Question Number 64384 by aliesam last updated on 17/Jul/19
Answered by MJS last updated on 17/Jul/19
![(y=x)∩(y=(x/8))= ((0),(0) ) (y=x)∩(y=(1/x^2 ))= ((1),(1) ) [x=(1/x^2 ) ⇒ x^3 =1 ⇒ x=1] (y=(x/8))∩(y=(1/x^2 ))= ((2),((1/4)) ) [(x/8)=(1/x^2 ) ⇒ x^3 =8 ⇒ x=2] ∫_0 ^1 xdx+∫_1 ^2 (dx/x^2 )−∫_0 ^2 (x/8)dx=[(x^2 /2)]_0 ^1 +[−(1/x)]_1 ^2 −[(x^2 /(16))]_0 ^2 = =((1/2)−0)+(−(1/2)+(1/1))−((1/2)−0)=(1/2)](https://www.tinkutara.com/question/Q64388.png)
$$\left({y}={x}\right)\cap\left({y}=\frac{{x}}{\mathrm{8}}\right)=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\left({y}={x}\right)\cap\left({y}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\left[{x}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow\:{x}^{\mathrm{3}} =\mathrm{1}\:\Rightarrow\:{x}=\mathrm{1}\right] \\ $$$$\left({y}=\frac{{x}}{\mathrm{8}}\right)\cap\left({y}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\begin{pmatrix}{\mathrm{2}}\\{\frac{\mathrm{1}}{\mathrm{4}}}\end{pmatrix} \\ $$$$\:\:\:\:\:\left[\frac{{x}}{\mathrm{8}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow\:{x}^{\mathrm{3}} =\mathrm{8}\:\Rightarrow\:{x}=\mathrm{2}\right] \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{xdx}+\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{{dx}}{{x}^{\mathrm{2}} }−\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\frac{{x}}{\mathrm{8}}{dx}=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} +\left[−\frac{\mathrm{1}}{{x}}\right]_{\mathrm{1}} ^{\mathrm{2}} −\left[\frac{{x}^{\mathrm{2}} }{\mathrm{16}}\right]_{\mathrm{0}} ^{\mathrm{2}} = \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0}\right)+\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}}\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$