Thread: Need take a look at integration culculus

$\displaystyle \int_{1}^{e} 2^l *\frac{1}{X} \, dx$.
U=1/x $\displaystyle dv=2^l$
du=lnxdx v=2^l/ln2
I=$\displaystyle \frac{2^l}{ln2}-\int 2^l*\frac{1}{x}$
Is it good?
"l" mean "lnx"

Originally Posted by Kristina

U=1/x du=lnxdx

That is not right: $\displaystyle du=-dx/x^2$. Better use the substitution $\displaystyle t=\ln x$ . Then:

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

Could you continue?.

Regards.

Fernando Revilla

Sorry, but I do not understand...
WHy I can not do like this?

because ln(x) is the integral of 1/x, not the derivative. If u= 1/x then $\displaystyle du= -1/x^2 dx$ 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.

Originally Posted by Kristina

$\displaystyle \int_{1}^{e} 2^l *\frac{1}{X} \, dx$.
U=1/x $\displaystyle dv=2^l$
du=lnxdx v=2^l/ln2
I=$\displaystyle \frac{2^l}{ln2}-\int 2^l*\frac{1}{x}$
Is it good?
"l" mean "lnx"

$\displaystyle 2^{\ln(x)}=e^{\ln(2)\ln(x)}=x^{\ln(2)}$

CB

$\displaystyle \displaystyle\int_0^e 2^{\ln x}\cdot \dfrac{dx}{x}=\displaystyle\int_0^12^tdt=2-1=1 $
t=lnx
t=ln1=0

Originally Posted by Kristina

$\displaystyle \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?

Originally Posted by Kristina

Is it good? or I'm wrong?

$\displaystyle \displaystyle\int_0^1 2^tdt=\left[\dfrac{2^t}{\ln 2}\right]_0^1=\dfrac{1}{\ln 2}$

Regards.

Fernando Revilla

You are collectivly making a meal of this:

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

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

CB