Today's calculation of Integral 390
Source: 1991 Sophia University entrance exam
November 13, 2008
calculusintegrationalgebrapolynomialtrigonometrycalculus computations
Problem Statement
Find the polynomials such that:
\frac{1}{\pi}\int_0^{\frac{1}{2}} \frac{tf'(t)\minus{}6g(t)}{\sqrt{1\minus{}t^2}}\ dt\equal{}f(x)\minus{}g(x)\plus{}x
\frac{6}{\pi}\int_0^{\frac{1}{2}} \frac{8f(t)\minus{}5g'(t)}{\sqrt{1\minus{}t^2}}\ dt\equal{}2f(x)\minus{}3g(x)\minus{}x^2\plus{}2x.