In a half range sine series f(x)=e^x,0<x<1, the Fourier coefficient bn is given by

The Fourier sine series for a function f(x) in the interval (0, L) is given by:

f(x) = Σ b_n * sin(nπx/L)

where the coefficients b_n are given by:

b_n = (2/L) * ∫_0^L f(x) * sin(nπx/L) dx

In this case, f(x) = e^x and L = 1, so we have:

b_n = 2 * ∫_0^1 e^x * sin(nπx) dx

This integral can be solved using integration by parts, where u = e^x and dv = sin(nπx) dx. The integral becomes:

b_n = 2 * [e^x * (-cos(nπx)/(nπ)) |_0^1 - ∫_0^1 e^x * cos(nπx)/(nπ) dx]

The second integral can be solved again using integration by parts, and the final result is:

b_n = 2 * [(e - 1)/(1 + (nπ)^2)]

This is the Fourier coefficient b_n for the half range sine series of f(x) = e^x in the interval (0, 1).