`f(x)=min({x,x^(2)}` and `g(x)=max{x,x^(2)}`

Options

A. `int_(-1)^(1)f(x)=-(1)/(6)`
B. `int_(-2)^(1)g(x)=(19)/(6)`
C. `int_(-1)^(1)f(x)=(19)/(6)`
D. `int_(-1)^(1)g(x)=-(1)/(6)`

Answer

A, B

Explanation

`f(x)=min{x,x^(2)}` `g(x)=max(x,x^(2)}`
`impliesf(x)={{:(x,xle0),(x^(2),00ltxne1),(x,-xgt1):}`
and `{{:(x^(2),xle0),(x,0ltxne1),(x^(2),xgt1):}`
(i). `int_(-1)^(1)f(x)=int_(-1)^(0)xdx+int_(0)^(1)xdx+int_(0)^(1)x^(2)dx`
`=-(1)/(2)+(1)/(3)=-(1)/(6)`
(ii). `int_(-2)^(1)g(x)=int_(-2)^(0)x^(2)dx+int_(0)^(1)xdx`
`=(8)/(3)+(1)/(2)=(19)/(6)`