What is the Correct Integral for Finding the Area Below y=0 and Above y=lnx?

[Arman777]

Homework Statement

Find the area Below $y=0$,above $y=lnx$, and to the right of $x=0$

Homework Equations

The Attempt at a Solution

I thought an integral like $\int_0^1 lnx \, dx$
then Its $-∞$ at $x=0$ So I used like $lim(a→0)=\int_a^1 lnx \, dx$ and from that it came
The integral result is $xlnx-x$ so $1(ln1-1)-a(lna-1)$ And if we take limit first term $1(ln1-1)$ is $-1$ but the other term bothers me.It will be $0(-∞-1)$. I can think like $lim (a→0)=a ln(a)$ and that gave me $0$ but there's also $+1$ so the answer turns $0$ but its impossible.Where I am doing wrong ?

 

[Ray Vickson]

Homework Statement

Find the area Below $y=0$,above $y=lnx$, and to the right of $x=0$

Homework Equations

The Attempt at a Solution

I thought an integral like $\int_0^1 lnx \, dx$
then Its $-∞$ at $x=0$ So I used like $lim(a→0)=\int_a^1 lnx \, dx$ and from that it came
The integral result is $xlnx-x$ so $1(ln1-1)-a(lna-1)$ And if we take limit first term $1(ln1-1)$ is $-1$ but the other term bothers me.It will be $0(-∞-1)$. I can think like $lim (a→0)=a ln(a)$ and that gave me $0$ but there's also $+1$ so the answer turns $0$ but its impossible.Where I am doing wrong ?

$$ \int_a^1 \ln x \, dx =\left. x \ln x -x \right|_a^1 = 1 \ln 1 - 1 - a \ln a + a$$
What is the limit of that as $a \to 0+$?

BTW; do not write $ln x$-- it is ugly and hard to read; instead, write $\ln x$. You do that by typing "\ln" instead of "ln". (Same for "log", "exp", "lim", "max", "min", all the trig functions and their inverses, and the hyperbolic functions---but not their inverses.)

 

[fresh_42]

Which area is described by the given conditions? Draw a picture of it or describe it with words.

 

[Arman777]

$$ \int_a^1 \ln x \, dx =\left. x \ln x -x \right|_a^1 = 1 \ln 1 - 1 - a \ln a + a$$
What is the limit of that as $a \to 0+$?

BTW; do not write $ln x$-- it is ugly and hard to read; instead, write $\ln x$. You do that by typing "\ln" instead of "ln". (Same for "log", "exp", "lim", "max", "min", all the trig functions and their inverses, and the hyperbolic functions---but not their inverses.)

oh ok I foıund thanks

 

[LCKurtz]

oh ok I foıund thanks

I doubt you have found the correct answer since your original integral is wrong. Remember area is$$
\int_a^b y_{upper} - y_{lower}~dx$$which is not what you have in your integrand.

 

[Arman777]

I doubt you have found the correct answer since your original integral is wrong. Remember area is$$
\int_a^b y_{upper} - y_{lower}~dx$$which is not what you have in your integrand.

Well that make sense...Hmm...ok thanks