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 158009 by zainaltanjung last updated on 30/Oct/21

Answered by puissant last updated on 30/Oct/21

$$\mathrm{\Omega }={\int }_2^4\frac{2x-1}{x^2-x}dx={[ln\mid x^2-x\mid ]}_2^4$$$$=ln12-ln2=ln\left(\frac{12}{2}\right)=ln6..$$

Commented by zainaltanjung last updated on 30/Oct/21

$$\mathrm{ok}.\mathrm{right}$$$$$$

Answered by Mr.D.N. last updated on 30/Oct/21

$${\int }_2^4\frac{2\mathrm{x}-1}{{\mathrm{x}}^2-\mathrm{x}}\mathrm{dx}$$$$\mathrm{Put},{\mathrm{x}}^2-\mathrm{x}=\mathrm{t}$$$$(2\mathrm{x}-1)\mathrm{dx}=\mathrm{dt}$$$$\mathrm{t}=12,2$$$${\int }_2^{12}\frac{\mathrm{dt}}{\mathrm{t}}={\left[\log \mathrm{t}\right]}_2^{12}$$$$\mathrm{or},\log 12-\log 2$$$$\mathrm{or},\log (2^2.3)-\log 2$$$$\mathrm{or},{\log 2}^2+\log 3-\log 2$$$$\mathrm{or},2\log 2-\log 2+\log 3$$$$\mathrm{or},\log 2+\log 3=\log (2.3)$$$$\therefore \log 6//.$$

Commented by zainaltanjung last updated on 30/Oct/21

$$\mathrm{ok}.{\mathrm{it}}^{\prime }\mathrm{right}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com