The work below leads to the correct answer for the computation of the integral 012xcos(x2+1)dx but it contains three notation mistakes. Rewrite this work, fixing all three notation mistakes.To compute 012xcos(x2+1)dx, we use substitution with u=x2+1 and du=2. The new bounds are u(0)=02+1=1 and u(1)-12+1=2. We get012xcos(x2+1)=01cos(u)du=[sin(u)]12=sin(2)-sin(1)