Answer:
[tex]\dfrac{6^2-3^2}{2}=\dfrac{27}{2}[/tex]
Step-by-step explanation:
You can go to the trouble to integrate y = x from 3 to 6, or you can figure the area geometrically.
[tex]\displaystyle\int_3^6{x}\,dx=\left.\dfrac{x^2}{2}\right|_3^6=\dfrac{6^2-3^2}{2}\quad\text{using integration}\\\\A=\dfrac{1}{2}(b_1+b_2)h=\dfrac{1}{2}(6+3)(6-3)=\dfrac{6^2-3^2}{2}\quad\text{using geometry}[/tex]
