Numerical Reasoning Consider a trapezoid of area 4 bounded...

Numerical Reasoning Consider a trapezoid of area 4 bounded by the graphs of $y=x, y=0, x=1,$ and $x=3 .$ (a) Sketch the region. (b) Divide the interval $[1,3]$ into $n$ subintervals of equal width $\quad$ and show that the endpoints are $\begin{aligned} 1<1+1\left(\frac{2}{n}\right) &<\cdots<1+(n-1)\left(\frac{2}{n}\right)<1+n\left(\frac{2}{n}\right) \\(\mathrm{c}) \text { Show that } s(n) &=\sum_{i=1}^{n}\left[1+(i-1)\left(\frac{2}{n}\right)\right]\left(\frac{2}{n}\right) \\ \text { (d) Show that } S(n) &=\sum_{i=1}^{n}\left[1+i\left(\frac{2}{n}\right)\right]\left(\frac{2}{n}\right) \end{aligned}$ (e) Find $s(n)$ and $S(n)$ for $n=5,10,50,$ and $100 .$ (f) Show that $\lim _{n \rightarrow \infty} s(n)=\lim _{n \rightarrow \infty} S(n)=4$

Numerical Reasoning Consider a trapezoid of area 4 bounded by the graphs of $y=x, y=0, x=1,$ and $x=3 .$
(a) Sketch the region.
(b) Divide the interval $[1,3]$ into $n$ subintervals of equal width $\quad$ and show that the endpoints are
$\begin{aligned} 1<1+1\left(\frac{2}{n}\right) &<\cdots<1+(n-1)\left(\frac{2}{n}\right)<1+n\left(\frac{2}{n}\right) \\(\mathrm{c}) \text { Show that } s(n) &=\sum_{i=1}^{n}\left[1+(i-1)\left(\frac{2}{n}\right)\right]\left(\frac{2}{n}\right) \\ \text { (d) Show that } S(n) &=\sum_{i=1}^{n}\left[1+i\left(\frac{2}{n}\right)\right]\left(\frac{2}{n}\right) \end{aligned}$
(e) Find $s(n)$ and $S(n)$ for $n=5,10,50,$ and $100 .$
(f) Show that $\lim _{n \rightarrow \infty} s(n)=\lim _{n \rightarrow \infty} S(n)=4$

Key Concepts

Geometric Visualization of Regions

Sketching or visualizing the bounded region is crucial for understanding the area that is being approximated or computed via integration. It involves interpreting graphs of functions and lines to identify the region of interest, which helps in setting up the proper limits of integration and understanding the configuration of the shapes (such as trapezoids) used in the numerical reasoning process.

Definite Integration as the Limit of Riemann Sums

This concept refers to the idea that the definite integral of a function over an interval is the limit of its Riemann sums as the number of subintervals approaches infinity. It provides the rigorous link between the sum of the approximated areas and the actual area under the curve, emphasizing the importance of limits in calculus.

Uniform Partitioning of an Interval

This concept involves dividing a closed interval into a number of subintervals of equal width. It lays the foundation for approximating areas under curves by creating manageable segments over which the function’s behavior is analyzed. In the context of numerical integration, this partitioning allows one to systematically generate endpoints that are used in summing approximations such as Riemann sums.

Riemann Sums – Left and Right Approximations

Riemann sums are a method for approximating the definite integral (or area under a curve) by summing the areas of simple geometric shapes, like rectangles. The left Riemann sum uses the left endpoint of each subinterval, while the right Riemann sum uses the right endpoint. These approximations become more accurate as the subintervals are made finer, reflecting key principles of numerical integration.

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