How can I use integration by parts to find the integral of a product?

In summary: It's less well-known than it should be, but it's pretty useful.In summary, the conversation discusses different methods for finding the integral of a product, specifically x*sqrt(1+x) dx. The methods mentioned include Integration by Parts (IBP), the use of substitution, and tabular integration. IBP is derived from the product rule for differentiation and involves the hope that one integral will be easier to solve than the other. The tabular method involves finding the consecutive derivatives of one function and the consecutive integrals of the other function. Both methods aim to simplify the integral and arrive at a known value.
  • #1
PrudensOptimus
641
0
How do you find the integral of a product, say x*sqrt(1+x) dx...


I am very confuzed with all the notation is involved, can someone explain and do a few examples? Thanks.
 
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  • #2
"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)1/2dx, let u=x and dv=(1+x)1/2dx.

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

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

And the integral on the right hand side is easy to do.
 
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  • #3
Yes, integration by parts is the "product rule in reverse".

However, for this example, I don't think that's what you would want to use. For the problem ∫x(x+1)1/2dx, it's not the product that is the problem: x(x1/2 would be easy enough: it's that "x+1" inside the root. I would make the easy substitution
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)1/2= (u+1)u1/2=
u3/2+ u1/2 so

∫x(x+1)1/2dx= ∫(u3/2+ u1/2)du= (2/5)u5/2+ (2/3)u3/2+ C= (2/5)(x+1)5/2+ (2/3)(x+1)3/2+ C which, I think you will find, is exactly what Hurkyl's method gives.
 
  • #4
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
 
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  • #5
Okay, so we have the tabular method, where [inte]vdu zero's out and we arrive at the the answer. Is there a name for the other case of IBP, where we will end up with some value 2([inte]udv)=(some known value ) ?
 
  • #6
Can't it be something related to

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

since integral of uv is F(x), d/dx F(x) = uv
 
  • #7
I think what you are looking for is to use "Integration by parts" to do this question. Tom has already demonstrated it.

originally posted by Tom
IBP is: ∫udv=uv-∫vdu. One does this with the hope that ∫vdu is easier than ∫udv.
IBP means integration by parts.
 
  • #8
"integration by parts" is basically the opposite of "the product rule" for differentiation.

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

What is the definition of the integral of a product?

The integral of a product is a mathematical operation that calculates the area under the curve of a product of two functions, represented by the symbol ∫(f(x)g(x))dx.

How is the integral of a product calculated?

The integral of a product is calculated using the product rule of integration, which states that the integral of a product is equal to the integral of one function multiplied by the other function, plus the integral of the derivative of one function multiplied by the integral of the other function.

What is the difference between the integral of a product and the product of integrals?

The integral of a product is a single integral that calculates the area under the curve of a product of two functions, while the product of integrals is the result of multiplying two separate integrals that calculate the areas under the curves of each individual function.

What is the significance of the integral of a product in mathematics?

The integral of a product is an important concept in calculus and is used to solve a variety of real-world problems, such as finding the total distance traveled by an object with varying velocity or the total amount of work done by a changing force.

Are there any special cases or exceptions when calculating the integral of a product?

Yes, there are certain special cases where the product rule of integration may not apply, such as when one function is a constant or when the two functions have a common factor. In these cases, different integration techniques may need to be used.

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