

- 5,500

- 8

"The Product Rule" for integration can be loosely said to be Integration by Parts. Actually, IBP is derived from the product rule for differentiation.

IBP is: ∫udv=uv-∫vdu. One does this with the hope that ∫vdu is easier than ∫udv.

To do ∫x(1+x)^{1/2}dx, let u=x and dv=(1+x)^{1/2}dx.

Then du=dx and v=(2/3)(1+x)^{3/2}, so and IBP says that:

∫x(1+x)^{1/2}dx=(2/3)x(1+x)^{3/2}-(2/3)∫(1+x)^{3/2}dx

And the integral on the right hand side is easy to do.

IBP is: ∫udv=uv-∫vdu. One does this with the hope that ∫vdu is easier than ∫udv.

To do ∫x(1+x)

Then du=dx and v=(2/3)(1+x)

∫x(1+x)

And the integral on the right hand side is easy to do.

Last edited:

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

However, for this example, I don't think that's what you would want to use. For the problem ∫x(x+1)

u= x+1. Then du= dx. Of course, x is now u- 1 but that's easy- it's to the first power. x(x+1)

u

∫x(x+1)

- #4

- 156

- 2

You could solve your integral by tabular integration.

[inte](x(1 + x)^{(1/2)}dx

Essentially, you have two functions, x and (1 + x)^{(1/2)}

Take the consecutive derivatives of x and consecutive integrals of

(1 + x)^{(1/2)}

Column A

row 1: x

row 2: 1

row 3: 0

Column B

row 1: (1 + x)^{(1/2)}

row 2: (2/3)(1 + x)^{(3/2)}

row 3: (4/15)(1 + x)^{(5/2)}

Now, I'll multiply column A, row 1, with column B, row 2. Then I'll subtract the product of column A, row 2 and column B and row 3 (subtraction simply because we alternate signs).

In tabular integration, you'll find the consecutive derivative of one function and the consecutive integrals of the other functions. It only works if you have a derivative that will eventually become zero and an integral that you can actually obtain.

So in our example, the integral will be:

x(2/3)(1 + x)^{(3/2)} - (4/15)(1 + x)^{(5/2)} + C

EDIT: syntax error

[inte](x(1 + x)

Essentially, you have two functions, x and (1 + x)

Take the consecutive derivatives of x and consecutive integrals of

(1 + x)

Column A

row 1: x

row 2: 1

row 3: 0

Column B

row 1: (1 + x)

row 2: (2/3)(1 + x)

row 3: (4/15)(1 + x)

Now, I'll multiply column A, row 1, with column B, row 2. Then I'll subtract the product of column A, row 2 and column B and row 3 (subtraction simply because we alternate signs).

In tabular integration, you'll find the consecutive derivative of one function and the consecutive integrals of the other functions. It only works if you have a derivative that will eventually become zero and an integral that you can actually obtain.

So in our example, the integral will be:

x(2/3)(1 + x)

EDIT: syntax error

Last edited:

- #5

- 136

- 0

- #6

- 635

- 0

d/dx (uv) = uv' + vu'??

since integral of uv is F(x), d/dx F(x) = uv

- #7

- 313

- 0

IBP means integration by parts.originally posted by Tom

IBP is: ∫udv=uv-∫vdu. One does this with the hope that ∫vdu is easier than ∫udv.

- #8

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

The "tabular integration" mentioned above is a method of regularizing repeated integration by parts.

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 933

- Last Post

- Replies
- 6

- Views
- 4K

- Replies
- 3

- Views
- 5K

- Replies
- 1

- Views
- 1K

- Replies
- 3

- Views
- 3K

- Replies
- 3

- Views
- 752

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 10

- Views
- 33K