Indian Statistical Institute B.Math & B.Stat : Polynomials

Let $0 < a_0 < a_1 < a_2 < \dots < a_n$ be real numbers. Suppose $p(t)$ is a real valued polynomial of degree n such that $$\int_{a_j}^{a_{j+1}} p(t) dt =0 \quad \forall \quad 0 \leq j \leq n-1.$$ Show that, for $0 \leq j \leq n-1$, the polynomial $p(t)$ has exactly one root in the interval $(a_j ,a_{j+1})$. Let $p(t) = b_0 t^n+b_1t^{n-1}+\dots+b_{n-1}t+b_n$ and let $g(t) = \frac{b_0}{n+1}t^{n+1}+\frac{b_1}{n}t^{n}+\dots+\frac{b_{n-1}}{2}t^{2}+\frac{b_n}{1}t^{1}$ Note that $g'(t) = p(t) \dots (A)$. Now consider the interval $[a_j,a_{j+1}] \quad where \quad 0 \leq j \leq n-1.$ $g(t)$ is continuous and differentiable in $[a_j,a_{j+1}]$ and $(a_j,a_{j+1})$ respectively ( $g(t)$ being a polynomial.) It is given that $\int_{a_j}^{a_{j+1}} p(t) dt =0 \implies g(a_{j+1})-g(a_j) = 0$ ( using $(A)$ ) $\implies g(a_{j+1}) =g(a_j)$ which in turn shows that $g(t)$ satisfies all the conditions of $Rolle's-Theorem$ on $[a_j,a_{j+1}]$. $\implies g'(t) = 0$ for at least one $t$ in $(a_j,a_{j+1})$ for all $j \in \{0,1,2,\dots,n-1\}$. $\implies p(t) = 0$ ( using $(A)$ )for at least one $t$ in $(a_j,a_{j+1})$ for all $j \in \{0,1,2,\dots,n-1\}$. $\implies p(t)$ has at least one real root in $(a_j,a_{j+1})$ for all $j \in \{0,1,2,\dots,n-1\}$. Since $degree(p(t))=n$ it has $n$ number of roots ( counting multiplicity ). Since there are $n$ interval of the form $(a_j,a_{j+1})$ and each of them contains at lest one root of $p(t)$, each of them must contain exactly one root of $p(t)$ otherwise number of roots will exceed $n$.