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 .

${\int }_1^2{x}^2\sin \left(5x^2\right)dx$

Short answer

The approximate value is -493.12 .

Step-by-step solution

Step 1. Given information .

Consider the given integral${\int }_1^2{x}^2\sin \left(5x^2\right)dx$.

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

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

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 .

$$\begin{gathered} {\int }_1^2{x}^2\sin \left(5x^2\right)dx={\int }_1^2\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}5x^{4\left(2k+1\right)}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}{\int }_1^{2}5x^{8k+4}dx \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}5{\left[\frac{x^{8k+5}}{8k+5}\right]}_1^2 \\ =\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k+1\right)!}\left[\frac{5·2^{8h+5}}{8k+5}\right] \end{gathered}$$

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

$$\begin{gathered} \Rightarrow 32-525·125+....... \\ \Rightarrow -493.12 \end{gathered}$$