The area of the closed figure bounded by `x=-1, y=0, y=x^(2)+x+1`, and the tangent to the curve `y=x^(2)+x+1` at A(1,3) is

Options

A. `4//3` sq. units
B. `7//3` sq. units
C. `7//6` sq. units
D. None of these

Answer

C

Explanation

`"Given "y=x^(2)+x+1=(x+(1)/(2))^(2)+(3)/(4)rArry-(3)/(4)=(x+(1)/(2))^(2).`

This is a parabola with vertex at `(-(1)/(2),(3)/(4))` and the curve is concave upwards.
`y=x^(2)+x+1or (dy)/(dx)=2x+1or ((dy)/(dx))_(((1,3)))=3`
Equation fo the tangent at `A(1,3) is y=3x`
Required (shaded) area = Area ABDMN-Area ONA
`"Now, area "ABDMN=overset(1)underset(-1)int(x^(2)+x+1)dx=2overset(1)underset(0)int(x^(2)+1)=(8)/(3)`
`"Area of "ONA=(1)/(2)xx1xx3=(3)/(2).`
`therefore" Required area "=(8)/(3)-(3)/(2)=(16-9)/(6)=(7)/(6)` sq. units