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Math Help - intergration, i m stuck.

  1. #1
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    intergration, i m stuck.

    can someone help me intergrate this step by step
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    Quote Originally Posted by Legendsn3verdie View Post
    can someone help me intergrate this step by step
    Two hints (sorry but I don't do step-by-step as a general rule):

    1. The integral can be written \frac{1}{\sqrt{2}} \int_{1}^{3} (-x + 3)^{1/2} \, dx.

    2. You should have been taught that \int (ax + b)^n \, dx = \frac{1}{a} \, \frac{1}{(n+1)} (ax + b)^{n+1} + C provided n \neq -1.
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    Quote Originally Posted by mr fantastic View Post
    Two hints (sorry but I don't do step-by-step as a general rule):

    1. The integral can be written \frac{1}{\sqrt{2}} \int_{1}^{3} (-x + 3)^{1/2} \, dx.

    2. You should have been taught that \int (ax + b)^n \, dx = \frac{1}{a} \, \frac{1}{(n+1)} (ax + b)^{n+1} + C provided n \neq -1.

    hmm dunno that proof u mentioned .. is there a way to do this with u subsitituion?> i keep seeming to juggle a 1/2 around.



    u = 3-x
    du = -dx/2?

    and just put a 2 outside of the integral so they cancel out.. is that a valid way?
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    Quote Originally Posted by Legendsn3verdie View Post
    hmm dunno that proof u mentioned .. is there a way to do this with u subsitituion?> i keep seeming to juggle a 1/2 around.

    [snip]
    u = 3-x
    du = -dx/2?

    and just put a 2 outside of the integral so they cancel out.. is that a valid way?
    Let  u = \frac{3 - x}{2}. Then \frac{du}{dx} = - \frac{1}{2} \Rightarrow dx = -2 du. Note also that x = 1  \Rightarrow u = 1 and  x = 3 \Rightarrow u = 0. Substitute all this in and solve the new integral.
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    Quote Originally Posted by mr fantastic View Post
    Let  u = \frac{3 - x}{2}. Then \frac{du}{dx} = - \frac{1}{2} \Rightarrow dx = -2 du. Note also that x = 1 \Rightarrow u = 1 and  x = 3 \Rightarrow u = 0. Substitute all this in and solve the new integral.

    ah i see i got 2 for a answer with that sub. and it is right, thanks, you are brillant.

    my only question to that is.. how would u know to sub U in for the whole (3-x)/2.. to me that kinda seems strange cause once u sub u for that, there is nothing really left!, and when u take there derivative of U to get du, u dont see a -1/2 or anything anywhere to relate it to.. so i guess my question summ'd up is, how do u know to sub U as 3-x/2?
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    is it because their is nothing else i can do? i remember my old teacher always saying that.
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    Quote Originally Posted by Legendsn3verdie View Post
    ah i see i got 2 for a answer with that sub. and it is right, thanks, you are brillant.

    my only question to that is.. how would u know to sub U in for the whole (3-x)/2.. to me that kinda seems strange cause once u sub u for that, there is nothing really left!, and when u take there derivative of U to get du, u dont see a -1/2 or anything anywhere to relate it to.. so i guess my question summ'd up is, how do u know to sub U as 3-x/2?
    The technique is called linear substitution. I knew to use it because \frac{3-x}{2} is a linear function of x.
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    Quote Originally Posted by mr fantastic View Post
    The technique is called linear substitution. I knew to use it because \frac{3-x}{2} is a linear function of x.
    but still what about 3-x over 2 makes it a linear function.. how would someonelike me (who isnt so great at math) know to just make u = 3 -x /2.. cause if i didnt come to this forum i d be tryin intgration by parts, u = 3- x, partial fractions all kinds of wastes of time..
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    Quote Originally Posted by Legendsn3verdie View Post
    but still what about 3-x over 2 makes it a linear function.. how would someonelike me (who isnt so great at math) know to just make u = 3 -x /2.. cause if i didnt come to this forum i d be tryin intgration by parts, u = 3- x, partial fractions all kinds of wastes of time..
    Experience.
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