Agebraic example of U-Substitution?

  • Thread starter tomas
  • Start date
  • Tags
    Example
In summary, to solve the indefinite integral of (2x+1)^3dx, we can use u-substitution by setting u = (2x+1) and finding the derivative du/dx = 2. Multiplying and dividing the integrand by this derivative, we can rewrite the integral as (1/2)u^3du. Using the chain rule for derivatives, we can then find the antiderivative to be (1/8)u^4, which when substituted back with u = (2x+1), gives us the final solution of (1/8)(2x+1)^4. Similarly, for the indefinite integral of xdx/(bx+a), we can use u-substitution by
  • #1
tomas
Could someone please show me a very easy agebraic example of U-Substitution?
 
Mathematics news on Phys.org
  • #2
Solve the following indefinite integral:
∫(2x+1)3dx

There are actually two ways for us to solve this problem:
(1) Expand the integrand and the integrate the polynomial.
(2) Perform a u-subtitution.

The easier method is clearly the second and it's what you want to know how to do anyway.

We find a subexpression to set u equal to (in this case it is fairly obvious):
u = (2x+1)

Then we find the derivative of this subexpression:
du/dx = 2

We multiply and divide the integrand by this derivative as follows (note that this simply means multiplying by one):
∫(2x+1)3 * (du/dx)/(2) dx
∫(1/2)(u)3(du/dx)dx

If you remember the chain rule for derivatives, then you'll also remember that the derivative of f(u(x)) = f'(u(x))u'(x). Well, Our integrand is already in this form now. To find its antiderivative, we can ignore the u'(x) and integrate f'(u) (w/ respect to u) to find f(u) and then simply change the formula back to f(u(x)) to obtain the final antiderivative.

Following through with this we obtain:
∫(1/2)u3du
(1/8)u4
(1/8)*(2x+1)4
 
  • #3
Well someone beat me to it but another example can't hurt. :smile:

Suppose you have:

[inte][2x(x2+ 1)]dx

You can solve this by u-substitution because if you let u = x2 + 1, then du = 2x.

[inte][u1/2]du = [(2u3/2)/3] + C

One thing about u substitution is that you should pick a u where du is in the integrand.
 
  • #4
change of variable

I like to treat "dx"s and "du"s like factors in the equation. It makes the problem seem a lot more algebra-y, and people like algebra. Some people throw du's and dx's on then end because they know they should, not because they think it means something.
Anyway, here's a fun example.

[inte] xdx/(bx+a)
let u = bx+a
then, du=bdx
since no "bdx" exists in the integrand, you have to solve the equation for something that does, ie, dx
(In my math class, the teacher taught us to rewrite the integrand to make it fit the du/dx equation. I feel that it's a lot clearer to rewrite the du/dx equation to make it fit the integrand.)
dx=du/b
replace dx with du/b and bx+a with u
[inte] x(du/b)/(u)
since u=bx+a
x=(u-a)/b
replace x with (u-a)/b
[inte] [(u-a)/b](du/b)/(u)
From there, just a little algebra and you get:
1/b2[inte] (1-a/u)du= 1/b2[u-alnu]+C
 
  • #5


Just in case anyone else falls upon this thread, i'll post one more problem.
(inte)(3x+4)^8 dx
u=3x+4
du=3dx
du/3=dx
plug your dx back into your original equation
(inte)(u)^8 (du/3)
1/3(inte)(u)^8 du
(1/3)(1/9)(u)^9 +c
(1/27)(3x+4)^9 + c
 

1. What is the concept behind U-substitution in algebra?

U-substitution, also known as the substitution method or the change of variables method, is a technique used to simplify integrals in algebra by substituting a more complex expression with a simpler one. This allows for easier integration and often leads to a more manageable result.

2. How do you identify when to use U-substitution in algebra?

U-substitution is typically used when the integrand contains a composite function, meaning that it is a function within a function. This can be recognized by identifying two parts of the integrand: the "outside" function and the "inside" function. The inside function is then substituted with a single variable, usually denoted by u, and the outside function is rewritten in terms of u.

3. Can you provide an example of U-substitution in algebra?

One example of u-substitution is solving the integral of 2x * (x^2 + 1)^3. In this case, the inside function is x^2 + 1, so we substitute it with u. This gives us the integral of 2x * u^3. We can then rewrite the outside function of 2x in terms of u as 2 * (u - 1) and substitute it back into the integral, giving us the integral of 2 * (u - 1) * u^3. This can then be solved using the power rule for integration.

4. What are the benefits of using U-substitution in algebra?

U-substitution allows for the simplification of integrals, making them easier to solve. It also allows for the integration of more complex functions that may not be solvable using other integration techniques. Additionally, it can be used to solve definite integrals and can be applied to a wide range of algebraic functions.

5. What are some common mistakes to avoid when using U-substitution in algebra?

Some common mistakes to avoid when using U-substitution include forgetting to substitute back in for the original variable, not properly identifying the inside and outside functions, and using an incorrect substitution for u. It is also important to check for any errors in algebraic manipulation when rewriting the integral in terms of u.

Similar threads

Replies
5
Views
275
  • General Math
Replies
6
Views
1K
Replies
17
Views
2K
Replies
4
Views
737
Replies
4
Views
492
Replies
1
Views
782
  • Calculus
Replies
6
Views
1K
Replies
8
Views
1K
Back
Top