MHB Why Does Substitution Change the Integral's Bounds?

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The discussion centers on the integral $\int_0^{\frac{1}{2}}\frac{4}{1+4t^2}dt$ and the impact of substitution on its bounds. A substitution of $u=2t$ transforms the integral into $2\int_0^1\frac{1}{1+u^2}du$, changing the limits from $[0, \frac{1}{2}]$ to $[0, 1]$. The integral evaluates to $\frac{\pi}{2}$, highlighting that the solution does not involve logarithmic functions but rather requires trigonometric substitution. Participants emphasize the importance of adjusting the limits of integration when substituting variables to maintain accuracy. Understanding these concepts is crucial for correctly solving integrals involving substitutions.
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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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