Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 109427 by mathdave last updated on 23/Aug/20

Answered by 1549442205PVT last updated on 23/Aug/20

$$\left|\begin{array}{cccccccccc}\mathrm{x}& 0& & 1& & 2& & 3& & 4\\ \mid \mathrm{x}-1\mid & & 1-\mathrm{x}& 0& \mathrm{x}-1& & \mathrm{x}-1& & \mathrm{x}-1& \\ \mid \mathrm{x}-2\mid & & 2-\mathrm{x}& & 2-\mathrm{x}& 9& \mathrm{x}-2& & \mathrm{x}-2& \\ \mid \mathrm{x}-3\mid & & 3-\mathrm{x}& & 3-\mathrm{x}& & 3-\mathrm{x}& & \mathrm{x}-3& \\ \mathrm{nom}& & 3-2\mathrm{x}& & 1& & 2\mathrm{x}-3& & 2\mathrm{x}-3& \\ \mathrm{deno}& & 8-4\mathrm{x}& & 6-2\mathrm{x}& & -2+2\mathrm{x}& & 4\mathrm{x}-8& \\ \mathrm{f}\left(\mathrm{x}\right)& & \frac{2\mathrm{x}-3}{4\mathrm{x}-8}& & \frac{1}{6-2\mathrm{x}}& & \frac{2\mathrm{x}-3}{2\mathrm{x}-2}& & \frac{2\mathrm{x}-3}{4\mathrm{x}-8}& \end{array}\right|$$$$\mathrm{From}\mathrm{above}\mathrm{tablet}\mathrm{we}\mathrm{have}$$$${\int }_0^4\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^1\frac{2\mathrm{x}-3}{4\mathrm{x}-8}\mathrm{dx}+{\int }_1^2\frac{1}{6-2\mathrm{x}}\mathrm{dx}+{\int }_2^3\frac{2\mathrm{x}-3}{2\mathrm{x}-2}\mathrm{dx}+{\int }_3^4\frac{2\mathrm{x}-3}{4\mathrm{x}-8}\mathrm{dx}$$$$\mathrm{A}=\frac{1}{4}{\int }_0^1\frac{2\mathrm{x}-3}{\mathrm{x}-2}\mathrm{dx}=\frac{1}{4}{\int }_0^1(2+\frac{1}{\mathrm{x}-2})\mathrm{dx}$$$$=\frac{1}{4}{[2\mathrm{x}+\ln (2-\mathrm{x})]}_0^1=\frac{1}{4}(2-\ln 2)\left(1\right)$$$$\mathrm{B}={\int }_1^2\frac{1}{6-2\mathrm{x}}\mathrm{dx}=\frac{1}{2}{\int }_1^2\frac{1}{3-\mathrm{x}}\mathrm{dx}=\frac{1}{2}\ln (3-\mathrm{x}){\mid }_1^2$$$$=\frac{1}{2}(0-\ln 2)=-0.5\ln 2\left(2\right)$$$$\mathrm{C}={\int }_2^3\frac{2\mathrm{x}-3}{2\mathrm{x}-2}\mathrm{dx}=\frac{1}{2}{\int }_2^3\frac{2\mathrm{x}-3}{\mathrm{x}-1}\mathrm{dx}$$$$=\frac{1}{2}{\int }_2^3(2-\frac{1}{\mathrm{x}-1})\mathrm{dx}=\frac{1}{2}{[2\mathrm{x}-\ln (\mathrm{x}-1)]}_2^3$$$$=\frac{1}{2}(6-\ln 2-4)=1-0.5\ln 2\left(3\right)$$$$\mathrm{D}={\int }_3^4\frac{2\mathrm{x}-3}{4\mathrm{x}-8}\mathrm{dx}=\frac{1}{4}(2\mathrm{x}-\ln \mid \mathrm{x}-2)\mid {\mid }_3^4$$$$=\frac{1}{4}(8-\ln 2-6)=0.5-0.25\ln 2\left(4\right)$$$$\mathrm{Adding}\mathrm{up}\mathrm{four}\mathrm{the}\mathrm{equalities}\mathrm{we}\mathrm{get}$$$$\mathrm{I}={\int }_0^4\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=0.5-0.25\ln 2-0.5\ln 2$$$$+1-0.5\ln 2+0.5-0.25\ln 2=2-\frac{3}{2}\ln 2$$$$\mathbf{Thus},\mathbf{I}=2-\frac{3}{2}\mathbf{ln}2$$

Commented by mathdave last updated on 23/Aug/20

$$youhadreallydoneagreatwork,kudostoyou$$

Commented by 1549442205PVT last updated on 24/Aug/20

$$\mathrm{Thank}\mathrm{you},\mathrm{sir}.\mathrm{You}\mathrm{are}\mathrm{welcome}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com