Integrating the square root of a linear function.

In summary, the problem is to find the integral of sqrt(2x-1) and the instructor suggested using a substitution of u = sqrt(2x-1) which leads to u^2. However, this does not match the instructor's suggestion of using u^2 directly as the integrand. It is possible that the teacher may have meant using u = 2x-1 instead, which would lead to an easier integration.
  • #1
cptrsn
1
0

Homework Statement


[tex]\int\sqrt{2*x-1}[/tex]


Homework Equations





The Attempt at a Solution


Homework Statement


Obviously this can be fixed with the antiderivative of a linear function with proper constants i.e. [tex]\int A\times f(bx + c) = \frac{A}{b} F(bx + c)[/tex], however my instructor provided me with a "clue" which I somehow can't seem to work out.

He claims that [tex]\int\sqrt{2*x-1}[/tex] is equivilant to [tex]\int u^2[/tex] using "the right" substitution. My idea usually is:
1) [tex]u = \sqrt{2*x - 1}[/tex]
2) [tex]x = u^2 [/tex]
3) [tex]dx = 2u du[/tex]

Which would lead to [tex]\int u dx = \int u 2u du = \int 2u^2 du[/tex].
Obviously this is NOT identical to what he suggests, so can anyone point me in the right direction? I'm quite keen to know how he see such a problem.
 
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  • #2
If u = sqrt(2x-1)
then how x = u^2?
 
  • #3
I don't think he quite knew what he was talking about; there are some problems with that which I wouldn't even try to work with or fix.
What part of the integrand can u be that will easily let you integrate with respect to u?
 
  • #4
The teacher not being here to defend himself, I might suggest that cptrsn did not quite remember what the teacher had suggested. Any time you have a linear term, ax+ b, an obvious substitution is just u= ax+ b so that du= adx.
 

1. What does it mean to integrate the square root of a linear function?

Integrating the square root of a linear function involves finding the antiderivative of the function, which is the original function before it was differentiated. In simpler terms, it is the process of finding the function that, when differentiated, gives the original function.

2. Why is it important to integrate the square root of a linear function?

Integrating the square root of a linear function is important in many areas of science and mathematics, particularly in physics and engineering. It allows us to calculate the displacement, velocity, and acceleration of an object in motion, as well as the work done by a force.

3. What are the steps to integrate the square root of a linear function?

The steps to integrate the square root of a linear function depend on the specific function being integrated. However, in general, the steps involve using techniques such as substitution, integration by parts, or partial fractions to simplify the function and then applying the power rule for integration.

4. Can the square root of a linear function be integrated using basic calculus techniques?

Yes, the square root of a linear function can be integrated using basic calculus techniques. However, for more complex functions, more advanced techniques such as trigonometric substitution or integration by parts may be necessary.

5. Are there any real-life applications of integrating the square root of a linear function?

Yes, integrating the square root of a linear function has many real-life applications. For example, it is used in calculating the time it takes for an object to reach a certain height when thrown upwards, determining the speed of a falling object, and finding the volume of a cone or pyramid.

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