Why Does Substitution Change the Integral's Bounds?

In summary, the given integral can be simplified to $\displaystyle 2\int_0^1\frac{1}{1+u^2}du$ by using the substitution $u=2t$. Then, using the trigonometric substitution $\displaystyle u=\tan(v)$, we can rewrite the integral as $\displaystyle 2\int_0^{\frac{\pi}{4}}\frac{1}{1+\tan^2(v)}\,dv$. Finally, using the trigonometric identity $\displaystyle \tan^2(v)=\sec^2(v)-1$, we can simplify the integral to $\displaystyle 2\int_0^{\frac{\pi}{4}}\
  • #1
karush
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$\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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  • #2
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!
 
  • #3
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.
 
  • #4
ok saw using trig substitution...
 
  • #5
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.
 
  • #6
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
 
  • #7
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.
 

FAQ: Why Does Substitution Change the Integral's Bounds?

What is an integral?

An integral is a mathematical concept that represents the total accumulated value of a function over a given interval or area. It is used to find the area under a curve, the volume of a solid, and other applications in calculus and physics.

What is the meaning of "1\f(x)" in an integral?

The "1\f(x)" in an integral represents the integrand, which is the function that is being integrated. It is a placeholder for any function that can be integrated, such as x^2, sin(x), or e^x.

How do you solve an integral in the form of 1\f(x)?

To solve an integral in the form of 1\f(x), you can use integration techniques such as substitution, integration by parts, or partial fractions. You can also use a graphing calculator or online integral calculator to find the solution.

What is the difference between a definite and indefinite integral of 1\f(x)?

A definite integral of 1\f(x) has specific limits of integration, which means that the integral is evaluated over a specific interval. An indefinite integral of 1\f(x) does not have limits of integration and represents the general antiderivative of the integrand.

Why is the integral in the form of 1\f(x) important in science?

The integral in the form of 1\f(x) is important in science because it allows us to calculate important physical quantities such as velocity, acceleration, and work. It also plays a crucial role in solving differential equations, which are used to model various natural phenomena in science.

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