Question

Use the results from Exercises 51–60 and Theorem 7.38 to approximate the values of the definite integrals in Exercises 61–70 to within 0.001 of their values.

${\int }_{-1}^1{e}^{-x^2/3}dx$

Short answer

The approximate value is$-5·978$.

Step-by-step solution

Step 1. Given information .

Consider the given integral${\int }_{-1}^1{e}^{-x^2/3}dx$.

Step 2. Using the result of sin x from question 51 - 60  and theorem 7·38 . 

The result of $e^x=\sum _{k=1}^{\infty }\frac{1}{k!}{x}^k$.

Theorem 7.38 - Let L be the sum of an alternating series satisfying the

hypotheses of the alternating series test. For any term Sn in the sequence of partial sums, $\left|L-S_n\right|<a_{n+1}$. Furthermore, the sign of the difference L − Sn is the sign of the coefficient of the term ..

Step 3. Find the value .

$$\begin{gathered} {\int }_{-1}^1{e}^{-x^2/3}dx={\int }_{-1}^1\sum _{k=0}^{\infty }\frac{{\left(-x^2/3\right)}^k}{k!} \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{3k!}{\int }_{-1}^1\left(x^{2k}\right) \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{3k!}{\left[\frac{x^{2k+1}}{2k+1}\right]}_{-1}^1 \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{3k!}\left(\frac{2}{2k+1}\right) \end{gathered}$$

Substitute $k=0,1,2,3,...........\infty$

$$\begin{gathered} \Rightarrow 2-\frac{1}{9}+\frac{1}{180}-\frac{1}{1270080} \\ \Rightarrow 2-0·111+0·00555-7·873 \\ \Rightarrow -5·978 \end{gathered}$$