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 $\frac{36}{n^2}\left(\frac{n(n+1)}{2}\right)-30$.

Part(b) The approximation for $100$ is $6.435$ and for $1000$ is $6.135$.

Part(c) The exact value is .

Part(a) Step 1. Given Information.

We are given, 

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

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 $x_k=a+k\Delta x$

$$\begin{gathered} \Delta x=\frac{3+3}{n} \\ =\frac{6}{n} \end{gathered}$$

And,

$$\begin{gathered} x_k=-3+k\left(\frac{6}{n}\right) \\ =\frac{6k}{n}-3 \end{gathered}$$

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

The right sum is given by, 

$$\begin{gathered} \sum _{k=1}^n\left(2\left(\frac{6k}{n}-3\right)+1\right)\left(\frac{6}{n}\right)=\left(\frac{6}{n}\right)\sum _{k=1}^n\left(\frac{12k}{n}-5\right) \\ =\frac{6}{n^2}\sum _{k=1}^{n}12k-\frac{6}{n}\sum _{k=1}^{n}5 \\ =\frac{6}{n^2}\left(12\right)\left(\frac{n(n+1)}{2}\right)-\frac{6}{n}\left(5n\right) \\ =\frac{36}{n^2}\left(\frac{n(n+1)}{2}\right)-30 \end{gathered}$$

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

The right sum is,

$\frac{36}{n^2}\left(\frac{n(n+1)}{2}\right)-30$

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

$$\begin{gathered} =\frac{36}{{100}^2}\left(\frac{100(100+1)}{2}\right)-30 \\ =6.435 \end{gathered}$$

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

$$\begin{gathered} =\frac{36}{{1000}^2}\left(\frac{1000(1000+1)}{2}\right)-30 \\ =6.135 \end{gathered}$$

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

The limit is given by,

The exact value is .