A quick question on in logs












2












$begingroup$


I was solving indefinite integrals



$$int_0^22^x x ,dx$$



I use ILATE as a clue to consider the first function and second function.
$2^x$ is a algebraic function or logarithmic function?






calculus indefinite-integrals







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$endgroup$








  • 2




    $begingroup$
    It is an exponential function? $2^x=e^{xln 2}.$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 23 at 15:10






  • 1




    $begingroup$
    wolframalpha.com/input/?i=integral+from+0+to+2+x+*2%5Ex
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 23 at 15:18
















2












$begingroup$


I was solving indefinite integrals



$$int_0^22^x x ,dx$$



I use ILATE as a clue to consider the first function and second function.
$2^x$ is a algebraic function or logarithmic function?






calculus indefinite-integrals







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    It is an exponential function? $2^x=e^{xln 2}.$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 23 at 15:10






  • 1




    $begingroup$
    wolframalpha.com/input/?i=integral+from+0+to+2+x+*2%5Ex
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 23 at 15:18














2












2








2





$begingroup$


I was solving indefinite integrals



$$int_0^22^x x ,dx$$



I use ILATE as a clue to consider the first function and second function.
$2^x$ is a algebraic function or logarithmic function?






calculus indefinite-integrals







share|cite|improve this question









$endgroup$




I was solving indefinite integrals



$$int_0^22^x x ,dx$$



I use ILATE as a clue to consider the first function and second function.
$2^x$ is a algebraic function or logarithmic function?






calculus indefinite-integrals




calculus indefinite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 23 at 15:06









LuciferLucifer

303




303








  • 2




    $begingroup$
    It is an exponential function? $2^x=e^{xln 2}.$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 23 at 15:10






  • 1




    $begingroup$
    wolframalpha.com/input/?i=integral+from+0+to+2+x+*2%5Ex
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 23 at 15:18














  • 2




    $begingroup$
    It is an exponential function? $2^x=e^{xln 2}.$
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 23 at 15:10






  • 1




    $begingroup$
    wolframalpha.com/input/?i=integral+from+0+to+2+x+*2%5Ex
    $endgroup$
    – Mohammad Zuhair Khan
    Jan 23 at 15:18








2




2




$begingroup$
It is an exponential function? $2^x=e^{xln 2}.$
$endgroup$
– Mohammad Zuhair Khan
Jan 23 at 15:10




$begingroup$
It is an exponential function? $2^x=e^{xln 2}.$
$endgroup$
– Mohammad Zuhair Khan
Jan 23 at 15:10




1




1




$begingroup$
wolframalpha.com/input/?i=integral+from+0+to+2+x+*2%5Ex
$endgroup$
– Mohammad Zuhair Khan
Jan 23 at 15:18




$begingroup$
wolframalpha.com/input/?i=integral+from+0+to+2+x+*2%5Ex
$endgroup$
– Mohammad Zuhair Khan
Jan 23 at 15:18










1 Answer
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$begingroup$

For:
$$intlimits_{0}^{2} xcdot 2^x:dx$$
Take $2^x =t$, then we get $2^x cdot ln(2) cdot dx=dt$, and, $x=log_2(t)$. Thus:
begin{align}
intlimits_{0}^{2} x cdot 2^x:dx&=frac{1}{ln(2)}intlimits_{1}^{4}log_2(t): dt \
&=frac{1}{(ln(2))^2}intlimits_{1}^{4}ln(t) : dt
end{align}



Now, I leave it for you. Use by parts (Hint: $ln(2)= 1cdotln(2)$).






share|cite|improve this answer











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    $begingroup$

    For:
    $$intlimits_{0}^{2} xcdot 2^x:dx$$
    Take $2^x =t$, then we get $2^x cdot ln(2) cdot dx=dt$, and, $x=log_2(t)$. Thus:
    begin{align}
    intlimits_{0}^{2} x cdot 2^x:dx&=frac{1}{ln(2)}intlimits_{1}^{4}log_2(t): dt \
    &=frac{1}{(ln(2))^2}intlimits_{1}^{4}ln(t) : dt
    end{align}



    Now, I leave it for you. Use by parts (Hint: $ln(2)= 1cdotln(2)$).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      For:
      $$intlimits_{0}^{2} xcdot 2^x:dx$$
      Take $2^x =t$, then we get $2^x cdot ln(2) cdot dx=dt$, and, $x=log_2(t)$. Thus:
      begin{align}
      intlimits_{0}^{2} x cdot 2^x:dx&=frac{1}{ln(2)}intlimits_{1}^{4}log_2(t): dt \
      &=frac{1}{(ln(2))^2}intlimits_{1}^{4}ln(t) : dt
      end{align}



      Now, I leave it for you. Use by parts (Hint: $ln(2)= 1cdotln(2)$).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        For:
        $$intlimits_{0}^{2} xcdot 2^x:dx$$
        Take $2^x =t$, then we get $2^x cdot ln(2) cdot dx=dt$, and, $x=log_2(t)$. Thus:
        begin{align}
        intlimits_{0}^{2} x cdot 2^x:dx&=frac{1}{ln(2)}intlimits_{1}^{4}log_2(t): dt \
        &=frac{1}{(ln(2))^2}intlimits_{1}^{4}ln(t) : dt
        end{align}



        Now, I leave it for you. Use by parts (Hint: $ln(2)= 1cdotln(2)$).






        share|cite|improve this answer











        $endgroup$



        For:
        $$intlimits_{0}^{2} xcdot 2^x:dx$$
        Take $2^x =t$, then we get $2^x cdot ln(2) cdot dx=dt$, and, $x=log_2(t)$. Thus:
        begin{align}
        intlimits_{0}^{2} x cdot 2^x:dx&=frac{1}{ln(2)}intlimits_{1}^{4}log_2(t): dt \
        &=frac{1}{(ln(2))^2}intlimits_{1}^{4}ln(t) : dt
        end{align}



        Now, I leave it for you. Use by parts (Hint: $ln(2)= 1cdotln(2)$).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 26 at 7:48









        DavidG

        2,5911726




        2,5911726










        answered Jan 23 at 15:14









        Mayank M.Mayank M.

        493413




        493413






























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