Question Number 32775 by NECx last updated on 02/Apr/18
$${Find}\:{the}\:{area}\:{bounded}\:{by}\:{the} \\ $$$${curve}\:{y}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3},{the}\:{x}\:{axis} \\ $$$${and}\:{the}\:{line}\:{x}=\mathrm{5}\:{and}\:{x}=\mathrm{2} \\ $$
Answered by MJS last updated on 02/Apr/18
![1. check the zeros x^2 +(2/3)x−1=0 x=−(1/3)±(√((1/9)+1))=−(1/3)±((√(10))/3) x_1 ≈−1.39 and x_2 ≈.72 are both inside [−5;2]⇒changes of sign 2. area: x<x_1 ⇒f(x)>0 x∈]x_1 ;x_2 [⇒f(x)<0 x>x_2 ⇒f(x)>0 ∫_(−5) ^x_1 f(x)dx−∫_x_1 ^x_2 f(x)dx+∫_x_2 ^2 f(x)dx= =F(x)∣_(−5) ^x_1 −F(x)∣_x_1 ^x_2 +F(x)∣_x_2 ^2 = [F(x)=x^3 +x^2 −3x] =91+((80(√(10)))/(27))](https://www.tinkutara.com/question/Q32779.png)
$$\mathrm{1}.\:\mathrm{check}\:\mathrm{the}\:\mathrm{zeros} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{3}}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{3}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{9}}+\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{3}}\pm\frac{\sqrt{\mathrm{10}}}{\mathrm{3}} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{1}.\mathrm{39}\:\mathrm{and}\:{x}_{\mathrm{2}} \approx.\mathrm{72}\:\mathrm{are}\:\mathrm{both} \\ $$$$\mathrm{inside}\:\left[−\mathrm{5};\mathrm{2}\right]\Rightarrow\mathrm{changes}\:\mathrm{of}\:\mathrm{sign} \\ $$$$ \\ $$$$\mathrm{2}.\:\mathrm{area}: \\ $$$${x}<{x}_{\mathrm{1}} \Rightarrow{f}\left({x}\right)>\mathrm{0} \\ $$$$\left.{x}\in\right]{x}_{\mathrm{1}} ;{x}_{\mathrm{2}} \left[\Rightarrow{f}\left({x}\right)<\mathrm{0}\right. \\ $$$${x}>{x}_{\mathrm{2}} \Rightarrow{f}\left({x}\right)>\mathrm{0} \\ $$$$\underset{−\mathrm{5}} {\overset{{x}_{\mathrm{1}} } {\int}}{f}\left({x}\right){dx}−\underset{{x}_{\mathrm{1}} } {\overset{{x}_{\mathrm{2}} } {\int}}{f}\left({x}\right){dx}+\underset{{x}_{\mathrm{2}} } {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}= \\ $$$$={F}\left({x}\right)\underset{−\mathrm{5}} {\overset{{x}_{\mathrm{1}} } {\mid}}−{F}\left({x}\right)\underset{{x}_{\mathrm{1}} } {\overset{{x}_{\mathrm{2}} } {\mid}}+{F}\left({x}\right)\underset{{x}_{\mathrm{2}} } {\overset{\mathrm{2}} {\mid}}= \\ $$$$\:\:\:\:\:\left[{F}\left({x}\right)={x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{3}{x}\right] \\ $$$$=\mathrm{91}+\frac{\mathrm{80}\sqrt{\mathrm{10}}}{\mathrm{27}} \\ $$
Commented by MJS last updated on 02/Apr/18
$$\mathrm{no}\:\mathrm{problem},\:\mathrm{posted}\:\mathrm{a}\:\mathrm{new}\:\mathrm{answer} \\ $$
Commented by NECx last updated on 02/Apr/18
$${I}'{m}\:{so}\:{sorry}…\:{I}\:{made}\:{a}\:{typo}\:{error} \\ $$$${the}\:{lines}\:{are}\:{x}=−\mathrm{5}\:{and}\:{x}=\mathrm{2}. \\ $$
Commented by Joel578 last updated on 03/Apr/18
$$\mathrm{You}\:\mathrm{just}\:\mathrm{set}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\mathrm{A}\:=\:\underset{−\mathrm{5}} {\overset{\frac{−\mathrm{1}\:−\:\sqrt{\mathrm{10}}}{\mathrm{3}}} {\int}}\:{f}\left({x}\right)\:{dx}\:+\:\underset{\frac{−\mathrm{1}\:+\:\sqrt{\mathrm{10}}}{\mathrm{3}}} {\overset{\frac{−\mathrm{1}\:−\:\sqrt{\mathrm{10}}}{\mathrm{3}}} {\int}}\:{f}\left({x}\right)\:{dx}\:+\:\underset{\frac{−\mathrm{1}\:+\:\sqrt{\mathrm{10}}}{\mathrm{3}}} {\overset{\mathrm{2}} {\int}}\:\:{f}\left({x}\right)\:{dx} \\ $$
Commented by MJS last updated on 02/Apr/18
$$\mathrm{the}\:\mathrm{middle}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{negative}, \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{area}\:\mathrm{you}\:\mathrm{need}\:\int\mid{f}\left({x}\right)\mid,\:\mathrm{so} \\ $$$$\mathrm{it}'\mathrm{s}\:\underset{−\mathrm{5}} {\overset{−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}} {\int}}{f}\left({x}\right){dx}−\underset{−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}} {\overset{−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}} {\int}}{f}\left({x}\right){dx}+\underset{−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\sqrt{\mathrm{10}}}{\mathrm{3}}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx} \\ $$$$\mathrm{and}\:\mathrm{that}'\mathrm{s}\:\mathrm{just}\:\mathrm{what}\:\mathrm{I}\:\mathrm{did} \\ $$
Commented by Joel578 last updated on 03/Apr/18
$$\mathrm{You}\:\mathrm{are}\:\mathrm{right}.\:\mathrm{It}'\mathrm{s}\:\mathrm{typo}.\:\mathrm{I}\:\mathrm{have}\:\mathrm{corrected}\:\mathrm{it} \\ $$