Answer
a) $ f(x) =\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}- ....$
$| Error| \lt 0.00052$
(b) $ f(x) \approx\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}+ ....+(-1)^5\dfrac{x^{32}}{31 \cdot 22}$
and $| Error| \lt 0.00089$
Work Step by Step
a) We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $1$.
$ f(x)=\int_0^{x} \tan^{-1} t dt=\int_0^{x} [t-\dfrac{t^3}{3}+\dfrac{t^{5}}{5}-...] dt $
or, $ f(x) =\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}- ....$
So, the first two terms of the series would give the accuracy.
And the error can be calculated as:
$| Error| \lt \dfrac{(0.5)^6}{30} \approx 0.00052$
(b) We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $1$.
$ f(x)=\int_0^{x} \tan^{-1} t dt=\int_0^{x} [t-\dfrac{t^3}{3}+\dfrac{t^{5}}{5}-...] dt $
or, $ f(x) =\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}- ....$
So, the first 16th term of the series would give the accuracy.
$ f(x) \approx\dfrac{x^2}{2}-\dfrac{x^{4}}{12}+\dfrac{x^{6}}{30}+ ....+(-1)^5\dfrac{x^{32}}{31 \cdot 22}$
And the error can be calculated as:
$| Error| \lt \dfrac{1}{33 \cdot 34} \approx 0.00089$
