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

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The area of a circle with radius 1 is π. If f(x) = 1x2\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? p2(x)=112x2π211(112x2)dx=53 so π103\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. 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}


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