Question

I NEED HELP PLEASE, THANKS! :)

Answer 1

`(6^2-3^2)/(2) =(27)/(2)`

(the third answer in your list of options)

Explanation:

The area of the region between y = x, and the x axis, between x= 3 and x=6 is the area of the trapezoid shown in the attached image.

It therefore can be calculated via the area of a trapezoid:

`(B+b)\,H/2=(6+3)*3/2=27/2`

or doing the integral (which it seems is what they want you to do):

`\int\limits^6_3 {x} \, dx =(x^2)/(2) ]\limits^6_3=(6^2-3^2)/(2) =(27)/(2)`