In an examination, I was asked to calculate $\int_0^{1}\frac{x^{300}}{1+x^2+x^3}dx$. Options were gives as

a - 0.00
b - 0.02
c - 0.10
d - 0.33
e - 1.00

Just look at the questions I felt the integral $I \geq \sum_0^1{}\frac{x^{300}}{1+x^2+x^3} = 0.33$. I felt, since numerator is very small as compared to denominator therefore, for value $\epsilon<1$, $1.00 $ isn't possible. So, I chose option d. But I am not sure whether its correct or not as I didn't follow standard procedure.

what is the correct answer and How can it be solved using a standard procedure?


Since $1+x^2+x^3 \ge 1$ for $x \ge 0$, we have $$ \int_0^{1}\frac{x^{300}}{1+x^2+x^3} \, dx \le \int_0^{1}x^{300} \, dx =\frac{1}{301} <\frac{1}{300} =0.00333\cdots $$ This is enough to answer the question.

A little more work gives a good estimate of the integral.

Since $1+x^2+x^3 \le 3$ for $x \ge 0$, we have $$ \int_0^{1}\frac{x^{300}}{1+x^2+x^3} \, dx \ge \int_0^{1} \frac{1}{3} x^{300} \, dx =\frac{1}{903} $$ Thus $$ 0.001107 < \frac{1}{903} \le \int_0^{1}\frac{x^{300}}{1+x^2+x^3} \, dx \le \frac{1}{301} < 0.003323 $$

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    $\begingroup$ The integral is approximately $0.00111357$ according to WA. $\endgroup$ – lhf Dec 19 '18 at 11:27
  • $\begingroup$ I liked the way you solved. Thanks for the help. :) $\endgroup$ – Mr.Sigma. Dec 19 '18 at 11:46

Since $\frac{x^{300}}{1+x^2+x^3}$ has a zero of order $300$ at the origin, most of the mass of the integral $\int_{0}^{1}\frac{x^{300}}{1+x^2+x^3}\,dx$ comes from a neighbourhood of the right endpoint of the integration range. Obviously $\int_{0}^{1}\frac{x^{300}}{3}\,dx=\frac{1}{903}$, and

$$ I-\frac{1}{903}=\int_{0}^{1}x^{300}\left(\frac{1}{1+x^2+x^3}-\frac{1}{3}\right)\,dx =\int_{0}^{1}x^{300}(1-x)\frac{2+2x+x^2}{3(1+x^2+x^3)}\,dx$$ is bounded by $$ \int_{0}^{1}x^{300}(1-x)\,dx = \frac{1}{301\cdot 302} $$ hence the wanted integral is $\color{green}{0.0011}$(unknown digits).


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