Question-213114 Tinku Tara October 30, 2024 Integration 0 Comments FacebookTweetPin Question Number 213114 by efronzo1 last updated on 30/Oct/24 ÷― Answered by issac last updated on 30/Oct/24 f(x)=−C(x−1)+iπ(x−1)5000ln(x−1)(thxwolframalpha!!)f(1)=0,cusf(1+0)+f(1−0)=05000=0f(1)(x+1)−f(1)(1−x)=5000x4999……??????????DamniwasfooledXD∫01f(x+1)dx+∫01f(1−x)dx=∫01x5000dx∫12f(u)−∫10f(u)du=∫02f(u)du=15001Explain.integralbothside∫01f(x+1)dx+∫01f(1−x)dx=∫01x5000dx∫01f(x+1)dx=∫12f(u)ducussubstitudex+1=udu=dxsimilarly∫01f(1−x)dx=∫01f(u)du1−x=udu=−dx∫ab+∫bc=∫ac∫02f(u)du=∫01x5000dx=15001Yeah!! Answered by mr W last updated on 30/Oct/24 ∫02f(x)dx=∫01f(x)dx+∫12f(x)dx=∫−10f(1+t)d(1+t)+∫0−1f(1−t)d(1−t)=∫−10f(1+t)dt+∫−10f(1−t)dt=∫−10[f(1+t)+f(1−t)]dt=∫−10t5000dt=[t50015001]−10=15001 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-213082Next Next post: determiner-a-et-b-par-AB-BC-AM-5-AC-16- Leave a Reply Cancel replyYour email address will not be published.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.