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 47376 by vajpaithegrate@gmail.com last updated on 09/Nov/18

Answered by MJS last updated on 09/Nov/18

f_1 : y=x^2 −5x+4  f_2 : y=1−x  f_3 : y=0  f_1 ∩f_2   x^2 −5x+4=1−x ⇒ x_1 =1; x_2 =3  f_1 ∩f_3   x^2 −5x+4=0 ⇒ x_3 =1; x_4 =4  ⇒ area=∣∫_3 ^4 f_1 dx+∫_1 ^3 f_2 dx∣=((19)/6)

$${f}_{\mathrm{1}} :\:{y}={x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4} \\ $$$${f}_{\mathrm{2}} :\:{y}=\mathrm{1}−{x} \\ $$$${f}_{\mathrm{3}} :\:{y}=\mathrm{0} \\ $$$${f}_{\mathrm{1}} \cap{f}_{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}=\mathrm{1}−{x}\:\Rightarrow\:{x}_{\mathrm{1}} =\mathrm{1};\:{x}_{\mathrm{2}} =\mathrm{3} \\ $$$${f}_{\mathrm{1}} \cap{f}_{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}=\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{3}} =\mathrm{1};\:{x}_{\mathrm{4}} =\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{area}=\mid\underset{\mathrm{3}} {\overset{\mathrm{4}} {\int}}{f}_{\mathrm{1}} {dx}+\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}{f}_{\mathrm{2}} {dx}\mid=\frac{\mathrm{19}}{\mathrm{6}} \\ $$

Commented by vajpaithegrate@gmail.com last updated on 09/Nov/18

thank u sir

$$\mathrm{thank}\:\mathrm{u}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com