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Question Number 72813 by Learner-123 last updated on 03/Nov/19

$$Findtheareaoftheregionenclosed$$$$bytheline5y=x+6andthecurve$$$$y=\sqrt{\mid x\mid }.$$

Commented by ajfour last updated on 03/Nov/19

Commented by ajfour last updated on 03/Nov/19

$$5y=x+6$$$$y=\sqrt{\mid x\mid }$$$$\Rightarrow forintersectionpoints$$$${(x+6)}^2=25\mid x\mid$$$$\Rightarrow x^2+12x-25\mid x\mid +36=0$$$$forx>0$$$$x^2-13x+36=0$$$$x=\frac{13}{2}\pm \sqrt{\frac{169}{4}-36}=\frac{13}{2}\pm \frac{5}{2}$$$$x_B=4,x_C=9$$$$forx<0$$$$x^2+37x+36=0$$$$(x+36)(x+1)=0$$$$x_A=-1$$$$AreaA=A_1+A_2+A_3$$$$A_1={\int }_{-1}^0(\frac{x+6}{5}-\sqrt{-x})dx$$$$={[\frac{x^2+12x}{10}-\frac{2}{3}{(-x)}^{3/2}]}_{-1}^0$$$$=\frac{2}{3}+\frac{11}{10}=\frac{53}{30}$$$$A_2={\int }_0^4(\frac{x+6}{5}-\sqrt{x})dx$$$$={[\frac{x^2+12x}{10}-\frac{2}{3}{x}^{3/2}]}_0^4$$$$=\frac{64}{10}-\frac{16}{3}=\frac{16}{15}$$$$A_3={\int }_4^9(\sqrt{x}-\frac{x+6}{5})dx$$$$=\frac{2}{3}{x}^{3/2}-\frac{x^2+12x}{10}$$$$=\frac{2}{3}(27-8)-\frac{1}{10}(81+108-16-48)$$$$=\frac{38}{3}-\frac{25}{2}=\frac{1}{6}$$$$A=\frac{53}{30}+\frac{32}{30}+\frac{5}{30}=3sq.units.$$

Commented by Learner-123 last updated on 03/Nov/19

$$thankssir.$$

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