In the following figure, the graphs of two functions y = f(x) and y = sin x are givne. They intersect at origin, `A(a,f(a)),B(pi,0) and C(2pi,0).A_(i)(i=1,2,3)` is the area bounded by the curves as shown in the figure, respectively, for `x in (0,a), x in (a, pi), x in (pi,2pi)`.
If `A_(1)=1+(a-1)cosa - sin` a, then ,the value of A 2 ?

Options

A. a) `(pi-1)"units"^(2)`
B. b) `(pi//2-1)"units"^(2)`
C. c) `(pi-sin1-1)"units"^(2)`
D. d)`pi//2"units"^(2)`

Answer

C

Explanation

`A_(1)=int_(0)^(a)(sinx-f(x))dx=1+(a-1)cosa-sina`
Differentiating w.r.t. 'a' both sides.
`therefore" "sina-f(a)=cosa-(a-1)sin a -cosa`
`therefore" "f(a)=asina`
`therefore" "f(x)=x sin x`
`"Solving y = sin x wity y = x sin x, we have x sin x = sin x"`
`therefore" "x=1 or x=pi,2pi`
`therefore" "A_(2)=int_(1)^(pi)(x sin x-sinx)dx`
`=(-x cosx)_(1)^(pi)-int_(1)^(pi)-cosxdx+(cosx)_(1)^(pi)`
`=(pi+cos1)+(0-sin1)+(-1-cos1)`
`=(pi-sin1-1)"sq. units"`