# Solve sqrt(16-x^2) between -4 and 0

Solve $\sqrt{16-{x}^{2}}$ between -4 and 0
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Definite integral of $\sqrt{16-{x}^{2}}$ between -4 and 0
Okay, you are looking at half of a circle.
Diameter is 4. So radius is 2.
You are finding an area, the formula is $A=\pi \cdot {r}^{2}$
$A=\pi {2}^{2}=4\pi$