Q. 46 - 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.

${\int }_{-1}^2\left(1-x^2\right)dx$

Step-by-Step Solution

Answer

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

Part (b) The right sums are $0.00 \& 0.00$

Part (c) The exact value is zero

Step 1. Given information

An expression is given as ${\int }_{-1}^2\left(1-x^2\right)dx$

Part (a) Step 1. General n-rectangle sum

The limits given are $[a, b]=[-1,2]$

Therefore,

$$\begin{gathered} \Delta x=\frac{h-a}{n}=\frac{2+1}{n}=\frac{3}{n} \\ {x}_k=a+k\Delta x \\ =-1+k(3 / n) \\ =-1+(3 k / n) \end{gathered}$$

Now the right sum is,

$$\begin{gathered} =\sum _{k=1}^{n}f\left(x_k\right)\Delta x \\ =\sum _{k=1}^n\left(1-x_k^2\right)\Delta x \\ =\sum _{k=1}^n\left(1-(-1+(3k/n){)}^2\right)(3/n) \\ =\frac{3}{n}\sum _{k=1}^n\left(1-1+\frac{9k^2}{n^2}-\frac{6k}{n}\right) \\ =\frac{3}{n}\sum _{k=1}^n\left(\frac{9k^2}{n^2}-\frac{6k}{n}\right) \\ =\frac{3}{n}\left(-\frac{6n(n+1)}{2n}+\frac{9n(n+1)(2n+1)}{6n^2}\right) \\ =\frac{9(n+1)(2n+1)}{2n^2}-\frac{9(n+1)}{n} \end{gathered}$$

Part (b) Step 1. Approximate integral

The right sum is,

$$\begin{gathered} \frac{9(n+1)(2n+1)}{2n^2}-\frac{9(n+1)}{n} \\ \Rightarrow n=100\Rightarrow =0.45 \\ \Rightarrow n=1000\Rightarrow =0.045 \end{gathered}$$

For both values of n we get zero.

Part (c) Step 1. Exact value

Put the limits in the right sum as,

$$\begin{gathered} =\frac{9(n+1)(2n+1)}{2n^2}-\frac{9(n+1)}{n} \\ \underset{n\to \infty }{\lim }{\int }_{-1}^2\left(1-x^2\right)dx=\underset{n\to w}{\lim }\left(\frac{9(n+1)(2n+1)}{2n^2}-\frac{9(n+1)}{n}\right) \\ =9-9=0 \end{gathered}$$