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Math Help - Need take a look at integration culculus

  1. #1
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    Need take a look at integration culculus

    Need take a look at integration culculus-43a17ab803f5e384197d31fe1dec41e4982.jpg
    \int_{1}^{e} 2^l *\frac{1}{X} \, dx.
    U=1/x dv=2^l
    du=lnxdx v=2^l/ln2
    I= \frac{2^l}{ln2}-\int 2^l*\frac{1}{x}
    Is it good?
    "l" mean "lnx"
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Kristina View Post
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    U=1/x du=lnxdx
    That is not right: du=-dx/x^2. Better use the substitution t=\ln x . Then:

    \displaystyle\int_0^e 2^{\ln x}\cdot \dfrac{dx}{x}=\displaystyle\int_0^12^tdt=\ldots

    Could you continue?.

    Regards.

    Fernando Revilla
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  3. #3
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    Sorry, but I do not understand...
    WHy I can not do like this?
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  4. #4
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    because ln(x) is the integral of 1/x, not the derivative. If u= 1/x then du= -1/x^2 dx not "ln x dx". That is precisely what FernandoRevilla said before. Instead of integration by parts, use the simple substitution t= ln x that FernandoRevilla suggested.
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  5. #5
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    Quote Originally Posted by Kristina View Post
    Click image for larger version. 

Name:	43a17ab803f5e384197d31fe1dec41e4982.jpg 
Views:	21 
Size:	7.5 KB 
ID:	19757
    \int_{1}^{e} 2^l *\frac{1}{X} \, dx.
    U=1/x dv=2^l
    du=lnxdx v=2^l/ln2
    I= \frac{2^l}{ln2}-\int 2^l*\frac{1}{x}
    Is it good?
    "l" mean "lnx"
    2^{\ln(x)}=e^{\ln(2)\ln(x)}=x^{\ln(2)}

    CB
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  6. #6
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    \displaystyle\int_0^e 2^{\ln x}\cdot \dfrac{dx}{x}=\displaystyle\int_0^12^tdt=2-1=1
    t=lnx
    t=ln1=0
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  7. #7
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    Quote Originally Posted by Kristina View Post
    \displaystyle\int_0^e 2^{\ln x}\cdot \dfrac{dx}{x}=\displaystyle\int_0^12^tdt=2-1=1
    t=lnx
    t=ln1=0
    Is it good? or I'm wrong?
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Kristina View Post
    Is it good? or I'm wrong?
    \displaystyle\int_0^1 2^tdt=\left[\dfrac{2^t}{\ln 2}\right]_0^1=\dfrac{1}{\ln 2}

    Regards.

    Fernando Revilla
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  9. #9
    Grand Panjandrum
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    You are collectivly making a meal of this:

    \displaystyle \int_1^e \frac{2^{\ln(x)}}{x}\;dx=\int_1^e x^{\ln(2)-1}\;dx=\left[ \frac{1}{\ln(2)} x^{\ln(2)}\right]_1^e

    ............. \displaystyle =\frac{1}{\ln(2)}(e^{\ln(2)}-1^{\ln(2)})=\frac{1}{\ln(2)}(2-1)=\frac{1}{\ln(2)}

    CB
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