Find the area bounded by the curve `y = x^(2) + x` and the lines x = 0 and x = a. Hence, fnd the greatest or least area which is applicable.

Options

A. Area = `(a^(3))/(3)+(a^(2))/(2)`, least are a = 0 sq. units, greatest area `= (1)/(6)` sq units
B. Area `=a^(3)+a^(2)`, least area a = 0 sq. units, greatest area `=(1)/(2)` sq units
C. Area `=(a^(3))/(3)+a`, least are a = 0 sq units greatest area = `(1)/(4)` sq units
D. Area `=a^(3)+1`, least are a = 1 sq units greatestarea = 6 sq units

Answer

A

Explanation

Given `y=x^(2)+x`
Required area `=underset(0)overset(a)intydx=underset(0)overset(a)int(x^(2)+x)dx=[(x^(3))/(3)+(x^(2))/(2)]_(0)^(a)`
`A=(a^(3))/(3)+(a^(2))/(2)`
For the area to be greatest or least
put `(dA)/(da)=0implies(1)/(3)a^(2)+(1)/(2).2a=0`
`impliesa^(2)+a=0impliesa(a+1)=0`
implies a=0,-1
Now, `(d^(2)A)/(da^(2))=2a+1`
At `a=0,(d^(2)A)/(da^(2))=1gt0`
Therefore, the area is least at a = 0
and the least area `=(0^(3))/(3)+(0^(2))/(2)=0` sq units
At `a=-1,(d^(2)A)/(da^(2))=-2+1=-1lt0`
Therefore, the area is greatest at a = - 1 And the greatest area
`=((-1)^(2))/(2)+((-1)^(2))/(2)=-(1)/(3)+(1)/(2)=(1)/(6)` sq units