Use the sixth-degree Taylor polynomial centered at zero for the function $f ( x ) = \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$ to approximate the integral $\int _ { 0 } ^ { 2 / 3 } \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$ . Round your answer to nearest ten thousandth.
A)0.6222
B)0.6322
C)0.8772
D)0.6272
E)0.3772

Question

Quizplus · Accepted Answer

To approximate the integral $\int _ { 0 } ^ { 2 / 3 } \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$ using the sixth-degree Taylor polynomial centered at zero for the function $f ( x ) = \frac { 1 } { \sqrt { 1 + x ^ { 2 } } }$, we first need to find the Taylor polynomial. The Taylor series of $f(x)$ around 0 is given by $f(x) = 1 - \frac{1}{2}x^2 + \frac{3}{8}x^4 - \frac{5}{16}x^6 + \cdots$. The sixth-degree Taylor polynomial is $P_6(x) = 1 - \frac{1}{2}x^2 + \frac{3}{8}x^4 - \frac{5}{16}x^6$. Integrating $P_6(x)$ from 0 to $2/3$ gives the approximation of the integral. The integral of $P_6(x)$ is $\int_0^{2/3} (1 - \frac{1}{2}x^2 + \frac{3}{8}x^4 - \frac{5}{16}x^6) dx$, which evaluates to approximately 0.6272 when rounded to the nearest ten thousandth.

Use the Sixth-Degree Taylor Polynomial Centered at Zero for the Function