Given the Definite Integral:
[tex]\int_0^1\sqrt{1-x^2}dx[/tex]
You can identify that the interval is:
[tex]\lbrack0,1\rbrack[/tex]
By definition, if the function is continuous and positive in a closed interval, then:
[tex]\int_a^bf(x)dx=Area[/tex]
In this case, you can identify that the function is:
[tex]y=\sqrt{1-x^2}[/tex]
You can graph it using a graphic tool:
Since the closed interval goes from 0 to 1, you need to find this area:
You can identify that you have to find the area of a quarter circle. In order to do it, you can use this formula:
[tex]A=\frac{\pi r^2}{4}[/tex]
Where "r" is the radius of the circle.
In this case, you can identify that:
[tex]r=1[/tex]
Therefore, you get:
[tex]A=\frac{\pi(1)^2}{4}=\frac{\pi}{4}[/tex]
Then:
[tex]\int_0^1\sqrt{1-x^2}dx=\frac{\pi}{4}[/tex]
Hence, the answer is: Option D.
