Find the Area $A$

Question

My attempt:

$A=\ln {e^2}-\ln 1=2$

Is this a correct answer?

Answer 1

Yes, your answer is correct. The area asked is an elementary application of integration. Specifically :

$$A = \int_1^{e^2}\frac{1}{x}dx=\big[ \ln(x)\big]_1^{e^2}= \ln(e^2)-\ln(1)=2\ln(e)=2$$

Answer 2

Yes it is indeed

$$A=\int_a^b \frac1x dx = [\log x]_a^b \implies A=\log e^2-\log1=2-0=2$$