1. The problem states two integrals: $$I(a,n)=\int_0^1 x^a (1-x)^n dx$$ and $$I(a,n)=\int_a^b f(x) dx$$. We want to understand or relate these integrals. 2. The first integral $$I(a,n)=\int_0^1 x^a (1-x)^n dx$$ is a Beta function integral form, where $a > -1$ and $n > -1$ for convergence. 3. The Beta function is defined as $$B(x,y) = \int_0^1 t^{x-1} (1-t)^{y-1} dt$$ for $x,y > 0$. 4. Comparing, we see $$I(a,n) = \int_0^1 x^a (1-x)^n dx = B(a+1, n+1) = \frac{\Gamma(a+1) \Gamma(n+1)}{\Gamma(a+n+2)}$$ where $\Gamma$ is the Gamma function. 5. The second integral $$I(a,n) = \int_a^b f(x) dx$$ is a general integral with limits $a$ and $b$ and integrand $f(x)$, which is unrelated to the first unless $f(x) = x^a (1-x)^n$ and limits are 0 to 1. 6. To summarize, the first integral is a Beta function integral with a closed form involving Gamma functions. Final answer: $$I(a,n) = \int_0^1 x^a (1-x)^n dx = \frac{\Gamma(a+1) \Gamma(n+1)}{\Gamma(a+n+2)}$$