Q. 17 - Calculus

Question

Consider the definite integral ${\int }_0^3{x}^{2}dx$.

1. Write down an $n$-rectangle right sum for ${\int }_0^3{x}^{2}dx$,and use algebra and a sum formula to write this sum as a formula in terms of $n$.
2. Write down an n-rectangle left sum for ${\int }_0^3{x}^{2}dx$, and use algebra and a sum formula to write this sum as a formula in terms of $n$.
3. Use your answers to $\left(a\right)$ and $\left(b\right)$ to show that the right

   sum and the left sum for ${\int }_0^3{x}^{2}dx$are different for $n=100$ and $n=1000$.

4. Use your answers to $\left(a\right)$ and $\left(b\right)$ to show that the right sum and the left sum for ${\int }_0^3{x}^{2}dx$ approach the same quantity as $n\to \infty$.What does this quantity represent geometrically?

Step-by-Step Solution

Answer

Part $\left(a\right)$: The right sum is, $\frac{27}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)$.

Part $\left(b\right)$: The right sum is, $\frac{27}{n^3}\left[\left(\frac{n(n+1)(2n+1)}{6}\right)-2\frac{n(n+1)}{2}+n\right]$.

Part $\left(c\right)$: The left sum and right sum are different for $n=100,n=1000$ is showed.

Part $\left(d\right)$: The left sum and right sum approach the same quantity as $n\to \infty$ is showed.

Part a Step 1 . Given information

${\int }_0^3{x}^{2}dx$.

Part a Step 2 . The right sum defined for n rectangles on a , b is, ∑ k = 1 n f x k Δ x .

where $\Delta x=\frac{b-a}{n},x_k=a+k\Delta x$.

The interval is, $\left[0,3\right]$.

Now,

$$\begin{gathered} \Delta x=\frac{3-0}{n} \\ =\frac{3}{n} \end{gathered}$$

And,

$$\begin{gathered} x_k=0+k\left(\frac{3}{n}\right) \\ =\frac{3k}{n} \end{gathered}$$

Part a Step 3 . The right sum is,

$$\begin{gathered} \sum _{k=1}^n{\left(\frac{3k}{n}\right)}^2\left(\frac{3}{n}\right)=\frac{27}{n^3}\sum _{k=1}^n\left(k^2\right) \\ =\frac{27}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right) \end{gathered}$$

Therefore, the right sum is, $\frac{27}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)$.

Part b Step 1 . The left sum defined for n rectangles on a , b is, ∑ k = 1 n f x k Δ x .

where $\Delta x=\frac{b-a}{n},x_k=a+k\Delta x$.

The interval is, $\left[0,3\right]$.

$$\begin{gathered} ∆x=\frac{3-0}{n} \\ =\frac{3}{n} \end{gathered}$$

And,

$$\begin{gathered} x_k=0+k\left(\frac{3}{n}\right) \\ =\frac{3k}{n} \\ {x}_{k-1}=\frac{3(k-1)}{n} \end{gathered}$$

Part b Step 2 . The left sum is,

$$\begin{gathered} \sum _{k=1}^n{\left(\frac{3(k-1)}{n}\right)}^2\left(\frac{3}{n}\right)=\frac{27}{n^3}\sum _{k=1}^n(k-1{)}^2 \\ =\frac{27}{n^3}\left(\sum _{k=1}^n{k}^2-2\sum _{k=1}^{n}k+\sum _{k=1}^{n}1\right) \\ =\frac{27}{n^3}\left[\left(\frac{n(n+1)(2n+1)}{6}\right)-2\frac{n(n+1)}{2}+n\right] \end{gathered}$$

Therefore, the left sum is, $\frac{27}{n^3}\left[\left(\frac{n(n+1)(2n+1)}{6}\right)-2\frac{n(n+1)}{2}+n\right]$.

Part c Step 1 . The objective is to show that the left sum and right sum are different for n = 100 , n = 1000 .

The right sum is,

$\frac{27}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)$

For $n=100$,its sum is,

$\frac{27}{{100}^3}\left(\frac{100(100+1)(200+1)}{6}\right)\approx 9.1345$

For $n=1000$,its sum is,

$\frac{27}{{1000}^3}\left(\frac{1000(1000+1)(2000+1)}{6}\right)\approx 9.0135$

Part c Step 2 . The left sum is,

$\frac{27}{n^3}\left[\left(\frac{n(n+1)(2n+1)}{6}\right)-2\frac{n(n+1)}{2}+n\right]$

For $n=100$,its sum is,

$\frac{27}{{100}^3}\left[\left(\frac{100(100+1)(200+1)}{6}\right)-2\frac{100(100+1)}{2}+100\right]\approx 8.8645$

For $n=1000$, its sum is,

$\frac{27}{{1000}^3}\left[\left(\frac{1000(1000+1)(2000+1)}{6}\right)-2\frac{1000(1000+1)}{2}+100\right]\approx 8.9865$

Therefore, the left sum and the right sum are different for $n=100,n=1000$.

Part d Step 1 . The right sum is,

$\frac{27}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)$

As $n\to \infty$,its sum is,

$$\begin{gathered} \underset{n\to \infty }{\lim }\frac{27}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)=\frac{54}{6} \\ =9 \end{gathered}$$

The left sum is,

$\frac{27}{n^3}\left[\left(\frac{n(n+1)(2n+1)}{6}\right)-2\frac{n(n+1)}{2}+n\right]$

As $n\to \infty$,its sum is,

$$\begin{gathered} \underset{n\to \infty }{\lim }\frac{27}{n^3}\left[\left(\frac{n(n+1)(2n+1)}{6}\right)-2\frac{n(n+1)}{2}+n\right]=\frac{54}{6}-\frac{27\left(2\right)}{2}+27 \\ =9+0 \\ =9 \end{gathered}$$

Therefore, the left sum and the right sum are the same quantity as $n\to \infty$.

This quantity represent geometrically that for the left and the right sum, the sum is $9$.