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890

Part of 2013 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 890

Source: 2013 Osaka Prefecture University entrance exam/Engineering

10/2/2013
A function fn(x) (n=1, 2, ⋯ )f_n(x)\ (n=1,\ 2,\ \cdots) is defined by f1(x)=xf_1(x)=x and fn(x)=x+e2∫01fn−1(t)ex−tdt (n=2, 3, ⋯ )f_n(x)=x+\frac{e}{2}\int_0^1 f_{n-1}(t)e^{x-t}dt\ (n=2,\ 3,\ \cdots). Find fn(x)f_n(x).
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