Sketch the region bounded by the curves `y=sqrt(5-x^2)` and `y=|x-1|` and find its area.

Options

A. `(5pi)/(4)-2`
B. `(5pi-2)/(4)`
C. `(5pi-2)/(2)`
D. `(pi)/(2)-5`

Answer

B

Explanation

The graphs of `y=|x-1|and y=sqrt(5-x^(2))` are showing in Fig. 15 and shaded region is the region bounded by th two curves.

Let A be the area bonded by the given curves. Then.
`A=underset(-1)overset(1)(int)(y_(2)-y_(1))dx+underset(1)overset(2)(int)(y_(4)-y_(3))dx`
`impliesA=underset(-1)overset(1)(int)(sqrt(5-x^(2))+x-1)dx+underset(1)overset(2)(int)(sqrt(5-x^(2))-x+1)dx`
`impliesAunderset(-1)overset(1)(int)sqrt(5-x^(2))dx+underset(1)overset(2)(int)sqrt(5-x^(2))dx+underset(-1)overset(1)(int)(x-1)dx+underset(1)overset(2)(int)(-x+1)dx`
`impliesA=underset(-1)overset(2)(int)sqrt(5-x^(2))dx+[(x^(2))/(2)-x]_(-1)^(1)+[-(x^(2))/(2)+x]_(1)^(2)`
`impliesA=[(1)/(2)xsqrt(5-x^(2))+5/2sin^(-1)""(x)/(sqrt2)]_(-1)^(2)-5/2`
`impliesA=1+5/2sin^(-1)""(2)/(sqrt5)+1+5/2sin^(-1)""(1)/(sqrt5)-5/2`
`impliesA=-1/2+5/2sin^(-1)((2)/(sqrt5)xxsqrt(1-1/5)+(1)/(sqrt5)sqrt(1-(4)/(5)))`
`A=-1/2+5/2sin^(-1)(1)=(5pi)/(4)-1/2=(5pi-2)/(4)`