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Question Number 116738 by Eric002 last updated on 06/Oct/20

$$determinetheareaoftheregionbounded$$$$byy={(2x+6)}^{0.5}andy=x-1$$

Commented by bobhans last updated on 06/Oct/20

Answered by 1549442205PVT last updated on 06/Oct/20

$$\mathrm{Question}\mathrm{should}\mathrm{be}:\mathrm{Calculate}\mathrm{the}\mathrm{area}$$$$\mathrm{of}\mathrm{region}\mathrm{bounded}\mathrm{by}\mathrm{curve}\mathrm{y}=\sqrt{2\mathrm{x}+6}$$$$,\mathrm{the}\mathrm{line}\mathrm{y}=\mathrm{x}-1\mathrm{and}\mathrm{the}\mathrm{Ox}$$$$\mathrm{We}\mathrm{find}\mathrm{the}\mathrm{intersection}\mathrm{of}\mathrm{line}\left(\mathrm{d}\right):$$$$\mathrm{y}=\mathrm{x}-1\mathrm{and}\mathrm{the}\mathrm{curve}\mathrm{C}:\mathrm{y}=\sqrt{2\mathrm{x}+6}$$$$\mathrm{x}-1=\sqrt{2\mathrm{x}+6}\iff 2\mathrm{x}+6={\mathrm{x}}^2-2\mathrm{x}+1(\mathrm{x}⩾1)$$$$\iff {\mathrm{x}}^2-4\mathrm{x}-5=0\iff (\mathrm{x}+1)(\mathrm{x}-5)=0$$$$\iff \mathrm{x}=5\Rightarrow \mathrm{y}=4.\Rightarrow \mathrm{C}\cap \mathrm{d}=\mathrm{B}(5;4)$$$$\mathrm{C}\cap \mathrm{Ox}=\mathrm{A}(-3;0),\mathrm{d}\cap \mathrm{Ox}=\mathrm{C}(1;0).\mathrm{Hence}$$$$\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{region}\mathrm{bounded}\mathrm{by}\mathrm{C},\mathrm{d}$$$$\mathrm{and}\mathrm{the}\mathrm{asix}\mathrm{Ox}\mathrm{equal}\mathrm{to}$$$$\mathrm{S}={\int }_{-3}^5\sqrt{2\mathrm{x}+6}\mathrm{dx}-{\int }_1^5(\mathrm{x}-1)\mathrm{dx}$$$$\mathrm{Put}\sqrt{2\mathrm{x}+6}=\mathrm{u}\Rightarrow 2\mathrm{x}+6={\mathrm{u}}^2$$$$\Rightarrow \mathrm{dx}=\mathrm{udu}\int \sqrt{2\mathrm{x}+6}\mathrm{dx}=\int {\mathrm{u}}^2\mathrm{du}=\frac{{\mathrm{u}}^3}{3}$$$$\Rightarrow \mathrm{S}=\frac{(2\mathrm{x}+6)\sqrt{2\mathrm{x}+6}}{3}{\mid }_{-3}^5-{[\frac{{\mathrm{x}}^2}{2}-\mathrm{x}]}_1^5=$$$$=\frac{64}{3}-7.5-0.5=\frac{64}{3}-8=\frac{40}{3}$$$$\mathrm{S}=\frac{40}{3}$$

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