Given:
a)
[tex]\int ^1_0g(x)dx[/tex]
Consider the shape included in the region from 0 to 1 of g(x).
The area is,
[tex]\int ^1_0g(x)dx=\frac{1}{2}\times1\times4=2[/tex]
b) From x = to x = 6 includes the semi-circle. Its area is calculated as,
[tex]\int ^6_2g(x)dx=-\frac{1}{2}(\pi\times r^2)=-\frac{1}{2}(\pi\times2^2)=-2\pi=-6.28[/tex]
