Calculus problem #2198

Differentiating $f(x)$ , we get $f'(x)=a\ln{x}+a.$ $f'(e)=a+a=2a=4,$ so $a=2$ . Evaluating the integral, we have $\int_{1}^{e} \left(2x\ln{x}+b \right) \, dx$ $=2\int_{1}^{e} x\ln{x} \, dx + b\int_1^e dx.$ Using integration by parts with $u = \ln x \Rightarrow du = \frac{1}{x} \, dx$ $dv = x \, dx \Rightarrow v=\frac{x^2}{2}$ and evaluating the right-hand integral, we get $2\left(\left[\frac{x^2}{2}\ln x \right]_1^e - \frac{1}{2}\int_1^e x \, dx \right) + b(e-1)$ $=e^2-\frac{1}{2} \left[x^2 \right]_1^e + b(e-1)$ $=\frac{e^2}{2} - \frac{1}{2} +b(e-1).$ We know this integral is equal to $\frac{1}{2}e(e+1)=\frac{e^2}{2}+\frac{e}{2}$ , so $\frac{e^2}{2} - \frac{1}{2} +b(e-1)=\frac{e^2}{2}+\frac{e}{2}.$ Solving for $b$ yields $b=\frac{1}{2},$ so $a+b=2+\frac{1}{2}=\fbox{2.5}.$