Question Number 141189 by Eric002 last updated on 16/May/21
$${find}\:{the}\:{area}\:{of}\:{the}\:{shaded}\:{region} \\ $$$${shown}\:{below}\:{which}\:{is}\:{boinded}\:{by}\:{to}\:{functions}\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} \:\:{g}\left({x}\right)=\mathrm{2}−{x}\:{and}\:{the}\:{x}-{axis} \\ $$$$ \\ $$
Commented by Eric002 last updated on 16/May/21
Commented by bramlexs22 last updated on 16/May/21
$${Area}\:=\:\frac{\mathrm{1}}{\mathrm{3}}.\mathrm{1}.\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{1}.\mathrm{1}\:=\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$
Commented by Dwaipayan Shikari last updated on 16/May/21
$${x}^{\mathrm{2}} =\mathrm{2}−{x}\Rightarrow{x}=\mathrm{1}\:,{y}=\mathrm{1}\:\:\left({Meeting}\:{point}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} {dx}+\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{1}.\left(\mathrm{2}−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$