Solving the Integral of sqrt(4x-1) | Math Help

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In summary: I forgot that the constant multiple could be put outside the integral originally.In summary, a user is struggling with taking the integral of sqrt(4x-1) with respect to x and is seeking an explanation. Another user suggests using a u-substitution and provides the steps to solve the integral. A third user suggests using a sight integral, where a multiple of the differential of the bracket appears outside the bracket. The correct answer is determined to be (2/3)(√(4x-1))^3 + C.
  • #1
PrudensOptimus
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I try to take the integral of sqrt(4x-1) with respect to x...


The correct answer is sqrt((4x-1)^3)/6, but I always get

2sqrt((4x-1)^3)/3... Can someone explain how to solve that integral pls.
 
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  • #2
Originally posted by PrudensOptimus
I try to take the integral of sqrt(4x-1) with respect to x...

OK

The correct answer is sqrt((4x-1)^3)/6,

No, it isn't.

Edit: My mistake; yes it is.

but I always get 2sqrt((4x-1)^3)/3...

That's not right either.

Can someone explain how to solve that integral pls.

Yes, you do a u-substitution. Let u=4x-1, so du=4dx. You then have:

(1/4)∫u1/2du,

which is elementary.
 
Last edited:
  • #3
If you didn't understand what the last person said when you take the integral of squrt(4x-1) with respect to xdx you must first find u and u prime (or the derivative of u), where u is what is in the parentheses, in this case (4x-1) so u prime (or the derivative of (4x-1)) = 4x, but you still have to deal with the remaining x and the 4 left by the derivative. To get rid of the 4 in the derivative you divide the integral by 4, and subistute for the x so. x=(u-1)/4, I'm you understand the rest. You multiply u^(1/2)*((u-1)/4) then take the antiderivative of that.
 
  • #4
The problem, by the way, is NOT the square root! You can easily integrate ∫ √(x) dx. √(x)= x1/2 and you can use the "power rule".
The problem is that "4x-1" inside the square root. To get rid of that you make the substitution mentioned earlier: u= 4x-1 so that
√(4x-1)= &radi;(u)= u1/2. Of course, you have to convert from "dx" to "du". Because 4x-1 is linear, that's easy
du/dx= 4 so du= 4 dx or (1/4)du= dx.

∫ &radic(4x-1)dx= (1/4)&int u1/2du.

The power rule says that an anti-derivative of un is
1/(n+1) un+1. In this case, n= 1/2 and n+1= 3/2. The anti-derivative is (2/3)u3/2+ C . Replacing u by 4x-1 again, ∫ &radic(4x-1)dx= (2/3)(4x-1)3/2+ C.

Since the original problem was given in terms of √ rather than a 1/2 power, it might be a good idea to set the answer in those terms: ∫ &radic(4x-1)dx= (2/3)(√(4x-1))3+ C.
 
  • #5
Why can't you just int. the bracket like you would int. x^0.5?

Code:
∫ (4x-1)^0.5 dx

increase the power and divied by it:

(4x-1)^1.5 + c = 2(√(4x-1))^3 + c
----------       ------------
   1.5                3
 
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  • #6
Why can't you just int. the bracket like you would int. x^0.5?

Because if you differentiate your result ((2/3)(4x-1)^(3/2) + C) you get:

4(4x - 1)(1/2)

which is not what you integrated.
 
  • #7
Originally posted by lavalamp
Why can't you just int. the bracket like you would int. x^0.5?

Code:
∫ (4x-1)^0.5 dx

increase the power and divied by it:

(4x-1)^1.5 + c = 2(√(4x-1))^3 + c
----------       ------------
   1.5                3

As Sting said, when you differentiate your result, you do not get the original integrand. That is because (4x-1)1/2 is a composition of functions.

f(u)=u1/2
u(x)=4x-1

In general, a composition of functions does not satisfy the same basic integration rule as the simple function f(u).

edit: typo
 
  • #8
Sorry, I meant to integrate it as a sight integral, where a multiple of the differential of the bracket appears outside the bracket, ie:

∫ f'(x) * (f(x))^n dx

in this case f(x)=4x-1 and f'(x)=4. The mulitple in this case is 0.25 which you can take out of the integral and put it back in at the end to get the right answer:

Code:
1 * (4x-1)^1.5 + c = (√(4x-1))^3 + c
-   ----------       ------------
4      1.5                6
 
  • #9
Lavalamp,
That's how I learned to do it, way back in '87. I think it's much quicker and easier.
Aaron
 
  • #10
I only learned about integration at the end of 2002, and then I only learned about sight integrals earlier this year. That's why I messed up the integration the first time round.
 

Related to Solving the Integral of sqrt(4x-1) | Math Help

What is the integral of sqrt(4x-1)?

The integral of sqrt(4x-1) is equal to (2/3)(4x-1)^(3/2) + C, where C is the constant of integration.

What is the process for solving the integral of sqrt(4x-1)?

To solve the integral of sqrt(4x-1), you can use the substitution method. Let u = 4x-1, then du = 4dx and the integral becomes (1/4)∫√u du. Using the power rule for integration, the integral becomes (1/4)((2/3)u^(3/2)) + C. Finally, substitute back in u = 4x-1 to get the final answer.

What are the limits of integration when solving the integral of sqrt(4x-1)?

The limits of integration will depend on the specific problem. Generally, you will be given the limits of integration as part of the integral and you will need to substitute them in the final answer.

What is the importance of solving the integral of sqrt(4x-1)?

Solving integrals is a fundamental skill in calculus and is used in many real-world applications, such as calculating areas and volumes. Understanding how to solve different types of integrals, including the integral of sqrt(4x-1), is essential for students pursuing careers in science, engineering, and mathematics.

What are some tips for solving the integral of sqrt(4x-1)?

One helpful tip is to always check your work by differentiating the answer to make sure it is equivalent to the original function. Another tip is to practice using different techniques, such as substitution and integration by parts, to solve integrals. Additionally, breaking the integral into smaller parts and using algebraic manipulation can make the solving process easier.

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