Let `b!=0` and for `j=0,1,2,....,n`. Let `S_(j)` be the area of the region bounded by Y_axis and the curve `x cdot e^(ay)=sin by, (jpi)/bleyle((j+1)pi)/(b)`. Show that `S_(0),S_(1),S_(2),...S_(n)` are in geometric progression. Also, find their sum for a=-1 and `b=pi`.

Answer

`[(pi(1+e))/((1+pi^(2)))*((e^(n+1)-1))/(e-1)]`

Explanation

Given, ` x=(sin by)e^(-ay)`
Now, `-1 le sin by le 1`
`rArr -e^(-ay) le e^(-ay) sin by le e^(-ay)`
`rArr -e^(ay) le x le e^(ay)`

In this case, if we take a and b positive, the values `-e^(-ay) and e^(-ay)` become left bond and right bond of the curve and due to oscillating nature of sin by , it will oscillate between `x=e^(-ay) and x= -e^(-ay)`
Now, `S_(j)=int_(j pi b)^((j+1)pi b) sin by * e^(-ay) dy`
`[("since, " I=intsinby *e^(-ay) dy),(I=(-e^(-ay))/(a^(2)+b^(2))(a sin by + b cos by))]`
`therefore S_(j) |(-1)/(a^(2)-b^(2))[e^((-a(j+1)pi)/(b)){a sin(j+1)pi+b cos (j+1)pi}-e^(-(a j pi)/(b)) (a sin j pi+ b cos j pi)]|`
`S_(j)=|-(1)/(a^(2)+b^(2))[e^(-(a)/(b)(j+1)pi){0+b(-1)^(j+1)}-e^(-a j pi//b){0+b(-1)^(j)}]|`
`=|(b(-1)^(j)e^(-(a)/(b)jpi))/(a^(2)+b^(2))(e^(-(a)/(b)pi)+1)|`
`[because (-1)^(j+2)=(-1)^(2)(-1)^(j)=(-1)^(j)]`
`=(be^(-(a)/(b) jpi))/(a^(2)+b^(2))(e^(-(a)/(b)pi)+1)`
`therefore (S_(j))/(S_(j-1))=((be^(-(a)/(b)jpi)(e^(-(a pi)/(b))+1))/(a^(2)+b^(2)))/((be^(-(a)/(b)(j-1)pi)(e^(-(a pi)/(b))+1))/(a^(2)+b^(2)))=(e^(-(a)/(b)jpi))/(e^(-(a)/(b)(j-1)pi))`
`=e^(-(a)/(b)pi)=` constant
`rArr S_(0),S_(1),S_(2), ..., S_(j)` form a GP.
For `a = -1 and b =pi`
`S_(j)=(pi*e^((1)/(pi)pij))/(1+pi^(2))(e^((1)/(pi)pi)+1)=(pi*e^(j))/((1+pi^(2)))(1+e)`
`rArr sum_(j=0)^(n)S_(j)=(pi*(1+e))/((1+pi)^(2))sum_(j=0)^(n)e^(j)=(pi(1+e))/((1+pi^(2)))(e^(0)+e^(1)+ ... + e^(n))`
`=(pi(1+e))/((1+pi^(2)))*((e^(n+1)-1))/(e-1)`