The numbers P, Q and R for which the function f(x)=Pe2x+Qex+Rx satisfies f(0)=–1 f′(log⁡2)=31 and ∫0log⁡4 [f(x)−Rx]dx=39/2 are given by

We have f′(x)=2Pe2x+Qex+R, so that 31=f′(log⁡2)=8P+2Q+RAlso, −1=f(0)=P+Q. Further,392=∫0log⁡4 [f(x)−Rx]dx=∫0log⁡4 Pe2x+Qexdx=P2e2x+Qex0log⁡4=P2×16+4Q−P2−Q=15P2+3QSolving the above equations, we get P=5,Q=−6 and R=3