For $-2 \leq x \leq 2$ , define $F(x)=\int_{-2}^{x} \sqrt{4-t^{2}} d t$ .What is the value of $F(0)$ ? Round to 2 decimal places.

Name: For $-2 \leq x \leq 2$ , define $F(x)=\int_{-2}^{x} \sqrt{4-t^{2}} d t$ .What is the value of $F(0)$ ? Round to 2 decimal places.
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The Answer of For $-2 \leq x \leq 2$ , define $F(x)=\int_{-2}^{x} \sqrt{4-t^{2}} d t$ .What is the value of $F(0)$ ? Round to 2 decimal places.

For −2≤x≤2-2 \leq x \leq 2−2≤x≤2 , Define F(x)=∫−2x4−t2dtF(x)=\int_{-2}^{x} \sqrt{4-t^{2}} d tF(x)=∫−2x​4−t2​dt

For $-2 \leq x \leq 2$ , Define $F(x)=\int_{-2}^{x} \sqrt{4-t^{2}} d t$