Question Number 146784 by tabata last updated on 15/Jul/21
Commented by tabata last updated on 15/Jul/21
$$??????? \\ $$
Commented by tabata last updated on 15/Jul/21
$${help}\:{me}\:{please} \\ $$
Answered by Cyriille last updated on 15/Jul/21
![Q2\ y=(√x) y=2x−1, x=0 at point of intersecrion, (√x) = 2x−1 ⇒x=(2x−1)^2 ⇒x=4x^2 −4x+1 ⇒4x^2 −5x+1=0 ⇒x=1 or x=(1/4) at x=1, y=1 at x=(1/4), y=-(1/2) y=(√x) ⇒x=y^2 y=2x−1 ⇒x=((y+1)/2) A=∫_(-(1/2)) ^1 y^2 dy −∫_(-(1/2)) ^1 ((y+1)/2)dy ⇒[(y^3 /3)]_(-(1/2)) ^1 −[(y^2 /4)+(y/2)]_(-(1/2)) ^1 ⇒[(1/3)−((((−(1/2))^3 )/3))]−[((1/4)+(1/2))−((((-(1/2))^2 )/4)+((-(1/2))/2))] ⇒(3/8)−((15)/(16))=-(9/(16))](https://www.tinkutara.com/question/Q146795.png)
$$\boldsymbol{\mathrm{Q}}\mathrm{2}\backslash \\ $$$${y}=\sqrt{{x}}\:\:\:\:{y}=\mathrm{2}{x}−\mathrm{1},\:\:{x}=\mathrm{0} \\ $$$${at}\:{point}\:{of}\:{intersecrion}, \\ $$$$\sqrt{{x}}\:=\:\mathrm{2}{x}−\mathrm{1} \\ $$$$\Rightarrow{x}=\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1}\:{or}\:{x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${at}\:{x}=\mathrm{1},\:{y}=\mathrm{1} \\ $$$${at}\:{x}=\frac{\mathrm{1}}{\mathrm{4}},\:{y}=-\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}=\sqrt{{x}}\:\:\Rightarrow{x}={y}^{\mathrm{2}} \\ $$$${y}=\mathrm{2}{x}−\mathrm{1}\:\:\Rightarrow{x}=\frac{{y}+\mathrm{1}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{A}}=\int_{-\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} {y}^{\mathrm{2}} {dy}\:−\int_{-\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{y}+\mathrm{1}}{\mathrm{2}}{dy} \\ $$$$\:\:\:\Rightarrow\left[\frac{{y}^{\mathrm{3}} }{\mathrm{3}}\right]_{-\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} −\left[\frac{{y}^{\mathrm{2}} }{\mathrm{4}}+\frac{{y}}{\mathrm{2}}\right]_{-\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\Rightarrow\left[\frac{\mathrm{1}}{\mathrm{3}}−\left(\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }{\mathrm{3}}\right)\right]−\left[\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\left(\frac{\left(-\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{-\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}}\right)\right] \\ $$$$\:\:\:\:\:\Rightarrow\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{15}}{\mathrm{16}}=-\frac{\mathrm{9}}{\mathrm{16}} \\ $$
Commented by tabata last updated on 15/Jul/21
$${sir}\:{how}\:{the}\:{area}\:{in}\:{negitive} \\ $$
Commented by Cyriille last updated on 01/Oct/21
$${take}\:{abosolute}\:{value} \\ $$
Answered by Cyriille last updated on 15/Jul/21
$$\boldsymbol{\mathrm{Q}}\mathrm{4}\backslash \\ $$$${y}=\int_{\mathrm{1}} ^{{x}} \sqrt{{t}^{\mathrm{2}} −\mathrm{1}}{dt},\:\:\boldsymbol{{A}}=\int_{{a}} ^{{b}} {ydx} \\ $$$${y}=\int_{\mathrm{1}} ^{{x}} \sqrt{{t}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$\Rightarrow{y}=\int_{{a}} ^{{b}} {tan}\theta\left({tan}\theta\right){sec}\theta{d}\theta \\ $$$$\:\:\:\:\:\:\:\:=\int_{{a}} ^{{b}} \frac{{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{3}} \theta}{d}\theta \\ $$$$\:\:\:\:\:\:\:\:\:=\int_{{a}} ^{{b}} {sin}\theta.\frac{{sin}\theta}{{cos}^{\mathrm{3}} \theta}{d}\theta \\ $$$$\boldsymbol{\mathrm{I}}\:\mathrm{can}\:\mathrm{not}\:\mathrm{continue}\:\mathrm{but}\:\mathrm{I}\:\mathrm{hope}\:\mathrm{my}\:\mathrm{idea}\:\mathrm{was}\:\mathrm{correct}! \\ $$