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 57992 by Tinkutara last updated on 15/Apr/19

Answered by MJS last updated on 16/Apr/19

∫_1 ^c (−x^5 +8x^2 )dx=((16)/3)  −(1/6)c^6 +(8/3)c^3 −(5/2)=((16)/3)  c^6 −16c^3 +47=0  t=c^3   t^2 −16t+47=0  t_1 =8−(√(17)); t_2 =8+(√(17))  c_1 =((8−(√(17))))^(1/3) ; c_2 =((8+(√(17))))^(1/3) ; all other solutions ∉R  but  −x^5 +8x^2 =0 ⇒ x_1 =x_2 =0; x_3 =2; all other solutions ∉R  ⇒ area in ]−∞; 0[ negative, in ]0; 2[ positive        and in ]2; +∞[ negative  c_2 ≈2.3>2 ⇒ not valid  ⇒ c=c_1 =((8−(√(17))))^(1/3)   c+2=2+((8−(√(17))))^(1/3) ≈3.57095

$$\underset{\mathrm{1}} {\overset{{c}} {\int}}\left(−{x}^{\mathrm{5}} +\mathrm{8}{x}^{\mathrm{2}} \right){dx}=\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}{c}^{\mathrm{6}} +\frac{\mathrm{8}}{\mathrm{3}}{c}^{\mathrm{3}} −\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{16}}{\mathrm{3}} \\ $$$${c}^{\mathrm{6}} −\mathrm{16}{c}^{\mathrm{3}} +\mathrm{47}=\mathrm{0} \\ $$$${t}={c}^{\mathrm{3}} \\ $$$${t}^{\mathrm{2}} −\mathrm{16}{t}+\mathrm{47}=\mathrm{0} \\ $$$${t}_{\mathrm{1}} =\mathrm{8}−\sqrt{\mathrm{17}};\:{t}_{\mathrm{2}} =\mathrm{8}+\sqrt{\mathrm{17}} \\ $$$${c}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{\mathrm{8}−\sqrt{\mathrm{17}}};\:{c}_{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\mathrm{8}+\sqrt{\mathrm{17}}};\:\mathrm{all}\:\mathrm{other}\:\mathrm{solutions}\:\notin\mathbb{R} \\ $$$$\mathrm{but} \\ $$$$−{x}^{\mathrm{5}} +\mathrm{8}{x}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1}} ={x}_{\mathrm{2}} =\mathrm{0};\:{x}_{\mathrm{3}} =\mathrm{2};\:\mathrm{all}\:\mathrm{other}\:\mathrm{solutions}\:\notin\mathbb{R} \\ $$$$\left.\Rightarrow\:\mathrm{area}\:\mathrm{in}\:\right]−\infty;\:\mathrm{0}\left[\:\mathrm{negative},\:\mathrm{in}\:\right]\mathrm{0};\:\mathrm{2}\left[\:\mathrm{positive}\right. \\ $$$$\left.\:\:\:\:\:\:\mathrm{and}\:\mathrm{in}\:\right]\mathrm{2};\:+\infty\left[\:\mathrm{negative}\right. \\ $$$${c}_{\mathrm{2}} \approx\mathrm{2}.\mathrm{3}>\mathrm{2}\:\Rightarrow\:\mathrm{not}\:\mathrm{valid} \\ $$$$\Rightarrow\:{c}={c}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{\mathrm{8}−\sqrt{\mathrm{17}}} \\ $$$${c}+\mathrm{2}=\mathrm{2}+\sqrt[{\mathrm{3}}]{\mathrm{8}−\sqrt{\mathrm{17}}}\approx\mathrm{3}.\mathrm{57095} \\ $$

Commented by MJS last updated on 16/Apr/19

we should start with  ∫_1 ^2 (−x^5 +8x^2 )dx=((49)/6)>((16)/3) ⇒ 1<c<2

$$\mathrm{we}\:\mathrm{should}\:\mathrm{start}\:\mathrm{with} \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\left(−{x}^{\mathrm{5}} +\mathrm{8}{x}^{\mathrm{2}} \right){dx}=\frac{\mathrm{49}}{\mathrm{6}}>\frac{\mathrm{16}}{\mathrm{3}}\:\Rightarrow\:\mathrm{1}<{c}<\mathrm{2} \\ $$

Commented by Tinkutara last updated on 18/Apr/19

Thanks Sir! Maybe the answer 1 in book is wrong.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com