The area of a circle with radius 1 is π. If f(x) = $\sqrt { 1 - x ^ { 2 } }$ gives the top half of this circle, as illustrated below, use the second Taylor polynomial of f(x) at x = 0 to find an approximate value for π. Is the following correct? $$\begin{array} { l }
\mathrm { p } 2 ( \mathrm { x } ) = 1 - \frac { 1 } { 2 } \mathrm { x } ^ { 2 } \
\frac { \pi } { 2 } \approx \int _ { - 1 } ^ { 1 } \left( 1 - \frac { 1 } { 2 } \mathrm { x } ^ { 2 } \right) \mathrm { dx } = \frac { 5 } { 3 } \text { so } \pi \approx \frac { 10 } { 3 }
\end{array}$$

Question

The area of a circle with radius 1 is π. If f(x) = $\sqrt { 1 - x ^ { 2 } }$ gives the top half of this circle, as illustrated below, use the second Taylor polynomial of f(x) at x = 0 to find an approximate value for π. Is the following correct? $$\begin{array} { l } 
\mathrm { p } 2 ( \mathrm { x } ) = 1 - \frac { 1 } { 2 } \mathrm { x } ^ { 2 } \
\frac { \pi } { 2 } \approx \int _ { - 1 } ^ { 1 } \left( 1 - \frac { 1 } { 2 } \mathrm { x } ^ { 2 } \right) \mathrm { dx } = \frac { 5 } { 3 } \text { so } \pi \approx \frac { 10 } { 3 }
\end{array}$$

Quizplus · Accepted Answer

The Answer of The area of a circle with radius 1 is π. If f(x) = $\sqrt { 1 - x ^ { 2 } }$ gives the top half of this circle, as illustrated below, use the second Taylor polynomial of f(x) at x = 0 to find an approximate value for π. Is the following correct? $$\begin{array} { l } 
\mathrm { p } 2 ( \mathrm { x } ) = 1 - \frac { 1 } { 2 } \mathrm { x } ^ { 2 } \
\frac { \pi } { 2 } \approx \int _ { - 1 } ^ { 1 } \left( 1 - \frac { 1 } { 2 } \mathrm { x } ^ { 2 } \right) \mathrm { dx } = \frac { 5 } { 3 } \text { so } \pi \approx \frac { 10 } { 3 }
\end{array}$$

The Area of a Circle with Radius 1 Is π 1−x2\sqrt { 1 - x ^ { 2 } }1−x2​

The Area of a Circle with Radius 1 Is π $\sqrt { 1 - x ^ { 2 } }$