Question Number 151899 by liberty last updated on 24/Aug/21
Answered by john_santu last updated on 24/Aug/21
![A=∫_0 ^6 (((−x^3 )/6)+x^2 )dx=[−(x^4 /(24))+(x^3 /3)]_0 ^6 A=−(6^4 /(24))+(6^3 /3)=18 xA=∫_0 ^6 (−(x^4 /6)+x^3 )dx=[−(x^5 /(30))+(x^4 /4)]_0 ^6 xA=−(6^5 /(30))+(6^4 /4)=((324)/5) x^− = ((xA)/A)=((324)/(90))=((18)/5) yA=∫_0 ^6 (1/2)(((−x^3 )/6)+x^2 )^2 dx yA=(1/2)∫_0 ^6 ((x^6 /(36))−(x^5 /3)+x^4 )dx yA=(1/2)[ (x^7 /(252))−(x^6 /(18))+(x^5 /5)]_0 ^6 yA=(6^7 /(504))−(6^6 /(36))+(6^5 /(10))=((1296)/(35)) y^− =((yA)/y)= ((1296)/(630))=((72)/(35)) therefore centroid C(((18)/5),((72)/(35)))](https://www.tinkutara.com/question/Q151901.png)
$$\mathrm{A}=\int_{\mathrm{0}} ^{\mathrm{6}} \left(\frac{−\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}=\left[−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{24}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{6}} \\ $$$$\mathrm{A}=−\frac{\mathrm{6}^{\mathrm{4}} }{\mathrm{24}}+\frac{\mathrm{6}^{\mathrm{3}} }{\mathrm{3}}=\mathrm{18} \\ $$$$\mathrm{xA}=\int_{\mathrm{0}} ^{\mathrm{6}} \left(−\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{6}}+\mathrm{x}^{\mathrm{3}} \right)\mathrm{dx}=\left[−\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{30}}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{0}} ^{\mathrm{6}} \\ $$$$\mathrm{xA}=−\frac{\mathrm{6}^{\mathrm{5}} }{\mathrm{30}}+\frac{\mathrm{6}^{\mathrm{4}} }{\mathrm{4}}=\frac{\mathrm{324}}{\mathrm{5}} \\ $$$$\overset{−} {\mathrm{x}}=\:\frac{\mathrm{xA}}{\mathrm{A}}=\frac{\mathrm{324}}{\mathrm{90}}=\frac{\mathrm{18}}{\mathrm{5}} \\ $$$$\mathrm{yA}=\int_{\mathrm{0}} ^{\mathrm{6}} \frac{\mathrm{1}}{\mathrm{2}}\left(\frac{−\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{yA}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{6}} \left(\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{36}}−\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{3}}+\mathrm{x}^{\mathrm{4}} \right)\mathrm{dx} \\ $$$$\mathrm{yA}=\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{\mathrm{x}^{\mathrm{7}} }{\mathrm{252}}−\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{18}}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}}\right]_{\mathrm{0}} ^{\mathrm{6}} \\ $$$$\mathrm{yA}=\frac{\mathrm{6}^{\mathrm{7}} }{\mathrm{504}}−\frac{\mathrm{6}^{\mathrm{6}} }{\mathrm{36}}+\frac{\mathrm{6}^{\mathrm{5}} }{\mathrm{10}}=\frac{\mathrm{1296}}{\mathrm{35}} \\ $$$$\overset{−} {\mathrm{y}}=\frac{\mathrm{yA}}{\mathrm{y}}=\:\frac{\mathrm{1296}}{\mathrm{630}}=\frac{\mathrm{72}}{\mathrm{35}}\: \\ $$$$\mathrm{therefore}\:\mathrm{centroid}\:\mathrm{C}\left(\frac{\mathrm{18}}{\mathrm{5}},\frac{\mathrm{72}}{\mathrm{35}}\right) \\ $$