Question Number 9822 by ridwan balatif last updated on 06/Jan/17
Answered by sandy_suhendra last updated on 06/Jan/17
![f ′(x)=k(3x^2 −12x+9)=0 x^2 −4x+3=0 (x−3)(x−1)=0 x=3(min) or x=1(max) ∫_0 ^( 3) k(x^3 −6x^2 +9x) dx = 27 k[(1/4)x^4 −2x^3 +4(1/2)x^2 ]_0 ^3 = 27 k(((81)/4)−54+((81)/2)) = 27 k(((27)/4)) = 27 k=4](https://www.tinkutara.com/question/Q9823.png)
$$\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{k}\left(\mathrm{3x}^{\mathrm{2}} −\mathrm{12x}+\mathrm{9}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{3}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{3}\left(\mathrm{min}\right)\:\mathrm{or}\:\mathrm{x}=\mathrm{1}\left(\mathrm{max}\right) \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\:\mathrm{3}} \mathrm{k}\left(\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} +\mathrm{9x}\right)\:\mathrm{dx}\:=\:\mathrm{27} \\ $$$$\mathrm{k}\left[\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{3}} +\mathrm{4}\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{3}} =\:\mathrm{27} \\ $$$$\mathrm{k}\left(\frac{\mathrm{81}}{\mathrm{4}}−\mathrm{54}+\frac{\mathrm{81}}{\mathrm{2}}\right)\:=\:\mathrm{27} \\ $$$$\mathrm{k}\left(\frac{\mathrm{27}}{\mathrm{4}}\right)\:=\:\mathrm{27} \\ $$$$\mathrm{k}=\mathrm{4} \\ $$
Commented by ridwan balatif last updated on 07/Jan/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$