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Math Help - Change of integral form

  1. #1
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    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 ] .


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  2. #2
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    Re: Change of integral form

    Quote Originally Posted by maxgunn555 View Post
    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.
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  3. #3
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    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.
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    Re: Change of integral form

    Quote Originally Posted by maxgunn555 View Post
    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.
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  5. #5
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    Re: Change of integral form

    Quote Originally Posted by maxgunn555 View Post
    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.)
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