∫(1+4x2)^1/2 when x runs from 0 to 2

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Homework Help Overview

The discussion revolves around evaluating the integral ∫(1+4x²)^(1/2) from 0 to 2. Participants are exploring methods for solving this integral, which involves calculus concepts.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts u-substitution but encounters difficulties with the remaining variable x. Some participants suggest trying trigonometric substitution, while others question how to apply it given the form of the integral. There is also a discussion about the necessity of including the differential in the integral.

Discussion Status

The discussion is active, with participants providing suggestions and raising questions about the appropriateness of different substitution methods. There is recognition that the initial approach may not be sufficient, and participants are exploring alternative strategies.

Contextual Notes

Participants note the importance of including the differential in the integral and express uncertainty about the applicability of trigonometric substitution in this context. There is a sense of urgency as the original poster mentions the approaching end of the school year.

lude1
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Homework Statement


∫(1+4x2)1/2 when x runs from 0 to 2.

The answer is 4.647


Homework Equations





The Attempt at a Solution


I tried u substitution, giving me
u = 1+4x2
du = 8x​

where 8 can be pulled out as a constant. But I still have x left over. I'm sure I'm doing something silly or forgetting something basic, cause I'm pretty sure this is suppose to come to me fairly easy, especially since school is almost over... :frown:
 
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Try a trig substitution. The simple one you tried won't work because of that extra factor of x.
 
But how would you turn it into a trig substitution problem if the radical is not on the bottom of a fraction?
 
lude1 said:

Homework Statement


∫(1+4x2)1/2 when x runs from 0 to 2.

The answer is 4.647


Homework Equations





The Attempt at a Solution


I tried u substitution, giving me
u = 1+4x2
du = 8x​
du = 8x dx

You omitted dx in your original integral. It is crucial to include the differential, especially when using trig substitution.

After you do a few of these types of problems you will probably start to realize that an ordinary substitution is not going to work at all. The reason for this is a vela already said - you are missing the x factor needed to get du, and there isn't anything you can do to get it.
lude1 said:
where 8 can be pulled out as a constant. But I still have x left over. I'm sure I'm doing something silly or forgetting something basic, cause I'm pretty sure this is suppose to come to me fairly easy, especially since school is almost over... :frown:
 
lude1 said:
But how would you turn it into a trig substitution problem if the radical is not on the bottom of a fraction?
Why does it need to be on the bottom?
 

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