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.

${\int }_{0·1}^{0·2}{\tan }^{-1}\left(x^2\right)dx$

Short answer

The approximate value is 0.238 .

Step-by-step solution

Step 1. Given information .

Consider the given integral${\int }_{0·1}^{0·2}\tan {}^{-1}\left(x^2\right)dx$.

Step 2. Using the result of tan-1x from question 51 - 60  and theorem 7.38 . 

The result of $\tan {}^{-1}\left(x\right)=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{2k+1}$.

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,. Furthermore, the sign of the difference L − Sn is the sign of the coefficient of the term .

Step 3. Find the value .

$$\begin{gathered} {\int }_{0·1}^{0·2}{\tan }^{-1}\left(x^2\right)dx={\int }_{0·1}^{0·2}\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{x}^{2\left(2k+1\right)}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{\int }_{0·1}^{0·2}{x}^{4k+2}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}{\left[\frac{x^{4k+3}}{4k+3}\right]}_{0·1}^{0·2} \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{2k+1}\left[\frac{0·2^{4k+3}-0·1^{4k+3}}{4k+3}\right] \end{gathered}$$

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

role="math" localid="1649681729809" $$\begin{gathered} \Rightarrow \frac{7}{30}+\frac{0·0142}{3} \\ \Rightarrow 0·2333+0·0047 \\ \Rightarrow 0·238 \end{gathered}$$