TIFR CSE 2019 | Part A | Question: 13

Question

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?

• A. $0.00$
• B. $0.02$
• C. $0.10$
• D. $0.33$
• E. $1.00$

Comments

@amit166 

yeah I attempted the exam.

​​​​​

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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...

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see here

Answer 1

$\displaystyle \int^{1}_{0} \frac{x^{300}}{1+x^2+x^3} dx \leq \int^{1}_{0} {x^{300}} dx (\because 1+x^2+x^3 \geq 1)$

$\qquad \qquad \leq \left[\frac{x^{301}}{301}\right]_0^1\leq \frac{1}{301} \leq  0.0033$

Only option matching is Option A.

Comments

That was really a good trick.

Thanks

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thanks sir

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Wow sir, you are just awesome

Answer 2

(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

 

Comments

@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. 

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Ok, don't worry

this question is really good.

thanks

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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. :-)