Answer
$\dfrac{38}{15}$
Work Step by Step
In order to integrate this function, we have to use an «u» substitution.
Choose an u, derivate it:
$ u=5x-1$
$ du = 5dx$
$ \dfrac{1}{5}du=dx$
Change limits:
$x_1 =1$
$u_1=5(1)-1=4$
$x_2 =2$
$u_2=5(2)-1=9$
Then substitute:
\[ \dfrac{1}{5} \int_{4}^{9} \sqrt{u} du = \dfrac{1}{5} \int_{4}^{9} u^{1/2}du \]
\[ \]
Then integrate
\[ \dfrac{1}{5} \bigg [\dfrac{2}{3}u^{2/3} \bigg ]_{4}^{9} = \dfrac{2}{15} \bigg [u^{2/3} \bigg ]_{4}^{9} \]
Apply calculus 2nd theorem and reduce:
\[ \dfrac{2}{15} [ 9^{3/2} -4^{3/2}] = \dfrac{2}{15}[27-8]=\dfrac{38}{15}\]
