Answer:
[tex]A=128[/tex]
Step-by-step explanation:
First of all we need to graph f(x)=8x, (First picture)
Now we have to calculate the area enclosed by the graph of the function, the horizontal axis, and vertical lines at [tex]x_{1}=2[/tex] and [tex]x_{2}=6[/tex] ,
The area that we have to calculate is in pink (second picture).
The function is positive and the domain is [tex][2,6][/tex] then we can calculate the area with this formula:
[tex]A=\int\limits^b_a {f(x)} \, dx[/tex],
In this case [tex]b=x_{2} , a=x_{1}[/tex]
[tex]A=\int\limits^6_2 {8x} \, dx = 8\int\limits^6_2 {x} \, dx[/tex]
The result of the integral is,
[tex]A=8\frac{x^{2}}{2}[/tex], but the integral is defined in [2,6] so we have to apply Barrow's rule,
Barrow's rule:
If f is continuous in [a,b] and F is a primitive of f in [a,b], then:
[tex]\int\limits^b_a {f(x)} \, dx =F(b)-F(a)[/tex]
Applying Barrow's rule the result is:
[tex]A=8.\frac{6^{2} }{2}-8.\frac{2^{2} }{2}[/tex]
[tex]A=8.\frac{36}{2} -8.\frac{4}{2}[/tex]
[tex]A=144-16[/tex]
[tex]A=128[/tex]
