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

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

Answered by mrW1 last updated on 15/Apr/17

$$3x^3-x^2-10x=-x^3+2x$$$$4x^3-x^2-12x=0$$$$x(4x^2-x-12)=0$$$$x=0$$$$4x^2-x-12=0$$$$x=\frac{1\pm \sqrt{1+4\times 4\times 12}}{2\times 4}=\frac{1\pm \sqrt{193}}{8}$$$$x_1=\frac{1-\sqrt{193}}{8}$$$$x_2=0$$$$x_3=\frac{1+\sqrt{193}}{8}$$$$$$$$A={\int }_a^b[f\left(x\right)-g\left(x\right)]dx$$$$={\int }_a^b(3x^3-x^2-10x+x^3-2x)dx$$$$={\int }_a^b(4x^3-x^2-12x)dx$$$$={[x^4-\frac{x^3}{3}-6x^2]}_a^b$$$$$$$$A_1=\mid {[x^4-\frac{x^3}{3}-6x^2]}_{x_1}^0\mid$$$$A_1=\mid {\left(\frac{1-\sqrt{193}}{8}\right)}^4-\frac{1}{3}{\left(\frac{1-\sqrt{193}}{8}\right)}^3-6{\left(\frac{1-\sqrt{193}}{8}\right)}^2\mid$$$$=\mid {\left(\frac{1-\sqrt{193}}{8}\right)}^2[{\left(\frac{1-\sqrt{193}}{8}\right)}^2-\frac{1}{3}\left(\frac{1-\sqrt{193}}{8}\right)-6]\mid$$$$$$$$A_2=\mid {[x^4-\frac{x^3}{3}-6x^2]}_0^{x_2}\mid$$$$A_2=\mid {\left(\frac{1+\sqrt{193}}{8}\right)}^4-\frac{1}{3}{\left(\frac{1+\sqrt{193}}{8}\right)}^3-6{\left(\frac{1+\sqrt{193}}{8}\right)}^2\mid$$$$=\mid {\left(\frac{1+\sqrt{193}}{8}\right)}^2[{\left(\frac{1+\sqrt{193}}{8}\right)}^2-\frac{1}{3}\left(\frac{1+\sqrt{193}}{8}\right)-6]\mid$$$$$$$$A_1+A_2=\frac{14113}{768}\approx 18.376$$

Commented by tawa last updated on 14/Apr/17

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

Commented by tawa last updated on 14/Apr/17

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

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

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

Commented by chux last updated on 15/Apr/17

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

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

$$hello.FSCcalculator.$$

Commented by chux last updated on 16/Apr/17

$$\mathrm{thanks}$$

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