So here is the question: Suppose Q(f) = $\sum_{j = 0}^n A_{j}f(x_{j})$ is a quadrature formula to approximate I(f) = $\int_{-1}^1 f(x) dx$ where $\sum_{j = 0}^n A_{j} = 2$. Let $Q_{N}$ be the composite formula corresponding to Q applied to [a,b] and suppose that g $\in$ C[a,b]. Using upper and lower Riemann sums to show that $Q_{N}(g) \to \int_a^b g(z) dz$ . $A_{j}\ge$ 0 for $0\le j\le n$.
I don't where to start and why we have that condition of Riemann sums of 2. Are we going to use the error of composite rules?
Appreciate any help!!!
To prove that $Q_N(g) \to \int_a^b g(z) dz$, we will be using the property of upper and lower Riemann sums. Let $x_j$ be the evenly spaced points between $a$ and $b$ with a distance of $\frac{b-a}{n+1}$ and let $f_j$ be the value of the function at those points. Then we can express the quadrature formula as: $Q_N(g)=A_0 g(x_0)+A_1 g(x_1)+A_2 g(x_2)+...+A_n g(x_n)$
Now, define the upper and lower Riemann sums to be: $U=\sum_{j=0}^n A_j\,sup\{g(x): x_j\le x\le x_{j+1}\}$ $L=\sum_{j=0}^n A_j\,inf \{g(x): x_j\le x\le x_{j+1}\}$
We know that $Q_N(g)$ is a linear combination of these values, so the sum of the combinations can be written as: $Q_N(g) =A_0(U-L)+A_1(U_1-L_1)+...+A_n(U_n-L_n)$
Since $A_1+A_2+...+A_n=2$, we can rewrite the above equation as: $Q_N(g) =2(U-L)$
Now, from the definition of the upper and lower Riemann sums, we have: $U-L=\sum_{j=0}^n \Bigg[sup\{g(x): x_j\le x\le x_{j+1}\} - inf\{g(x): x_j\le x\le x_{j+1}\}\Bigg]$
From this, it follows that as $n$ approaches infinity, then U−L approaches $\int_a^b g(z)dz$. Hence, as $n\to \infty$: $Q_N(g)\to 2\int_a^b g(z)dz$
Therefore, when $n\to \infty$, $Q_N(g)$ converges to $\int_a^b g(z)dz$.