The value of $2^{2010} \frac{\int_0^1 x^{1004}(1-x)^{1004} dx}{\int_0^1 x^{1004}\left(1-x^{2010}\right)^{1004} dx}$ is ____.
Step-by-Step Solution
Key Concept: Use Beta function symmetry property B(p,q) = B(q,p) and the identity ∫₀¹x^m(1-x)^n dx = B(m+1,n+1) to relate the two integrals. The denominator integral ∫₀¹x^1004(1-x^2010)^1004 dx requires substitution recognition that (1-x^2010) creates a different measure structure than (1-x)^1004.
Split $I_1 = ∫_0^1x^{1004}(1-x)^{1004}dx = 2∫_0^{1/2}x^{1004}(1-x)^{1004}dx$ using symmetry. For $I_2 = ∫_0^1x^{1004}(1-2010)^{1004}dx$, substitute $u = 1005x$ to get $I_2 = \frac{1}{1005}∫_0^{1005}(1-t)^{1004}t^{1004}dt$. Through substitution $t = 2y$ and integration, establish that $\frac{I_1}{I_2} = \frac{2^{2010}}{2^{2008}} = 4$.
Correct Answer: 4