Question Number 35325 by Cheyboy last updated on 17/May/18
$${Sketch}\:{the}\:{region}\:{enclosed}\:{by}\:{the} \\ $$$${curves}\:{of}\:{y}=\mathrm{1}/{x}\:{and}\:{y}=\mathrm{1}/{x}^{\mathrm{2}} \:{and} \\ $$$${find}\:{the}\:{area}\:{of}\:{the}\:{region}. \\ $$$${plzz}\:{help}\:{me} \\ $$
Answered by MJS last updated on 17/May/18
![we need the points of intersection (1/x)=(1/x^2 ) ⇒ x≠0 x^2 −x=0 x(x−1)=0 ⇒ x=1 so we have only one common point in R: I_1 = ((1),(1) ) there′s no such region if we consider I_2 = ((∞),(0) ) we can try to calculate ∫_1 ^∞ ((1/x)−(1/x^2 ))dx=[ln x+(1/x)]_1 ^∞ =ln ∞+(1/∞)−(ln 1+(1/1))= =∞+0−(0+1)=∞ so no area exists](https://www.tinkutara.com/question/Q35326.png)
$$\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{points}\:\mathrm{of}\:\mathrm{intersection} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow\:{x}\neq\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}=\mathrm{0} \\ $$$${x}\left({x}−\mathrm{1}\right)=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{1} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{only}\:\mathrm{one}\:\mathrm{common}\:\mathrm{point}\:\mathrm{in}\:\mathbb{R}: \\ $$$${I}_{\mathrm{1}} =\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{such}\:\mathrm{region} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{consider}\:{I}_{\mathrm{2}} =\begin{pmatrix}{\infty}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{try}\:\mathrm{to}\:\mathrm{calculate} \\ $$$$\underset{\mathrm{1}} {\overset{\infty} {\int}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}=\left[\mathrm{ln}\:{x}+\frac{\mathrm{1}}{{x}}\overset{\infty} {\right]}_{\mathrm{1}} =\mathrm{ln}\:\infty+\frac{\mathrm{1}}{\infty}−\left(\mathrm{ln}\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}\right)= \\ $$$$=\infty+\mathrm{0}−\left(\mathrm{0}+\mathrm{1}\right)=\infty \\ $$$$\mathrm{so}\:\mathrm{no}\:\mathrm{area}\:\mathrm{exists} \\ $$
Commented by Cheyboy last updated on 18/May/18
$${Thank}\:{you}\:{very}\:{much}\:{sir} \\ $$