Question

Find the definite integral to six decimal places using the power series.

Short answer

The definite integral's six decimal place are\(\int_0^{0.3} {\frac{{{x^2}}}{{1 + {x^4}}}} dx = 0.008969.\)

Step-by-step solution

Concept of Geometric series.

The sum of the geometric series with initial term \(a\) and common ratio \(r\) is \(S = \sum\limits_{n = 0}^\infty a {r^n} = \frac{a}{{1 - r}}\).

Calculate and modify the definite integral.

The definite integral is \(\int_0^{0.3} {\frac{{{x^2}}}{{1 + {x^4}}}} dx\)

Let \(f(x) = \int_0^{0.3} {\frac{{{x^2}}}{{1 + {x^4}}}} dx\)

Modify the function

\(\begin{aligned}f(x) &= \int_0^{0.3} {\frac{{{x^2}}}{{1 + {x^4}}}} dx\\f(x) &= {x^2}\int_0^{0.3} {\frac{1}{{1 + {x^4}}}} dx\\f(x) &= {x^2}\int_0^{0.3} {\frac{1}{{1 - \left( { - {x^4}} \right)}}} dx\end{aligned}\)

Show the function by using result (1)

\(\begin{aligned}f(x) &= {x^2}\int_0^{0.3} {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} } \left( {{x^{4n}}} \right)dx\\f(x) &= \int_0^{0.3} {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} } \left( {{x^{4n + 2}}} \right)dx\\f(x) &= \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \left( {\frac{{{x^{4n + 3}}}}{{4n + 3}}} \right)_0^{0.3}\end{aligned}\)

Apply the limit and simplify further.

\(f(x) = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \left( {\frac{{{x^{4n + 3}}}}{{4n + 3}}} \right)_0^{0.3}\)

\(f(x) = \left[ {\begin{aligned}{\left( {\frac{{{{( - 1)}^0}{{(0.3)}^{4(0) + 3}}}}{{4(0) + 3}} - \frac{{{{( - 1)}^0}{{(0)}^{4(0) + 3}}}}{{4(0) + 3}}} \right)}\\{ + \left( {\frac{{{{( - 1)}^1}{{(0.3)}^{4(0) + 3}}}}{{4(1) + 3}} - \frac{{{{( - 1)}^1}{{(0)}^{4(1) + 3}}}}{{4(1) + 3}}} \right)}\\{ + \left( {\frac{{{{( - 1)}^2}{{(0.3)}^{4(2) + 3}}}}{{4(2) + 3}} - \frac{{{{( - 1)}^2}{{(0)}^{4(2) + 3}}}}{{4(2) + 3}}} \right) + \cdots }\end{aligned}} \right]\)

\(f(x) = \left( {\frac{{{{(0.3)}^3}}}{3} - \frac{{{{(0.3)}^7}}}{7} + \frac{{{{(0.3)}^{11}}}}{{11}} - \cdots } \right)\)

\(f(x) = \left( {\frac{{{{(0.3)}^3}}}{3} - \frac{{{{(0.3)}^7}}}{7} + \frac{{{{(0.3)}^{11}}}}{{11}} - \cdots } \right)\)

\(f(x) = (0.009 - 0.000031 + 0.000000)\)

\(f(x) = 0.008969\)

Calculate the two terms of the summation.

Due to the fact that the fourth term is smaller than 0.000001

Therefore, the definite integral's six decimal place are \(\int_0^{0.3} {\frac{{{x^2}}}{{1 + {x^4}}}} dx = 0.008969\).