TITLE: Area of the unit circle using integrals
QUESTION [2 upvotes]: In Calculus by Michael Spivak, the area of the unit circle is explored using integrals:

If $0 \le x \le 1$, this area can be expressed as the sum of the area
  of a triangle and the area of a region under the unit circle:
$\dfrac{x\sqrt{1 - x^2}}{2} + \int_{x}^{1} (\sqrt{1 - t^2}) dt$

I am struggling to understand the expression $\dfrac{x\sqrt{1 - x^2}}{2} + \int_{x}^{1} (1 - t^2) dt$.
As I understand it, the term $\sqrt{1 - x^2}$ is a semicircle, and the term $\dfrac{x}{2}$ in $\dfrac{x\sqrt{1 - x^2}}{2}$ is the area of a sector of the unit circle. I also understand the $\int_{x}^{1} (\sqrt{1 - t^2}) dt$ term.
But why do we have the term $\dfrac{x\sqrt{1 - x^2}}{2}$? How does it contribute to, as the author says, finding the area of a unit circle if $0 \le x \le 1$?
I would greatly appreciate it if people could please take the time to break this expression down for me and clarify its components.

REPLY [5 votes]: "This area" is seen here:

It is a sector of a unit circle, the top corner has a given $x$ coordinate.
The unit circle has equation $x^2+y^2=1$.  The top half of the circle has equation $y = \sqrt{1-x^2}$.
The green triangle has area
$$
\frac{x\sqrt{1-x^2}}{2}
$$
(base times height divided by $2$).  And the blue region has area
$$
\int_x^1 \sqrt{1-t^2}\;dt
$$
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