Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

Question Number 54378 by ngo last updated on 02/Feb/19

∫_(π/3) ^(3π/2)   [ 2 cos x ] dx =

3π/2π/3[2cosx]dx=

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

with the help of graph trying to solve...  ∫_(π/3) ^(π/2)  [2cosx]dx+∫_(π/2) ^((2π)/3) [2cosx]dx+∫_((2π)/3) ^((4π)/3) [2cosx]dx+∫_((4π)/3) ^((3π)/2)  [2cosx]dx  =0+(−1)(((2π)/3)−(π/2))+(−2)(((4π)/3)−((2π)/3))+(−1)(((3π)/2)−((4π)/3))  =(−1)((π/6))+(−2)(((2π)/3))+(−1)((π/6))  =−(π/3)−((4π)/3)=((−5π)/3)  pls check....

withthehelpofgraphtryingtosolve...π3π2[2cosx]dx+π22π3[2cosx]dx+2π34π3[2cosx]dx+4π33π2[2cosx]dx=0+(1)(2π3π2)+(2)(4π32π3)+(1)(3π24π3)=(1)(π6)+(2)(2π3)+(1)(π6)=π34π3=5π3plscheck....

Terms of Service

Privacy Policy

Contact: info@tinkutara.com