Given:
[tex]F(x)=\int_0^x\sqrt{36-t^2}dt[/tex]
Required:
To find the range of the given function.
Explanation:
The graph of the function
[tex]y=\sqrt{36-t^2}[/tex]
is upper semicircle with center (0,0) and radius 6, with
[tex]-6\leq t\leq6[/tex]
So,
[tex]\int_0^x\sqrt{36-t^2}dt[/tex]
is the area of the portion of the right half of the semicircle that lies between
t=0 and t=x.
When x=0, the value of the integral is also 0.
When x=6, the value of the integral is the area of the quarter circle, which is
[tex]\frac{36\pi}{4}=9\pi[/tex]
Therefore, the range is
[tex][0,9\pi][/tex]
Final Answer:
The range of the function is,
[tex][0,9\pi][/tex]
