Question Number 59829 by ANTARES VY last updated on 15/May/19
Commented by ANTARES VY last updated on 15/May/19
$$\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{painted}}\:\:\boldsymbol{\mathrm{area}} \\ $$
Commented by MJS last updated on 15/May/19
$$\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:−{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{7}\:\mathrm{or}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{is}\:\mathrm{totally} \\ $$$$\mathrm{wrong} \\ $$
Answered by ajfour last updated on 15/May/19
$$\mathrm{y}=\:−\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\left(?\right)\mathrm{7} \\ $$
Commented by ANTARES VY last updated on 15/May/19
$$+ \\ $$
Commented by ajfour last updated on 15/May/19
$$\mathrm{then}\:\mathrm{for}\:\mathrm{x}=\mathrm{0}\:\:,\:\mathrm{y}=\mathrm{7} \\ $$$$\mathrm{its}\:\mathrm{dont}\:\mathrm{appear}\:\mathrm{so}\:\mathrm{in}\:\mathrm{the}\:\mathrm{graph}.. \\ $$
Answered by MJS last updated on 15/May/19
![f(x)=−x^2 −6x−7 [=−(x+3−(√2))(x+3+(√2))] g(x)=−x^2 −2x+5 [=−(x+1−(√6))(x+1+(√6))] h(x)=2x f(x)=g(x) ⇒ x=−3 f(x)=h(x) ⇒ x=−7 ∨ x=−1 g(x)=h(x) ⇒ x=−5 ∨ x=1 area=∫_(−3) ^1 g(x)dx−∫_(−3) ^(−1) f(x)dx−∫_(−1) ^1 h(x)dx=20](https://www.tinkutara.com/question/Q59869.png)
$${f}\left({x}\right)=−{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{7}\:\:\:\:\:\left[=−\left({x}+\mathrm{3}−\sqrt{\mathrm{2}}\right)\left({x}+\mathrm{3}+\sqrt{\mathrm{2}}\right)\right] \\ $$$${g}\left({x}\right)=−{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}\:\:\:\:\:\left[=−\left({x}+\mathrm{1}−\sqrt{\mathrm{6}}\right)\left({x}+\mathrm{1}+\sqrt{\mathrm{6}}\right)\right] \\ $$$${h}\left({x}\right)=\mathrm{2}{x} \\ $$$${f}\left({x}\right)={g}\left({x}\right)\:\Rightarrow\:{x}=−\mathrm{3} \\ $$$${f}\left({x}\right)={h}\left({x}\right)\:\Rightarrow\:{x}=−\mathrm{7}\:\vee\:{x}=−\mathrm{1} \\ $$$${g}\left({x}\right)={h}\left({x}\right)\:\Rightarrow\:{x}=−\mathrm{5}\:\vee\:{x}=\mathrm{1} \\ $$$$\mathrm{area}=\underset{−\mathrm{3}} {\overset{\mathrm{1}} {\int}}{g}\left({x}\right){dx}−\underset{−\mathrm{3}} {\overset{−\mathrm{1}} {\int}}{f}\left({x}\right){dx}−\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}{h}\left({x}\right){dx}=\mathrm{20} \\ $$