# Thread: Change of integral form

1. ## Change of integral form

By making use of the change of variable y = x + 3 write the integral below in a form where the range of integration is $[0,\infty ]$

$\int^{\infty}_{3}e^{-y}cos^2ydy$

All I know to do is to substitute x + 3 into the integral where y is. I'd guess minusing 3 from the x + 3 could be what's required to have the integral in the form where the range is $[0,\infty ]$ .

2. ## Re: Change of integral form

Originally Posted by maxgunn555
By making use of the change of variable y = x + 3 write the integral below in a form where the range of integration is $[0,\infty ]$
$\int^{\infty}_{3}e^{-y}cos^2ydy$
All I know to do is to substitute x + 3 into the integral where y is. I'd guess minusing 3 from the x + 3 could be what's required to have the integral in the form where the range is $[0,\infty ]$ .

You don't need a substitution. See here.

3. ## Re: Change of integral form

Thanks, though I'm not sure what the new ranged integral is from looking at the link. It was an indefinite integral in the wolfram alpha - I'm confused still sorry.

4. ## Re: Change of integral form

Originally Posted by maxgunn555
Thanks, though I'm not sure what the new ranged integral is from looking at the link. It was an indefinite integral in the wolfram alpha - I'm confused still sorry.
Then make it into a definite integral.

5. ## Re: Change of integral form

Originally Posted by maxgunn555
By making use of the change of variable y = x + 3 write the integral below in a form where the range of integration is $[0,\infty ]$

$\int^{\infty}_{3}e^{-y}cos^2ydy$

All I know to do is to substitute x + 3 into the integral where y is. I'd guess minusing 3 from the x + 3 could be what's required to have the integral in the form where the range is $[0,\infty ]$ .

Yes, replace every y in the integrand with x+ 3: $e^y= e^{x+3}= e^3e^x$ and $cos(y)= cos(x+3)= cos(3)cos(x)- sin(3)sin(x)$. When y= 3, x= 0 so the lower limit is 0. As y goes to infinity so does x= y+ 3, so the upper limit is infinity. Of course, with y= x+ 3, dy= dx.

(By the way "minusing" is not a word in the English language. Use "subtracting" instead.)