91Ó°ÊÓ

We are given,

${∫}_2^5(5-x)dx$

The right sum defined for n rectangles on $[a,b]$is $\underset{k=1}{\overset{n}{∑}}f\left(x_k\right)\mathrm{Δ}x$.

Where, $\mathrm{Δ}x=\frac{b-a}{n},$

and $x_k=a+k\mathrm{Δ}x$

$$\begin{gathered} \mathrm{Δ}x=\frac{5-2}{n} \\ \mathrm{Δ}x=\frac{3}{n} \end{gathered}$$

role="math" localid="1648648381261" $$\begin{gathered} x_k=2+k\left(\frac{3}{n}\right) \\ =2+\frac{3k}{n} \end{gathered}$$

The right sum is given by,

$$\begin{gathered} \underset{k=1}{\overset{n}{∑}}\left(5-\left(2-\frac{3k}{n}\right)\right)\left(\frac{3}{n}\right)=\left(\frac{3}{n}\right)\underset{k=1}{\overset{n}{∑}}\left(3-\frac{3k}{n}\right) \\ =\frac{3}{n}\underset{k=1}{\overset{n}{∑}}3-\frac{3}{n}\underset{k=1}{\overset{n}{∑}}\frac{3k}{n} \\ =\frac{3}{n}\left(3n\right)-\left(\frac{3}{n}\right)\left(\frac{3}{n}\right)\left(\frac{n(n+1)}{2}\right) \\ =\frac{3}{n}\left(3n\right)-\frac{9}{n^2}\left(\frac{n(n+1)}{2}\right) \end{gathered}$$

The right sum is,

$\frac{3}{n}\left(3n\right)-\frac{9}{n^2}\left(\frac{n(n+1)}{2}\right)$

For, $n=100$the approximation will be,

$\frac{3}{100}(3×100)-\frac{9}{(100{)}^2}\left(\frac{100(100+1)}{2}\right)=4.455$

For, $n=1000$the approximation will be,

$\frac{3}{1000}(3×1000)-\frac{9}{(1000{)}^2}\left(\frac{1000(1000+1)}{2}\right)=4.4955$

The graph is as follows,

From the graph it can be seen that the area is given by,

$$\begin{gathered} A=\frac{1}{2}×3×3 \\ A=4.5 \end{gathered}$$

Hence the approximation is correct.

The limit is given by,

$$\begin{gathered} {∫}_2^5(5-x)dx=\underset{n→∞}{\lim }\left(\frac{3}{n}\left(3n\right)-\frac{9}{n^2}\left(\frac{n(n+1)}{2}\right)\right) \\ =\underset{n→∞}{\lim }\left(9-\frac{9}{n^2}\left(\frac{n^2+n}{2}\right)\right) \\ =\underset{n→∞}{\lim }\left(\frac{18-9n^2-9n}{2n^2}\right) \\ =\frac{9}{2} \\ =4.5 \end{gathered}$$

The exact value is $4.5$.