Question

Use series to approximate the definite integral to within the indicated accuracy.

\(\int_0^1 {sin} \left( {{x^4}} \right)dx\)Four decimal places

Short answer

The approximate value of the integral is about \(0.1876\)

Step-by-step solution

Step 1: Given information

The given value is:

\(\int_0^1 {sin} \left( {{x^4}} \right)dx\)

Step 2: Convert to series form

Using the series for\(\sin x,\)convert to series form.

\(\sin x = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{2n + 1}}}}{{(2n + 1)!}} = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \frac{{{x^7}}}{{7!}} + \cdots \)\(\sin \left( {{x^4}} \right) = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{{\left( {{x^4}} \right)}^{2n + 1}}}}{{(2n + 1)!}} = {x^4} - \frac{{{{\left( {{x^4}} \right)}^3}}}{{3!}} + \frac{{{{\left( {{x^4}} \right)}^5}}}{{5!}} - \frac{{{{\left( {{x^4}} \right)}^7}}}{{7!}} + \cdots \)\( = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{8n + 4}}}}{{(2n + 1)!}} = {x^4} - \frac{{{x^{12}}}}{{3!}} + \frac{{{x^{20}}}}{{5!}} - \frac{{{x^{28}}}}{{7!}} + \cdots \)

Evaluate the Integrate

\(\int_0^1 {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} } \frac{{{x^{8n + 4}}}}{{(2n + 1)!}}dx = \left[ {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{8n + 5}}}}{{(8n + 5)(2n + 1)!}}} \right]_0^1\)

This is an alternating series so we look for the term that is\( < 0.0001\)(at\(x = 1\)since that makes the largest value). We can start at\(n = 1\)

\(\begin{array}{l}(n = 1):\;\;\;\frac{1}{{(8(1) + 5)(2(1) + 1)!}}\\ \approx 0.013\end{array}\)\(\begin{array}{l}(n = 2):\;\;\;\frac{1}{{(8(2) + 5)(2(2) + 1)!}}\\ \approx 0.0004\end{array}\)\(\begin{array}{l}(n = 3):\;\;\;\frac{1}{{(8(3) + 5)(2(3) + 1)!}}\\ \approx 0.0000068 < 0.0001\end{array}\)

This means we only need up to\(n = 2\)(3 terms).

Write 3 terms

\(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{8n + 5}}}}{{(8n + 5)(2n + 1)!}} \approx \frac{{{x^5}}}{5} - \frac{{{x^{13}}}}{{13(3!)}} + \frac{{{x^{21}}}}{{21(5!)}}\)\(\int_0^1 {\sin } \left( {{x^4}} \right)dx \approx \left[ {\frac{{{x^5}}}{5} - \frac{{{x^{13}}}}{{13(3!)}} + \frac{{{x^{21}}}}{{21(5!)}}} \right]_0^1\)\( \approx \frac{1}{5} - \frac{1}{{13(3!)}} + \frac{1}{{21(5!)}} - 0\)\( \approx 0.187576\)

Rounded to 4 decimal places, it would be\(0.1876\)

The approximate value of the integral is about \(0.1876\)