Recall that the area between the graphs of two functions on the interval [a,b] is:
[tex]\int ^b_a|f(x)-h(x)|dx\text{.}[/tex]
Therefore, the area between f(x)=4√x and the x-axis with equation y=0, on the interval [4,9] is:
[tex]\int ^9_4|4\sqrt[]{x}-0|dx=\int ^9_4|4\sqrt[]{x}|dx\text{.}[/tex]
Notice that
[tex]4\sqrt[]{x}>0,[/tex]
for all x in the given interval, therefore:
[tex]\int ^9_4|4\sqrt[]{x}|dx=\int ^9_44\sqrt[]{x}dx=4\int ^9_4x^{\frac{1}{2}}dx=4(\frac{2x^{\frac{3}{2}}}{3})|^9_4=\frac{8}{3}(9^{\frac{3}{2}}-4^{\frac{3}{2}})=\frac{8}{3}(19)=\frac{152}{3}.[/tex]
Answer: 152/3.
