Evaluate the integral below by interpreting it in terms of areas ∫1 −1 √ 1^2 − x ^2 dx

Answer 1

If what you need is ∫1-√(1-x²) dx, a graph would be very helpful.

1 - √(1-x²) is the bottom half of a circle [ -√(1-x²) ] that has been moved up 1 unit.

Here is a picture from desmos. This is -√(1-x²)

Then the 1 + [ -√(1-x²) ] moves it up 1, like in the picture below.

And the integral is the area between the curve and the x-axis. This area is shaded in the figure below.

The easiest way to find this area is to find the area of the rectangle from -1< x < 1 and up to y=1. Then subtract off the area of the semi-circle.

Arectangle = 2

Ahalf-circle = 0.5(π)r²

So this integral evaluates to 2-(0.5)(π)(1)² = 2 - 0.5π