To find the area between the [tex]$x$[/tex]-axis and the graph of the function [tex]$y=x$[/tex] from [tex]$x=3$[/tex] to [tex]$x=6$[/tex], we integrate the function [tex]$y=x$[/tex] with respect to [tex]$x$[/tex] over the interval [tex]$[3, 6]$[/tex].
The integral of [tex]$y = x$[/tex] from [tex]$x=3$[/tex] to [tex]$x=6$[/tex] is given by:
[tex]\[
\int_{3}^{6} x \, dx
\][/tex]
Using the power rule for integration, we have:
[tex]\[
\int x \, dx = \frac{x^2}{2}
\][/tex]
Evaluating this integral from [tex]$x=3$[/tex] to [tex]$x=6$[/tex], we get:
[tex]\[
\left[ \frac{x^2}{2} \right]_{3}^{6} = \frac{6^2}{2} - \frac{3^2}{2}
\][/tex]
Now we calculate the values:
[tex]\[
\frac{6^2}{2} = \frac{36}{2} = 18
\][/tex]
[tex]\[
\frac{3^2}{2} = \frac{9}{2} = 4.5
\][/tex]
Subtracting these results gives:
[tex]\[
18 - 4.5 = 13.5
\][/tex]
Thus, the area between the [tex]$x$[/tex]-axis and the graph of [tex]$y = x$[/tex] from [tex]$x = 3$[/tex] to [tex]$x = 6$[/tex] is:
[tex]\[
\frac{6^2-3^2}{2}=\frac{27}{2}=13.5
\][/tex]
So the correct answer is [tex]\(13.5\)[/tex], confirming that:
[tex]\(\frac{6^2-3^2}{2} = \frac{27}{2} = 13.5\)[/tex].
