1. **Stating the problem:** We are given the integral $$I(a,n)=\int_0^1 x^a (1-x)^n dx$$ and the special case $$I(a,0)=\int_0^1 x^a dx$$. We need to use integration by parts to establish two relations between $$I(a+1,n)$$ and $$I(a,n+1)$$ and then calculate $$I(a,n)-I(a,n+1)$$. 2. **Recall integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 3. **First relation (using integration by parts on $$I(a,n)$$):** Let $$u = (1-x)^n$$ and $$dv = x^a dx$$. Then $$du = -n(1-x)^{n-1} dx$$ and $$v = \frac{x^{a+1}}{a+1}$$. Applying integration by parts: $$I(a,n) = \int_0^1 x^a (1-x)^n dx = \left. (1-x)^n \frac{x^{a+1}}{a+1} \right|_0^1 + \frac{n}{a+1} \int_0^1 x^{a+1} (1-x)^{n-1} dx$$ Evaluating the boundary term: At $$x=1$$, $$ (1-1)^n \frac{1^{a+1}}{a+1} = 0$$. At $$x=0$$, $$ (1-0)^n \frac{0^{a+1}}{a+1} = 0$$. So boundary term is zero. Therefore: $$I(a,n) = \frac{n}{a+1} I(a+1,n-1)$$ Rearranged: $$I(a+1,n-1) = \frac{a+1}{n} I(a,n)$$ 4. **Second relation (using integration by parts on $$I(a,n)$$ with different choice):** Let $$u = x^a$$ and $$dv = (1-x)^n dx$$. Then $$du = a x^{a-1} dx$$ and $$v = -\frac{(1-x)^{n+1}}{n+1}$$. Applying integration by parts: $$I(a,n) = \left. -x^a \frac{(1-x)^{n+1}}{n+1} \right|_0^1 + \frac{a}{n+1} \int_0^1 x^{a-1} (1-x)^{n+1} dx$$ Evaluating boundary term: At $$x=1$$, $$-1^a \frac{(1-1)^{n+1}}{n+1} = 0$$. At $$x=0$$, $$-0^a \frac{(1-0)^{n+1}}{n+1} = 0$$. So boundary term is zero. Therefore: $$I(a,n) = \frac{a}{n+1} I(a-1,n+1)$$ Rearranged: $$I(a-1,n+1) = \frac{n+1}{a} I(a,n)$$ 5. **Calculate $$I(a,n) - I(a,n+1)$$:** Recall: $$I(a,n) = \int_0^1 x^a (1-x)^n dx$$ $$I(a,n+1) = \int_0^1 x^a (1-x)^{n+1} dx$$ Subtracting: $$I(a,n) - I(a,n+1) = \int_0^1 x^a (1-x)^n dx - \int_0^1 x^a (1-x)^{n+1} dx = \int_0^1 x^a (1-x)^n (1 - (1-x)) dx$$ Simplify inside the integral: $$1 - (1-x) = x$$ So: $$I(a,n) - I(a,n+1) = \int_0^1 x^a (1-x)^n x dx = \int_0^1 x^{a+1} (1-x)^n dx = I(a+1,n)$$ **Final answers:** - $$I(a+1,n-1) = \frac{a+1}{n} I(a,n)$$ - $$I(a-1,n+1) = \frac{n+1}{a} I(a,n)$$ - $$I(a,n) - I(a,n+1) = I(a+1,n)$$