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 80752 by bagjamath last updated on 06/Feb/20

Commented by john santu last updated on 06/Feb/20

x−2010= t x−2009=t+1 ∫_(−1) ^0 (t^2 +t)dt = (1/3)t^3 +(1/2)t^2 ∣_(−1) ^0 = −(−(1/3)+(1/2))=(1/3)−(1/2)=−(1/6)

x−2010=tx−2009=t+1∫−10(t2+t)dt=13t3+12t2∣−10=−(−13+12)=13−12=−16

Answered by MJS last updated on 06/Feb/20

∫_a ^b (x−a)(x−b)dx= =∫_a ^b (x^2 −(a+b)x+ab)dx= =[(1/3)x^3 −((a+b)/2)x^2 +abx]_a ^b = =(1/6)(a^3 −b^3 )−(1/2)ab(a−b)= [b=a+c] =(1/6)(a^3 −(a+c)^3 )+(1/2)a(a+c)c=−(c^3 /6)= [c=1] =−(1/6)

∫ba(x−a)(x−b)dx==∫ba(x2−(a+b)x+ab)dx==[13x3−a+b2x2+abx]ab==16(a3−b3)−12ab(a−b)=[b=a+c]=16(a3−(a+c)3)+12a(a+c)c=−c36=[c=1]=−16

Commented by bagjamath last updated on 06/Feb/20

thank you Sir

thankyouSir

Terms of Service

Privacy Policy

Contact: info@tinkutara.com