Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

Find the area of the region between the graphs of f(x) = 3x^3  − x^2  − 10x  and g(x) = − x^3  + 2x

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

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

Wow, i really appreciate sir.

$$\mathrm{Wow},\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

Commented by tawa last updated on 14/Apr/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

Commented by tawa last updated on 14/Apr/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by chux last updated on 15/Apr/17

wow! please which calculator did  this?

$$\mathrm{wow}!\:\mathrm{please}\:\mathrm{which}\:\mathrm{calculator}\:\mathrm{did} \\ $$$$\mathrm{this}? \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17

hello. FSC calculator.

$${hello}.\:{FSC}\:{calculator}. \\ $$

Commented by chux last updated on 16/Apr/17

thanks

$$\mathrm{thanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com