Find-the-area-of-the-region-between-the-graphs-of-f-x-3x-3-x-2-10x-and-g-x-x-3-2x-

Question Number 12139 by tawa last updated on 14/Apr/17

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{between}\:\mathrm{the}\:\mathrm{graphs}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{3x}^{\mathrm{3}} \:−\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{10x} \\ $$$$\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right)\:=\:−\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{2x} \\ $$

Answered by mrW1 last updated on 15/Apr/17

3x^3 −x^2 −10x=−x^3 +2x 4x^3 −x^2 −12x=0 x(4x^2 −x−12)=0 x=0 4x^2 −x−12=0 x=((1±(√(1+4×4×12)))/(2×4))=((1±(√(193)))/8) x_1 =((1−(√(193)))/8) x_2 =0 x_3 =((1+(√(193)))/8) A=∫_a ^b [f(x)−g(x)]dx =∫_a ^b (3x^3 −x^2 −10x+x^3 −2x)dx =∫_a ^b (4x^3 −x^2 −12x)dx =[x^4 −(x^3 /3)−6x^2 ]_a ^b A_1 =∣[x^4 −(x^3 /3)−6x^2 ]_x_1 ^0 ∣ A_1 =∣(((1−(√(193)))/8))^4 −(1/3)(((1−(√(193)))/8))^3 −6(((1−(√(193)))/8))^2 ∣ =∣(((1−(√(193)))/8))^2 [(((1−(√(193)))/8))^2 −(1/3)(((1−(√(193)))/8))−6]∣ A_2 =∣[x^4 −(x^3 /3)−6x^2 ]_0 ^x_2 ∣ A_2 =∣(((1+(√(193)))/8))^4 −(1/3)(((1+(√(193)))/8))^3 −6(((1+(√(193)))/8))^2 ∣ =∣(((1+(√(193)))/8))^2 [(((1+(√(193)))/8))^2 −(1/3)(((1+(√(193)))/8))−6]∣ A_1 +A_2 =((14113)/(768))≈18.376

$$\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{10}{x}=−{x}^{\mathrm{3}} +\mathrm{2}{x} \\ $$$$\mathrm{4}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{12}{x}=\mathrm{0} \\ $$$${x}\left(\mathrm{4}{x}^{\mathrm{2}} −{x}−\mathrm{12}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −{x}−\mathrm{12}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}×\mathrm{4}×\mathrm{12}}}{\mathrm{2}×\mathrm{4}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{193}}}{\mathrm{8}} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}−\sqrt{\mathrm{193}}}{\mathrm{8}} \\ $$$${x}_{\mathrm{2}} =\mathrm{0} \\ $$$${x}_{\mathrm{3}} =\frac{\mathrm{1}+\sqrt{\mathrm{193}}}{\mathrm{8}} \\ $$$$ \\ $$$${A}=\int_{{a}} ^{{b}} \left[{f}\left({x}\right)−{g}\left({x}\right)\right]{dx} \\ $$$$=\int_{{a}} ^{{b}} \left(\mathrm{3}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{10}{x}+{x}^{\mathrm{3}} −\mathrm{2}{x}\right){dx} \\ $$$$=\int_{{a}} ^{{b}} \left(\mathrm{4}{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{12}{x}\right){dx} \\ $$$$=\left[{x}^{\mathrm{4}} −\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{6}{x}^{\mathrm{2}} \right]_{{a}} ^{{b}} \\ $$$$ \\ $$$${A}_{\mathrm{1}} =\mid\left[{x}^{\mathrm{4}} −\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{6}{x}^{\mathrm{2}} \right]_{{x}_{\mathrm{1}} } ^{\mathrm{0}} \mid \\ $$$${A}_{\mathrm{1}} =\mid\left(\frac{\mathrm{1}−\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{3}} −\mathrm{6}\left(\frac{\mathrm{1}−\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{2}} \mid \\ $$$$=\mid\left(\frac{\mathrm{1}−\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{2}} \left[\left(\frac{\mathrm{1}−\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}−\sqrt{\mathrm{193}}}{\mathrm{8}}\right)−\mathrm{6}\right]\mid \\ $$$$ \\ $$$${A}_{\mathrm{2}} =\mid\left[{x}^{\mathrm{4}} −\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{6}{x}^{\mathrm{2}} \right]_{\mathrm{0}} ^{{x}_{\mathrm{2}} } \mid \\ $$$${A}_{\mathrm{2}} =\mid\left(\frac{\mathrm{1}+\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{3}} −\mathrm{6}\left(\frac{\mathrm{1}+\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{2}} \mid \\ $$$$=\mid\left(\frac{\mathrm{1}+\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{2}} \left[\left(\frac{\mathrm{1}+\sqrt{\mathrm{193}}}{\mathrm{8}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}+\sqrt{\mathrm{193}}}{\mathrm{8}}\right)−\mathrm{6}\right]\mid \\ $$$$ \\ $$$${A}_{\mathrm{1}} +{A}_{\mathrm{2}} =\frac{\mathrm{14113}}{\mathrm{768}}\approx\mathrm{18}.\mathrm{376} \\ $$

Commented by tawa last updated on 14/Apr/17