Q. 43 - 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.

${\int }_0^{1}2x^{2}dx$

Step-by-Step Solution

Answer

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

Part(b) The approximation for $n=100$ is $0.6767$ and for $n=1000$ is $0.667667$.

Part(c) The exact value is $\frac{2}{3}$.

Part(a) Step 1. Given Information.

We are given, 

${\int }_0^{1}2x^{2}dx$

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 $x_k=a+k\Delta x$

$$\begin{gathered} \Delta x=\frac{1-0}{n} \\ =\frac{1}{n} \end{gathered}$$

And.

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

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

The right sum is given by,

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

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

The right sum is,

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

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

$\frac{2}{{100}^3}\left(\frac{100(100+1)(200+1)}{6}\right)=0.6767$

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

$\frac{2}{{1000}^3}\left(\frac{1000(1000+1)(2000+1)}{6}\right)=0.667667$

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

The limit is given by,

$$\begin{gathered} {\int }_0^{1}2x^{2}dx=\underset{n\to \infty }{\lim }\left(\frac{2}{n^3}\left(\frac{n(n+1)(2n+1)}{6}\right)\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{1}{n^3}\left(\frac{\left(n^2+n\right)(2n+1)}{3}\right)\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{2n^3+3n^2+n}{3n^3}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{2+\frac{3}{n}+\frac{n}{n^2}}{3}\right) \\ =\frac{2}{3} \end{gathered}$$

The exact value is $\frac{2}{3}$.