Question Number 226850 by cherokeesay last updated on 17/Dec/25
Answered by TonyCWX last updated on 17/Dec/25
![y = x(x−1) ⇒ x = 0 or x = 1 A>0, A=1 x(x−1) = (1/2)x x^2 −x−(1/2)x = 0 x^2 −(3/2)x = 0 x(x−(3/2)) = 0 x = 0 or x = (3/2) B>1, B=(3/2) y_C = (1/2)((3/2)) = (3/4) C = (1, (3/4)) Area shaded in blue = (1/2)((3/2))((3/4))−∫_1 ^(3/2) (x^2 −x)dx = (9/(16))−[(x^3 /3)−(x^2 /2)] = (9/(16))−[(((((3/2))^3 )/3)−((((3/2))^2 )/2))−((1^3 /3)−(1^2 /2))] = ((19)/(48))](https://www.tinkutara.com/question/Q226851.png)
$${y}\:=\:{x}\left({x}−\mathrm{1}\right)\:\Rightarrow\:{x}\:=\:\mathrm{0}\:\mathrm{or}\:{x}\:=\:\mathrm{1} \\ $$$$\mathrm{A}>\mathrm{0},\:\mathrm{A}=\mathrm{1} \\ $$$$ \\ $$$${x}\left({x}−\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x} \\ $$$${x}^{\mathrm{2}} −{x}−\frac{\mathrm{1}}{\mathrm{2}}{x}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{x}\:=\:\mathrm{0} \\ $$$${x}\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$${x}\:=\:\mathrm{0}\:\mathrm{or}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{B}>\mathrm{1},\:\mathrm{B}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$${y}_{\mathrm{C}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{C}\:=\:\left(\mathrm{1},\:\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$ \\ $$$$\mathrm{Area}\:\mathrm{shaded}\:\mathrm{in}\:\mathrm{blue} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\int_{\mathrm{1}} ^{\frac{\mathrm{3}}{\mathrm{2}}} \left({x}^{\mathrm{2}} −{x}\right)\mathrm{dx} \\ $$$$=\:\frac{\mathrm{9}}{\mathrm{16}}−\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right] \\ $$$$=\:\frac{\mathrm{9}}{\mathrm{16}}−\left[\left(\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} }{\mathrm{3}}−\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}\right)−\left(\frac{\mathrm{1}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}\right)\right] \\ $$$$=\:\frac{\mathrm{19}}{\mathrm{48}} \\ $$
Commented by cherokeesay last updated on 17/Dec/25
$${thank}\:{you}\:{so}\:{much}. \\ $$
Answered by som(math1967) last updated on 17/Dec/25
![y=x(x−1) ⇒y=0.5x x^2 −x=0.5x ⇒x(x−1.5x)=0⇒x=0,1.5 area of △OCB ∫_0 ^(1.5) 0.5xdx=[((0.5×x^2 )/2)]_0 ^(1.5) =0.5625squ area of ABCA ∫_1 ^(1.5) (x^2 −x)dx [intersect Xaxis at1 amd line1.5] [(x^3 /3) −(x^2 /2)]_1 ^(1.5) =(1.5)^2 (((1.5)/3) −(1/2))−((1/3)−(1/2)) =2.25×0+0.167=0.167sq u Blue area=0.5625−0.167 =0.3955squ](https://www.tinkutara.com/question/Q226852.png)
$${y}={x}\left({x}−\mathrm{1}\right)\: \\ $$$$\Rightarrow{y}=\mathrm{0}.\mathrm{5}{x} \\ $$$${x}^{\mathrm{2}} −{x}=\mathrm{0}.\mathrm{5}{x} \\ $$$$\Rightarrow{x}\left({x}−\mathrm{1}.\mathrm{5}{x}\right)=\mathrm{0}\Rightarrow{x}=\mathrm{0},\mathrm{1}.\mathrm{5} \\ $$$$\:{area}\:{of}\:\bigtriangleup{OCB} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}.\mathrm{5}} \mathrm{0}.\mathrm{5}{xdx}=\left[\frac{\mathrm{0}.\mathrm{5}×{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}.\mathrm{5}} =\mathrm{0}.\mathrm{5625}{squ} \\ $$$${area}\:{of}\:{ABCA} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{1}.\mathrm{5}} \:\left({x}^{\mathrm{2}} −{x}\right){dx}\:\left[{intersect}\:{Xaxis}\:{at}\mathrm{1}\:{amd}\:{line}\mathrm{1}.\mathrm{5}\right] \\ $$$$\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{1}.\mathrm{5}} =\left(\mathrm{1}.\mathrm{5}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}.\mathrm{5}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)−\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}.\mathrm{25}×\mathrm{0}+\mathrm{0}.\mathrm{167}=\mathrm{0}.\mathrm{167}{sq}\:{u} \\ $$$${Blue}\:{area}=\mathrm{0}.\mathrm{5625}−\mathrm{0}.\mathrm{167} \\ $$$$=\mathrm{0}.\mathrm{3955}{squ} \\ $$
Commented by cherokeesay last updated on 17/Dec/25
$${thank}\:{you}\:{sir}. \\ $$
Answered by mr W last updated on 17/Dec/25
$$\mathrm{0}.\mathrm{5}{x}={x}\left({x}−\mathrm{1}\right) \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\mathrm{0},\:{x}_{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${h}=\mathrm{0}.\mathrm{5}×\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}\right)=\frac{\mathrm{9}}{\mathrm{16}} \\ $$$${shaded}\:=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{9}}{\mathrm{16}}−\mathrm{1}×\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{19}}{\mathrm{48}}\:\checkmark \\ $$
Commented by cherokeesay last updated on 17/Dec/25
$${thank}\:{you}\:{master}\:! \\ $$