Use properties of the Bernoulli Polynomials to prove $\int_0^1 B_n(x)dx= 0$

Question

Use the properties of the Bernoulli Polynomials to prove: $$\int_0^1 P_n(x)dx= 0 \tag{for n>0}$$

I have these properties to work with below: $$P_0(x) = 1; \qquad P'_n(x) = n P_{n-1} (x) \qquad P_n (x+1) - P_n (x) = nx^{n-1} $$

I don't have work to show because I am questioning what step one should be? I believe I want to manipulate the derivative somehow so that the integral just undoes it, and end with the difference property, which is apart by 1 just like the bounds on the integral. How can I show this?

Answer 1

Okay, I see the concise answer. After a hint, I was able to construct a long-winded answer of my own:

$$P'_{n+1}(x)=(n+1)P_{n}(x)$$ $$P_{n}(x)= \frac{P'_{n+1}(x)}{n+1}$$ $$\int_0^1 \frac{P'_{n+1}(x)}{n+1}dx$$ $$P_{n+1} (x+1) - P_{n+1} (x) = \frac{(n+1)x^{n}}{(n+1)}$$ $$=x^n , ({x = 0})$$ $$=0$$

Answer 2

You will have to proceed by induction on $n$. You can manipulate the second identity to obtain $$\int_0^1B_n(x)dx = \frac{1}{n+1}\int_0^1B_{n+1}'(x)dx,$$ which you can evaluate with the Fundamental Theorem of Calculus. Can you handle it from here?

Answer 3

$$I_n=\int_0^1 B_{n-1}(x)dx=\frac1{n}\left[B_n(1)-B_n(0)\right]$$ Since $$B_n(x+1)-B_n(x)=nx^{n-1}$$ we evaluate at $x=0$: $$I_n=\frac1n[n\cdot 0^{n-1}]=0$$ QED