Question

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

Short answer

The definite integral's six decimal places are\(\int_0^{0.4} {\ln } \left( {1 + {x^4}} \right)dx = 0.002034\).

Step-by-step solution

Differentiate the function with regard to \(x\).

Let \(f(x) = \ln \left( {1 + {x^4}} \right)\).

\(\begin{aligned}d(f(x)) &= \frac{1}{{1 + {x^4}}} \cdot 4{x^3}\quad \\d(f(x)) &= 4{x^3} \cdot \frac{1}{{1 - \left( { - {x^4}} \right)}}\end{aligned}\)

\(\begin{aligned}4{x^3} \cdot \frac{1}{{1 - \left( { - {x^4}} \right)}} &= 4{x^3} \cdot \sum\limits_{n = 0}^\infty 1 \cdot {\left( { - {x^4}} \right)^n}\\4{x^3} \cdot \frac{1}{{1 - \left( { - {x^4}} \right)}} &= 4{x^3} \cdot \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} {x^{4n}}\\4{x^3} \cdot \frac{1}{{1 - \left( { - {x^4}} \right)}} &= \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} 4{x^{4n + 3}}\end{aligned}\)

Integrate the function.

\(\begin{aligned}\ln \left( {1 + {x^4}} \right) &= 4\int {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} } {x^{4n + 3}}dx\\\ln \left( {1 + {x^4}} \right) &= 4{( - 1)^n}\left( {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{4n + 4}}}}{{4n + 4}}} \right) + C\\\ln \left( {1 + {x^4}} \right) &= 4\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{{(x)}^{4n + 4}}}}{{4n + 4}} + C\\\ln \left( {1 + {x^4}} \right) &= \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{{(x)}^{4n + 4}}}}{{n + 1}}} + C\end{aligned}\)

Substitute \(x = 0\) to find the value of \({\rm{C}}\) in the function \(\ln \left( {1 + {x^4}} \right)\).

\(\begin{aligned}\ln \left( {1 + {{(0)}^4}} \right) &= \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{{(0)}^{4n + 4}}}}{{n + 1}} + C\ln (1 + 0)\\\ln \left( {1 + {{(0)}^4}} \right) &= 0 + C0\\\ln \left( {1 + {{(0)}^4}} \right) &= 0 + CC\\\ln \left( {1 + {{(0)}^4}} \right) &= 0\end{aligned}\)

Substitute the value and integrate the function.

Substitute \(C = 0\) in the above function.

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

Integrate the function.

\(\begin{aligned}\int_0^{0.4} {\ln } \left( {1 + {x^4}} \right)dx &= \int_0^{0.4} {\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{{(x)}^{4n + 4}}}}{{n + 1}}} } dx\\\int_0^{0.4} {\ln } \left( {1 + {x^4}} \right)dx &= \left[ {\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{4n + 5}}}}{{(n + 1)(4n + 5)}}} \right]_0^{0.4}\\\int_0^{0.4} {\ln } \left( {1 + {x^4}} \right)dx &= \sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{{(0.4)}^{4n + 5}}}}{{(n + 1)(4n + 5)}}\end{aligned}\)

Simplify further,

\(\begin{aligned}\int_0^{0.4} {\ln } \left( {1 + {x^4}} \right)dx &= \frac{{{{( - 1)}^0}{{(0.4)}^5}}}{{1 \cdot 5}} + \frac{{{{( - 1)}^1}{{(0.4)}^9}}}{{2 \cdot 9}} + \frac{{{{( - 1)}^2}{{(0.4)}^{13}}}}{{3 \cdot 13}} + \cdots \\\int_0^{0.4} {\ln } \left( {1 + {x^4}} \right)dx &= \frac{{{{(0.4)}^5}}}{5} - \frac{{{{(0.4)}^9}}}{{18}} + \frac{{{{(0.4)}^{13}}}}{{39}} - \cdots \\\int_0^{0.4} {\ln } \left( {1 + {x^4}} \right)dx &= 0.002034\end{aligned}\)

Calculate the three terms of the summation.

since the fourth term is smaller amount than 0.000001

Therefore, the definite integral's six decimal places are\(\int_0^{0.4} {\ln } \left( {1 + {x^4}} \right)dx = 0.002034\).