Q. 64 - Calculus

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.

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

Step-by-Step Solution

Answer

The approximate value is 0.238 .

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 - 1 x 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 $k=0,1,2,3...................\infty$

$$\begin{gathered} \Rightarrow \frac{7}{30}+\frac{0·0142}{3} \\ \Rightarrow 0·2333+0·0047 \\ \Rightarrow 0·238 \end{gathered}$$