## Calculus, 10th Edition (Anton)

In order to integrate this function, we have to use an «u» substitution. Choose an u, derivate it: $u=2x-1$ $du=2dx$ $\dfrac{1}{2}du=dx$ Change limits: $x_1 =1$ $u_1=2(1)-1=1$ $x_2 =5$ $u_2=2(5)-1=9$ Then substitute: $= \dfrac{1}{2}\int_{1}^{9} \dfrac{du}{\sqrt{u}} =\dfrac{1}{2}\int_{1}^{9} u^{-1/2} du$ Then integrate $\bigg [ u^{1/2} \bigg ]_{1}^{9}$ Apply calculus 2nd theorem and simplify: $9^{1/2}-1^{1/2}=3-1=2$