Calculus problem(Integrals)

Calculus Level 2

Use the first 4 non-zero terms of the power series of tan ⁡ − 1 ( x 2 ) \tan^{-1}(x^2) to approximate ∫ 0 1 tan ⁡ − 1 ( x 2 ) d x \displaystyle \int_0^1 \tan^{-1}(x^2)\ dx .

Clue : tan ⁡ − 1 ( x ) = x − x 3 3 + x 5 5 − x 7 7 + ⋯ \tan^{-1}(x)=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+\cdots

Use 3 significant digits.


The answer is 0.294.

∆Lex Palacin by ∆lex Palacin , 32, Spain

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Henry U
Dec 26, 2018

Using the given Taylor series

∫ 0 1 t a n − 1 ( x 2 ) d x ≈ ∫ 0 1 ( x 2 ) − ( x 2 ) 3 3 + ( x 2 ) 5 5 − ( x 2 ) 7 7 d x = ∫ 0 1 x 2 − x 6 3 + x 10 5 − x 14 7 d x = [ x 3 3 − x 7 21 + x 11 55 − x 15 105 ] 0 1 = [ 1 3 − 1 21 + 1 55 − 1 105 ] − [ 0 3 − 0 21 + 0 55 − 0 105 ] = 68 231 ≈ 0.294 \begin{aligned} \displaystyle \int_0^1 tan^{-1} \left( x^2 \right) \, dx & \approx \int_0^1 \left( x^2 \right) - \frac {\left( x^2 \right)^3}{3} + \frac {\left( x^2 \right)^5}{5} - \frac {\left( x^2 \right)^7}{7} \, dx \\ & = \int_0^1 x^2 - \frac {x^6}{3} + \frac {x^{10}}{5} - \frac {x^{14}}{7} \, dx \\ & = \left[ \frac {x^3}{3} - \frac {x^7}{21} + \frac {x^{11}}{55} - \frac {x^{15}}{105} \right]_0^1 \\ & = \left[ \frac {1}{3} - \frac {1}{21} + \frac {1}{55} - \frac {1}{105} \right] - \left[ \frac {0}{3} - \frac {0}{21} + \frac {0}{55} - \frac {0}{105} \right] \\ & = \frac {68}{231} \approx \boxed{0.294} \end{aligned}

3 Helpful 0 Interesting 1 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...