The area of the closed figure bounded by `y=(x^2)/2-2x+2` and the tangents to it at `(1,1/2)a n d(4,2)` is `9/8s qdotu n i t s` (b) `3/8s qdotu n i t s` `3/2s qdotu n i t s` (d) `9/4s qdotu n i t s`

Options

A. `9//8` sq. units
B. `3//8` sq. units
C. `3//2` sq. units
D. `9//4` sq. units

Answer

A

Explanation

`y=(x^(2))/(2)-2x+2=((x-2)^(2))/(2),`
`(dy)/(dx)=x-2, ((dy)/(dx))_(x=1)=-1, ((dy)/(dx))_(x=4)=2`
Thus, tangent at `(1,1//2)" is "y-1//2=-1(x-1)or 2x+2y-3=0`
Tangent at `(4,2)" is "y-2=2(x-4)or 2x-y-6=0`

`"Hence, "A=overset(5//2)underset(1)int((x^(2))/(2)-2x+2-(3-2x)/(2))dx+overset(4)underset(5//2)int((x^(2))/(2)-2x+2-(2x-6))dx`
`=overset(4)underset(1)int((x^(2))/(2)-2x+2)dx-overset(5//2)underset(1)int((3-2x)/(2))dx-overset(4)underset(5//2)int(2x-6)dx`
`=((x^(3))/(6)-x^(2)+2x)_(1)^(4)-(1)/(2)(3x-x^(2))_(1)^(5//2)-(x^(2)-6x)_(5//2)^(4)`
`=((63)/(3)-15+6)-(1)/(2)(3xx(3)/(2)-((25)/(4)-1))-((16-(25)/(4))-6(4-(5)/(2)))`
`=(3)/(2)-(1)/(2)((9)/(2)-(21)/(4))-((39)/(4)-6((3)/(2)))`
`=(9)/(8)` sq. units