IIT-JEE 2001Question 1
Let $b\neq0$ and for $j=0,1,2,\ldots,n$, let $S_j$ be the area of the region bounded by the $y$-axis and the curve $xe^{ay}=\sin by$, $\dfrac{j\pi}{b}\le y\le\dfrac{(j+1)\pi}{b}$. Show that $S_0,S_1,S_2,\ldots,S_n$ are in geometric progression. Also, find their sum for $a=-1$ and $b=\pi$.
Application of Integrals
AnswerCommon ratio $e^{-a\pi/b}$; for $a=-1,b=\pi$: $\dfrac{\pi(1+e)\left(e^{n+1}-1\right)}{(1+\pi^{2})(e-1)}$
Solution
With $y=t+\frac{j\pi}b$, $S_j=\int\left|e^{-ay}\sin by\right|dy=e^{-aj\pi/b}S_0$, so the $S_j$ are in G.P. with ratio $e^{-a\pi/b}$. For $a=-1,b=\pi$: $S_0=\int_0^1e^y\sin\pi y\,dy=\frac{\pi(1+e)}{1+\pi^2}$, ratio $e$, so the sum is $S_0\frac{e^{n+1}-1}{e-1}=\frac{\pi(1+e)(e^{n+1}-1)}{(1+\pi^2)(e-1)}$.