Question

Use series to approximate the definite integral to

within the indicated accuracy. 47. \(\int\limits_0^1 {xcos({x^3})dx} \)

Short answer

\(\int\limits_0^1 {xcos({x^3})dx = } 0.440\)

Step-by-step solution

Step 1: Using the fundamental theorem of calculus

We know that the series formed will be :

\(\int\limits_0^1 {x\cos ({x^3})dx} = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{x^{6n + 2}}}}{{(6n + 2)(2n)!}} + C} \)

Step 2: Using the fundamental theorem of calculus

Using the fundamental theorem of calculus, we now have:

\(\begin{array}{l}\int\limits_0^1 {x\cos ({x^3})dx} = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\left. {\frac{{{x^{6n + 2}}}}{{(6n + 2)(2n)!}}} \right|_0^1} \\ = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{1^{6n + 2}}}}{{(6n + 2)(2n)!}} - } \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{0^{6n + 2}}}}{{(6n + 2)(2n)!}}} \\ = \sum\limits_{n = 0}^\infty {{{( - 1)}^n}\frac{{{1^{6n + 2}}}}{{(6n + 2)(2n)!}}} \\ = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{(6n + 2)(2n)!}}} \end{array}\)

Step 3: Approximation

Now all we need to approximate the sum to the third decimal place:

\(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{(6n + 2)(2n)!}}} \approx 0.440\)