Q. 41 - Calculus

Question

For each definite integral in Exercises 41–46, (a) find the general n-rectangle right sum and simplify your answer with sum formulas. Then (b) use your answer to approximate the definite integral with $n = 100$ and $n = 1000$. Finally, (c) take the limit as $n \to \infty$ to find the exact value.

${\int }_2^5(5-x)dx$

Step-by-Step Solution

Answer

Part(a) The right sum is $\frac{3}{n}\left(3n\right)-\frac{9}{n^2}\left(\frac{n(n+1)}{2}\right)$.

Part(b) The approximation for $n=100$ is $4.455$ and for $n=1000$ is $4.4955$.

Part(c) The exact value is $4.5$.

Part(a) Step 1. Given Information.

We are given,

${\int }_2^5(5-x)dx$

Part(a) Step 2. Finding the right sum.

The right sum defined for n rectangles on $[a, b]$ is $\sum _{k=1}^{n}f\left(x_k\right)\Delta x$.

Where, $\Delta x=\frac{b-a}{n},$

and $x_k=a+k\Delta x$

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

$$\begin{gathered} x_k=2+k\left(\frac{3}{n}\right) \\ =2+\frac{3k}{n} \end{gathered}$$

Part(a) Step 3. Finding the right sum.

The right sum is given by,

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

Part(b) Step 1. Approximating the definite integral.

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,

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

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

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

Part(b) Step 2. Checking the approximation

The graph is as follows,

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

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

Hence the approximation is correct.

Part(c) Step 1. Finding the exact value.

The limit is given by,

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

The exact value is  $4.5$.