## University Calculus: Early Transcendentals (3rd Edition)

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 $| Error| \lt 0.00089$
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: $| Error| \lt \dfrac{1}{33 \cdot 34} \approx 0.00089$