Exists $a < c < b$ where $\int_a^b (b - x)^n f^{(n + 1)}(x)\,dx = {{f^{(n + 1)}(c)}\over{n + 1}}(b - a)^{n + 1}$? Let $\tilde{a} < a < b < \tilde{b}$ and let $n$ be a nonnegative integer. Suppose $f(x)$ is a real-valued function on $(\tilde{a}, \tilde{b})$ which is $(n + 1)$-times continuously differentiable on $(\tilde{a}, \tilde{b})$.
My question is this. Does there necessarily exist some $a < c < b$ such that$$\int_a^b (b - x)^n f^{(n + 1)}(x)\,dx = {{f^{(n + 1)}(c)}\over{n + 1}}(b - a)^{n + 1}?$$
 A: Since $f$ is $(n+1)$-times continuously differentiable on $(\tilde{a},\tilde{b})$ , $f^{(n+1)}$ is continuous on $(\tilde{a},\tilde{b})$. By the Mean Value Theorem for integrals, one has 
$$\int_a^b (b - x)^n f^{(n + 1)}(x)\,dx = f^{(n + 1)}(c)\int_a^b(b - x)^ndx={{f^{(n + 1)}(c)}\over{n + 1}}(b - a)^{n + 1}.$$
A: If $F(x)$ and $G(x)$ are $n + 1$-times continuously differentiable on $(\tilde{a}, \tilde{b})$ and $F^{(k)}(a) = G^{(k)}(a)$ for $0 \le k \le n$ and $G^{(k)}(x) \neq 0$ for $x \in (a, b)$ and $1 \le k \le n$, we can apply the generalized mean value theorem $n + 1$ times to get $a < c_{n + 1} < c_n < \ldots < c_1 < b$ such that\begin{align*} & {{F(b) - F(a)}\over{G(b) - G(a)}} = {{F'(c_1)}\over{G'(c_1)}} \\ & = {{F'(c_1) - F'(a)}\over{G'(c_1) - G'(a)}} = {{F^{\prime\prime}(c_2)}\over{G^{\prime\prime}(c_2)}} \\ & = \ldots \\ & = {{F^{(k)}(c_k)  - F^{(k)}(a)}\over{G^{(k)}(c_k) - G^{(k)}(a)}} = {{F^{(k + 1)}(c_{k + 1})}\over{G^{(k + 1)}(c_{k + 1})}} \\ & = \ldots \\ & = {{F^{(n)}(c_n)  - F^{(n)}(a)}\over{G^{(n)}(c_n) - G^{(n)}(a)}} = {{F^{(n + 1)}(c_{n + 1})}\over{G^{(n + 1)}(c_{n + 1})}}.\end{align*}We now apply this to the functions$$F(x) = \int_{t = a}^x (x - t)^n f^{(n + 1)}(t)\,dt$$and $G(x) = (x - a)^{n + 1}$. By induction on $k$ for $0 \le k \le n$ we get$$F^{(k)}(x) = n(n - 1) \ldots (n - k + 1) \int_{t = a}^x (x - t)^{n - k} f^{(n + 1)}(t)\,dt,$$because when $0 \le k < n$, according to the chain rule the differentiation of$$n(n - 1) \ldots (n - k + 1) \int_{t = a}^x (x - t)^{n - k} f^{(n + 1)}(t)\,dt$$with respect to $x$ yields


*

*the value of$$n(n - 1) \ldots (n - k + 1)(x - t)^{n - k}f^{(n + 1)}(t)$$at $t = x$ by the fundamental theorem of calculus (which is $0$) plus

*$$n(n - 1) \ldots (n - k + 1) \int_{t = a}^x {d\over{dx}}\left((x - t)^{n - k} f^{(n + 1)}(t)\right)\,dt = F^{(k + 1)}(x).$$


Moreover,$$F^{(n + 1)}(x) = n!\, f^{(n + 1)}(x),$$because$$F^{(n)}(x) = n! \int_{t = a}^x f^{(n + 1)}(t)\,dt.$$Clearly, $F^{(k)}(a) = 0$ for $0 \le k \le n$ and$$G^{(k)}(x) = (n + 1)n \ldots (n - k + 2)(x - a)^{n + 1 - k}$$for $0 \le k \le n + 1$ so that $G^{(k)}(a) = 0$ for $0 \le k \le n$ and $G^{(k)}(x) \neq 0$ for $a < x < b$ and $0 \le k \le n$. Thus we can find $a < c = c_{n + 1} < b$ such that$${{F(b) - F(a)}\over{G(b) - G(a)}} = {{F^{(n + 1)}(c_{n + 1})}\over{G^{(n + 1)}(c_{n + 1})}},$$which is the same as$${{\int_a^b (b - x)^n f^{(n + 1)}(x)\,dx}\over{(b - a)^{n + 1}}} = {{f^{(n + 1}(c)}\over{n + 1}}.$$
