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 2.7k views answer comment Share Follow See all 11 Comments See all 11 11 Comments reply Show 8 previous comments 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.