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Question Number 109674 by I want to learn more last updated on 24/Aug/20

Find the area of the region enclose by the curve y = 1 − x^2 , and the line y = 1 + 3x, x = 1.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{enclose}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve}\:\:\:\mathrm{y}\:\:=\:\:\mathrm{1}\:\:−\:\:\mathrm{x}^{\mathrm{2}} \:, \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:\:\:\:\mathrm{y}\:\:=\:\:\mathrm{1}\:\:+\:\:\mathrm{3x},\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{1}. \\ $$

Answered by bobhans last updated on 25/Aug/20

((⋰♭o♭⋰)/(hans)) Area 1 = ∫_0 ^1 (1+3x)−(1−x^2 )dx = ∫_0 ^1 (3x+x^2 )dx=[(3/2)x^2 +(1/3)x^3 ]_0 ^1 = (3/2)+(1/3)=((11)/6) Area 2 = ((Δ(√Δ))/(6a^2 ))= ((9(√9))/(6.1))=((9.3)/6)=((27)/6) where 1+3x=1−x^2 ⇒x^2 +3x=0; Δ=9−4.1.0=9 Totally area =((11+27)/6)=((38)/6)

$$\:\:\:\frac{\iddots\flat{o}\flat\iddots}{{hans}} \\ $$$${Area}\:\mathrm{1}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{1}+\mathrm{3}{x}\right)−\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{3}{x}+{x}^{\mathrm{2}} \right){dx}=\left[\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$${Area}\:\mathrm{2}\:=\:\frac{\Delta\sqrt{\Delta}}{\mathrm{6}{a}^{\mathrm{2}} }=\:\frac{\mathrm{9}\sqrt{\mathrm{9}}}{\mathrm{6}.\mathrm{1}}=\frac{\mathrm{9}.\mathrm{3}}{\mathrm{6}}=\frac{\mathrm{27}}{\mathrm{6}} \\ $$$${where}\:\mathrm{1}+\mathrm{3}{x}=\mathrm{1}−{x}^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} +\mathrm{3}{x}=\mathrm{0};\:\Delta=\mathrm{9}−\mathrm{4}.\mathrm{1}.\mathrm{0}=\mathrm{9} \\ $$$${Totally}\:{area}\:=\frac{\mathrm{11}+\mathrm{27}}{\mathrm{6}}=\frac{\mathrm{38}}{\mathrm{6}} \\ $$

Commented by I want to learn more last updated on 25/Aug/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Answered by 1549442205PVT last updated on 25/Aug/20

We need find intesection point of two graphs of the functions y=1−x^2 and y=3x+1. { ((y=1−x^2 )),((y=3x+1)) :}⇔ { ((1−x^2 =3x+1(1))),((y=3x+1)) :} (1)⇔x^2 +3x=0⇔x(x+3)=0⇔x∈{0;−3} y∈{1;−8}⇒A(0;1),B(−3;−8) S_1 =∫_0 ^( 1) ∣3x+1−(1−x^2 )∣dx=∫_0 ^1 (x^2 +3x)dx S_2 =∫_(−3) ^( 0) ∣(x^2 +3x)∣dx=−∫_(−3) ^( 0) (x^2 +3x)∣dx −(x^3 /3)−((3x^2 )/2)=−9+((27)/2)=(9/2) =((x^3 /3)+((3x^2 )/2))_0 ^1 =(1/3)+(3/2)=((11)/6) .From that S=S_1 +S_2 =(9/2)+((11)/6)=((19)/3)

$$\mathrm{We}\:\mathrm{need}\:\mathrm{find}\:\mathrm{intesection}\:\mathrm{point}\:\mathrm{of}\:\mathrm{two} \\ $$$$\mathrm{graphs}\:\mathrm{of}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{y}=\mathrm{1}−\mathrm{x}^{\mathrm{2}} \mathrm{and} \\ $$$$\mathrm{y}=\mathrm{3x}+\mathrm{1}. \\ $$$$\begin{cases}{\mathrm{y}=\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\\{\mathrm{y}=\mathrm{3x}+\mathrm{1}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} =\mathrm{3x}+\mathrm{1}\left(\mathrm{1}\right)}\\{\mathrm{y}=\mathrm{3x}+\mathrm{1}}\end{cases} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{3x}=\mathrm{0}\Leftrightarrow\mathrm{x}\left(\mathrm{x}+\mathrm{3}\right)=\mathrm{0}\Leftrightarrow\mathrm{x}\in\left\{\mathrm{0};−\mathrm{3}\right\} \\ $$$$\mathrm{y}\in\left\{\mathrm{1};−\mathrm{8}\right\}\Rightarrow\mathrm{A}\left(\mathrm{0};\mathrm{1}\right),\mathrm{B}\left(−\mathrm{3};−\mathrm{8}\right) \\ $$$$\mathrm{S}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} \mid\mathrm{3x}+\mathrm{1}−\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\mid\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}\right)\mathrm{dx} \\ $$$$\mathrm{S}_{\mathrm{2}} =\int_{−\mathrm{3}} ^{\:\mathrm{0}} \mid\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}\right)\mid\mathrm{dx}=−\int_{−\mathrm{3}} ^{\:\mathrm{0}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}\right)\mid\mathrm{dx} \\ $$$$−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{2}}=−\mathrm{9}+\frac{\mathrm{27}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$=\left(\frac{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{2}}\right)_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{6}}\:.\mathrm{From}\:\mathrm{that} \\ $$$$\boldsymbol{\mathrm{S}}=\boldsymbol{\mathrm{S}}_{\mathrm{1}} +\boldsymbol{\mathrm{S}}_{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{2}}+\frac{\mathrm{11}}{\mathrm{6}}=\frac{\mathrm{19}}{\mathrm{3}}\: \\ $$

Commented by I want to learn more last updated on 25/Aug/20

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Commented by 1549442205PVT last updated on 25/Aug/20

You are welcome!

$$\mathrm{You}\:\mathrm{are}\:\mathrm{welcome}! \\ $$

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