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Question Number 124675 by sdfg last updated on 05/Dec/20

∫_1 ^3 (x−1)^3 (3−x)^2 dx

$$\int_{\mathrm{1}} ^{\mathrm{3}} \left({x}−\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{3}−{x}\right)^{\mathrm{2}} {dx} \\ $$

Commented by mohammad17 last updated on 05/Dec/20

put:m=x−1⇒x=m+1⇒dx=dm x=3⇒m=2 , x=1⇒m=0 ∴∫_1 ^( 3) (x−1)^3 (3−x)^2 dx=∫_0 ^( 2) m^3 (2−m)^2 dm =∫_0 ^( 2) (4m^3 −4m^4 +m^5 )dm =(m^4 −((4m^5 )/5)+(m^6 /6))_0 ^2 =16−((128)/5)+((64)/6)=((16)/(15)) (mohammad Aldolaimy)

$${put}:{m}={x}−\mathrm{1}\Rightarrow{x}={m}+\mathrm{1}\Rightarrow{dx}={dm} \\ $$$$ \\ $$$${x}=\mathrm{3}\Rightarrow{m}=\mathrm{2}\:\:\:\:,\:\:\:\:{x}=\mathrm{1}\Rightarrow{m}=\mathrm{0} \\ $$$$ \\ $$$$\therefore\int_{\mathrm{1}} ^{\:\mathrm{3}} \left({x}−\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{3}−{x}\right)^{\mathrm{2}} {dx}=\int_{\mathrm{0}} ^{\:\mathrm{2}} {m}^{\mathrm{3}} \left(\mathrm{2}−{m}\right)^{\mathrm{2}} {dm} \\ $$$$ \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{2}} \left(\mathrm{4}{m}^{\mathrm{3}} −\mathrm{4}{m}^{\mathrm{4}} +{m}^{\mathrm{5}} \right){dm} \\ $$$$ \\ $$$$=\left({m}^{\mathrm{4}} −\frac{\mathrm{4}{m}^{\mathrm{5}} }{\mathrm{5}}+\frac{{m}^{\mathrm{6}} }{\mathrm{6}}\right)_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{16}−\frac{\mathrm{128}}{\mathrm{5}}+\frac{\mathrm{64}}{\mathrm{6}}=\frac{\mathrm{16}}{\mathrm{15}} \\ $$$$ \\ $$$$\left({mohammad}\:{Aldolaimy}\right) \\ $$

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