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 195900 by Calculusboy last updated on 12/Aug/23

Answered by cortano12 last updated on 13/Aug/23

$${[\frac{1}{3}{\mathrm{x}}^3+\frac{1}{2}{\mathrm{x}}^2]}_{-2}^{\mathrm{b}}=\frac{16}{3}$$$${\left[\frac{1}{2}{\mathrm{x}}^2(\frac{2}{3}\mathrm{x}+1)\right]}_{-2}^{\mathrm{b}}=\frac{16}{3}$$$$\frac{1}{2}{\mathrm{b}}^2(\frac{2}{3}\mathrm{b}+1)-2(-\frac{1}{3})=\frac{16}{3}$$$$\frac{1}{3}{\mathrm{b}}^3+\frac{1}{2}{\mathrm{b}}^2-\frac{14}{3}=0$$$${2\mathrm{b}}^3+{3\mathrm{b}}^2-28=0$$$$(\mathrm{b}-2)({2\mathrm{b}}^2+7\mathrm{b}+14)=0$$$$\mathrm{b}=2$$$${2\mathrm{b}}^2+7\mathrm{b}+28=0$$$$\mathrm{b}=\frac{-7+3\mathrm{i}\sqrt{7}}{4};\mathrm{b}=\frac{-7-3\mathrm{i}\sqrt{7}}{4}$$

Commented by Calculusboy last updated on 13/Aug/23

$$thanks$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com