What is $\int_0^{1}\frac{x^{300}}{1+x^2+x^3}dx$ upto $2$ decimal places? 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?
 A: 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).
A: 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
$$
0.001107
<
\frac{1}{903}
\le
\int_0^{1}\frac{x^{300}}{1+x^2+x^3} \, dx
\le
\frac{1}{301}
<
0.003323
$$
