Calculus problem 

\[0=\int_c^d (x+b)dx =(\frac{x^2}{2}+bx)|_c^{d}\]
\[(\frac{d^2}{2}+bd)-(\frac{c^2}{2}+bc)=0\]
\[\Rightarrow  \frac{d^2-c^2}{2}+b(d-c)=0\]
\[-b(d-c)=\frac{d^2-c^2}{2}=\frac{(d-c)(d+c)}{2}\]
\[\Rightarrow  -b(d-c)=\frac{(d-c)(d+c)}{2}\]
\[\Rightarrow b=-\frac{d+c}{2}.\]
The answer is 
\[b=-\frac{d+c}{2}.\]