find the area between $f(x)=\frac 1x$ and the $x$-axis for $1\le x\le 10$ Find the area between $f(x)=\frac 1x$ and the $x$-axis for $1\le x\le 10$

my work 
I know that 


hence, area $\Delta = \int\limits _{1}^{10}\frac 1x dx$  . stuck here not sure if my work correct or not 
 A: First draw the two curves. You will find that they intersect only once ( in the given domain ) and that point is $(1,1)$. Now, $y=x$ goes above $\frac{1}{x}$ after that point.
The area between $x$ and x-axis is $$A_1=\int_1^{10}xdx$$
The area between $\frac{1}{x}$ and x-axis is $$A_2=\int_1^{10}\frac{1}{x}dx$$
Next, shade the areas and you will find that the area enclosed is $A_1-A_2$
A: Edit: Here's a more descriptive explanation on the topic
If you want to calculate the surface $S$ you use following formula: $S = \int_a^b f(x)dx$. $b$ is your upper limit (10,100,$A\to\infty$) and $a$ is your lower limit ($1$). Be aware if $f(x)$ gives negative values you will also get negative surface. There is no such thing as negative surface so you have to avoid this. This can be realised by using following formula $S = \int_b^a f(x)dx$ (just switch a and b from position).
For example: You want to know the surface between $1$ and $10$ with $f(x)$ being positive from $1 \to 5$ and negative from $5 \to 10$ you will have to split you integral in two parts:
$$S= \int_1^5 f(x)dx + \int_{10}^5 f(x)dx$$
If you want to calculate volume V from the graph rotated around the x-axis you use following formula
$$V = \pi \int_a^b (f(x))^2dx$$ 
You don't have to worry about $f(x)$ being negative or positive. Because of the exponent $2$ you will always get positive volume.
$a = 1$, $b = \{10, 100,+\infty\}$
Anymore questions? If yes ask them or just use google. These are basic applications of a integral.
A: For part $A$ you have to find the area covered by $f(x)$ and $x$ acids where $f(x) = \frac{1}{x}$ . (given the interval $1 \le x \le 10$)
So you integrate $f(x)$ from $1 \to 10$.
when $\frac{1}{x}$ is integrated it becomes $ln|x|$ (limit from 1 to 10)
Which gives $ln|10| - ln|1| = ln|10|$ (as $ln|1$| is zero).
So the answer is $ln|10|$ square units. 
