The area enclosed by `y=x^(2)+ cos x" and its normal at "x=(pi)/(2)` in the first quadrant is

Options

A. A. `(pi^(5))/(32)-(pi^(4))/(64)+(pi^(3))/(32)+1`
B. B. `(pi^(5))/(16)-(pi^(4))/(32)+(pi^(3))/(24)-1`
C. C. `(pi^(5))/(32)-(pi^(4))/(32)+(pi^(3))/(16)`
D. D. `(pi^(5))/(32)-(pi^(4))/(32)+(pi^(3))/(24)+1`

Answer

D

Explanation

`f(x)=x^(2)+cos x`
`rArr" "f'(x)=2x- sin x`
`rArr" "f'((pi)/(4))=pi-1`
`rArr" Equation of normal at "x=(pi)/(2)` is
`(y-(pi^(4))/(4))=(1)/(1-pi)(x-(pi)/(2))`
`"It meets x-axis at "x=((pi-1)pi^(2))/(4)+(pi)/(2)`
Also `f'(x)=2x- sin x gt 0" for "x gt 0 and lt 0" for "xlt 0`
So f(x) increases for `x in (0,oo) and " decreases for "x in (-oo,0)`
Graph of the function is as shown in the followin figure.

Required area
=Area OABCO+Area of `DeltaBCD`
`=overset(pi//2)underset(0)int(x^(2)+cos x)dx+(1)/(2)[((pi-1)pi^(2))/(4)+(pi)/(2)-(pi)/(2)]xx(pi^(2))/(4)`
`=[(x^(3))/(3)+sin x]_(0)^(pi//2)+((pi-1)pi^(2))/(8)`
`=(pi^(3))/(24)+1+(pi^(4))/(32)(pi-1)`
`=(pi^(5))/(32)-(pi^(4))/(32)+(pi^(3))/(24)+1`