Question

HELP!!! URGENT!!! For -1 ≤ x ≤, F(x) = integral of the square root of
1 - t^2 dt from -1 to x.

a. What does F(1) represent geometrically?
b. Find F'(x).

Answer 1

\[\int\limits_{-1}^{x}\sqrt{1-t^{2}}dt\]

Answer 2

\[F(x)=\int_{-1}^x\sqrt{1-t^2}~dt\]
So,
\[F(1)=\int_{-1}^1\sqrt{1-t^2}~dt\]
This definite integral gives you the area under the curve \(f(t)=\sqrt{1-t^2}\). What does that give you?