1 Answer

Everlyne · Answer 1 · 2023-10-31T00:44:25+0000

Given:

a)

$\int ^1_0g(x)dx$

Consider the shape included in the region from 0 to 1 of g(x).

The area is,

$\int ^1_0g(x)dx=(1)/(2)*1*4=2$

b) From x = to x = 6 includes the semi-circle. Its area is calculated as,

$\int ^6_2g(x)dx=-(1)/(2)(\pi* r^2)=-(1)/(2)(\pi*2^2)=-2\pi=-6.28$

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