Why Does Substitution Change the Integral's Bounds?

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Discussion Overview

The discussion revolves around the process of changing the bounds of an integral when performing a substitution, specifically in the context of the integral $\displaystyle\int_0^\frac{1}{2}\frac{4}{1+4t^2}dt$. Participants explore different methods of substitution and the implications for the limits of integration.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest that the integral can be simplified using the substitution $u=2t$, which changes the bounds from $[0, \frac{1}{2}]$ to $[0, 1]$.
  • Others propose that the integral does not yield a logarithmic result and instead requires a trigonometric substitution, indicating that the solution may be more complex than initially thought.
  • A participant notes the importance of adjusting the limits of integration when changing variables, emphasizing that the new bounds must reflect the substitution accurately.
  • One participant references a known integral form involving $\tan^{-1}$, suggesting a connection to the problem at hand, but does not clarify how it directly applies.

Areas of Agreement / Disagreement

Participants generally agree that substitution affects the bounds of the integral, but there is no consensus on the best method to approach the integral or the nature of the resulting solution (logarithmic vs. trigonometric).

Contextual Notes

Some participants express uncertainty about the specific substitution methods and their implications for the integral's evaluation, indicating a reliance on different mathematical techniques that may not be universally familiar.

karush
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$\displaystyle\int_0^\frac{1}{2}\frac{4}{1+4t^2}dt$

my first step with this was

$\displaystyle4\int_0^\frac{1}{2}\frac{1}{1+4t^2}dt$

thot this could be a log rule. but doesn't seem to fit into that

the answer to this is $\frac{\pi}{2}$
 
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karush said:
$\displaystyle\int_0^\frac{1}{2}\frac{4}{1+4t^2}dt$

my first step with this was

$\displaystyle4\int_0^\frac{1}{2}\frac{1}{1+4t^2}dt$

thot this could be a log rule. but doesn't seem to fit into that

the answer to this is $\frac{\pi}{2}$

Note that

\[\int_0^{\frac{1}{2}}\frac{4}{1+4t^2}\,dt = 4\int_0^{\frac{1}{2}}\frac{1}{1+(2t)^2}\,dt\]

Make the substitution $u=2t$ to get

\[2\int_0^1\frac{1}{1+u^2}\,du\]

This should integrate to something familiar.

I hope this helps!
 
Just a note on top of what Chris L T521 wrote, the solution will not contain a logarithm. It requires trig substitution which you might not have seen yet. I don't want you to keep pursuing the idea on this that the answer will contain a logarithm.
 
ok saw using trig substitution...
 
As a further hint, observe that if we let:

$\displaystyle u=\tan(v)$

then:

$\displaystyle du=\sec^2(v)\,dv=(1+\tan^2(v))\,dv=(1+u^2)\,dv\, \therefore \,dv=\frac{1}{1+u^2}\,du$

Now, rewrite the definite integral in terms of v, making sure to also rewrite the limits of integration in terms of v.
 
presume we are referring to the base form of.

$\displaystyle \int\frac{dx}{x^2+a^2}=\frac{1}{a}tan^{-1}\left(\frac{x}{a}\right)$

BTW why is there a 1 on the interval
 
karush said:
BTW why is there a 1 on the interval

Chris made a substitution from the variable $t$ to the variable $u$. Since these are not the same, the bounds won't stay the same. If the interval for variable $t$ is $[0,1/2]$ and $u=2t$ then the interval in terms of $u$ is $[2*0,2*1/2]$ or simply $[0,1]$.

When doing substitution problems you can either rewrite the interval in terms of the new variable or solve the indefinite integral and then rewrite everything in terms of the first variable. Just make sure you apply the correct bounds.
 

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