find-find-I-1-3-x-2-x-2-4x-2-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 29162 by abdo imad last updated on 04/Feb/18 findfindI=∫13∣x−2∣(x2−4x)2dx. Commented by abdo imad last updated on 06/Feb/18 thechaslesrelationgiveI=∫12−x+2(x2−4x)2dx+∫23x−2(x2−4x)2dx∫12−x+2(x2−4x)2dx=−12∫122x−4(x2−4x)2dx=12[1x2−4x]12=12(−14+13)=124and∫23x−2(x2−4x)2dx=12∫132x−4(x2−4x)2dx=−12[1x2−4x]13=−12(−13+13)=0⇒I=124. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: let-give-f-x-x-1-2-x-2-x-1-2-x-2-1-simlify-f-x-2-solve-inside-N-2-the-equation-f-x-y-Next Next post: let-give-n-p-from-N-2-and-1-p-n-prove-that-k-0-p-C-n-k-C-n-k-p-k-2-p-C-n-p- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![the chasles relation give I=∫_1 ^2 ((−x+2)/((x^2 −4x)^2 ))dx +∫_2 ^3 ((x−2)/((x^2 −4x)^2 ))dx ∫_1 ^2 ((−x +2)/((x^2 −4x)^2 ))dx=−(1/2)∫_1 ^2 ((2x−4)/((x^2 −4x)^2 ))dx =(1/2)[ (1/(x^2 −4x))]_1 ^2 =(1/2)(−(1/4) +(1/3))=(1/(24)) and ∫_2 ^3 ((x−2)/((x^2 −4x)^2 ))dx=(1/2) ∫_1 ^3 ((2x−4)/((x^2 −4x)^2 ))dx =−(1/2)[ (1/(x^2 −4x))]_1 ^3 =−(1/2)( −(1/3) +(1/3))=0 ⇒ I=(1/(24)) .](https://www.tinkutara.com/question/Q29304.png)