Answer
$${\text{A}} = \frac{{13}}{3}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt {4x - 3} {\text{ on the interval }}\left[ {1,3} \right] \cr
& {\text{The area between two curves is defined by }}\int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{where }}f{\text{ and }}g{\text{ are continuous functions and }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right]{\text{for this exercise }} \cr
& g\left( x \right) = 0{\text{ }}\left( {x{\text{ - axis}}} \right){\text{ and the interval }}\left[ {1,3} \right]{\text{ implies that }}a = 1{\text{ and }}b = 3 \cr
& {\text{then}} \cr
& {\text{A}} = \int_1^3 {\left[ {\sqrt {4x - 3} - 0} \right]} dx \cr
& {\text{A}} = \int_1^3 {\sqrt {4x - 3} } dx \cr
& {\text{use substitution}}{\text{. Let }}u = 4x - 3,{\text{ so that }}du = 4dx,\,\,\,\,\frac{1}{4}du = dx \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 3,{\text{ then }}u = 4\left( 3 \right) - 3 = 9 \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = 4\left( 1 \right) - 3 = 1 \cr
& {\text{Then}} \cr
& {\text{A}} = \int_1^3 {\sqrt {4x - 3} } dx = \int_1^9 {\sqrt u \left( {\frac{1}{4}du} \right)} \cr
& {\text{A}} = \frac{1}{4}\int_1^9 {\sqrt u } du \cr
& {\text{write }}\sqrt u {\text{ as }}{u^{1/2}} \cr
& {\text{A}} = \frac{1}{4}\int_1^9 {{u^{1/2}}} du \cr
& {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& {\text{A}} = \frac{1}{4}\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)_1^9 \cr
& {\text{A}} = \frac{1}{4}\left( {\frac{{2{u^{3/2}}}}{3}} \right)_1^9 \cr
& {\text{A}} = \frac{1}{6}\left( {{u^{3/2}}} \right)_1^9 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& {\text{A}} = \frac{1}{6}\left( {{9^{3/2}} - {1^{3/2}}} \right) \cr
& {\text{simplifying}} \cr
& {\text{A}} = \frac{1}{6}\left( {26} \right) \cr
& {\text{A}} = \frac{{13}}{3} \cr} $$
