7 votes 7 votes Consider the integral $$\int^{1}_{0} \frac{x^{300}}{1+x^2+x^3} dx$$ What is the value of this integral correct up to two decimal places? $0.00$ $0.02$ $0.10$ $0.33$ $1.00$ Calculus tifr2019 engineering-mathematics calculus definite-integral + – Arjun asked Dec 18, 2018 • edited Nov 22, 2022 by Lakshman Bhaiya Arjun 2.9k views answer comment Share Follow See all 11 Comments See all 11 11 Comments reply arvin commented Dec 9, 2018 reply Follow Share dx is missing and is it yours question or from any book or test series? 1 votes 1 votes goxul commented Dec 9, 2018 reply Follow Share This question is from TIFR's paper held today. 0 votes 0 votes amit166 commented Dec 9, 2018 reply Follow Share yes 0 votes 0 votes Himanshu Kumar 1 commented Dec 9, 2018 reply Follow Share Waiting for answer 0 votes 0 votes Lakshman Bhaiya commented Dec 9, 2018 reply Follow Share $\infty (or)0?$ 0 votes 0 votes Sayan Bose commented Dec 9, 2018 reply Follow Share @Lakshman Patel RJIT Pls explain with solution 0 votes 0 votes HeadShot commented Dec 10, 2018 reply Follow Share I asked it on SO but they said its pretty tough and surely wouldn't have asked in exam. LOL!! 0 votes 0 votes amit166 commented Dec 10, 2018 reply Follow Share it's tifr ph.d question 0 votes 0 votes HeadShot commented Dec 10, 2018 reply Follow Share @amit166 yeah I attempted the exam. 0 votes 0 votes arvin commented Dec 10, 2018 i edited by akash.dinkar12 Jun 9, 2019 reply Follow Share i think you cannot directly break $x^3+x^2+1$ it can be done only by using $x= \tan\ y$ than range will be from $0\ to\ \pi$.. now applying definite integral formula over the new range... but it is getting more complex and it will really need a good time to solve... that's why its Ph.D. question... 0 votes 0 votes Lakshman Bhaiya commented Jun 8, 2019 reply Follow Share see here 2 votes 2 votes Please log in or register to add a comment.
Best answer 19 votes 19 votes $\displaystyle \int^{1}_{0} \frac{x^{300}}{1+x^2+x^3} dx \leq \int^{1}_{0} {x^{300}} dx (\because 1+x^2+x^3 \geq 1)$ $\qquad \qquad \leq \left[\frac{x^{301}}{301}\right]_0^1\leq \frac{1}{301} \leq 0.0033$ Only option matching is Option A. Arjun answered Jun 8, 2019 Arjun comment Share Follow See all 3 Comments See all 3 3 Comments reply `JEET commented Dec 6, 2019 reply Follow Share That was really a good trick. Thanks 1 votes 1 votes Nikhil gate 2020 commented Jan 18, 2020 reply Follow Share thanks sir 0 votes 0 votes shashankrustagi commented Dec 4, 2020 reply Follow Share Wow sir, you are just awesome 0 votes 0 votes Please log in or register to add a comment.
6 votes 6 votes (a) is answer. We can prove this to be less than 0.009 and select option a. https://drive.google.com/file/d/1Cr5cSe0nzq_RFruzPlVkTfGIDKPFFISz/view?usp=drivesdk See this image in above link rahulvats1996 answered Dec 10, 2018 • edited Dec 11, 2018 by rahulvats1996 rahulvats1996 comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments rahulvats1996 commented Dec 11, 2018 reply Follow Share @Lakshman Patel RJIT i have tried to get a upper bound of the answer. As it is not possible to get the exact value by conventional methods. (i hope someone would do that). i think essence of this question was this trick only as seen by options. 0 votes 0 votes Lakshman Bhaiya commented Dec 11, 2018 reply Follow Share Ok, don't worry this question is really good. thanks 0 votes 0 votes rahulvats1996 commented Dec 11, 2018 reply Follow Share I saw your previous comment on email. So i am new here dont know how to write equation neither image was uploading so solved at back of my notebook. Clicked that and shared drive link. :-) 1 votes 1 votes Please log in or register to add a comment.