Find the area of the region bounded by the graphs of y = x, y = 4 - 3x, and x = 0.

Question

Question options:

1)	0.375
2)	2
3)	4
4)	None of these

Answer 1

You can do it with calculus, or just a little geometry.

set the two equations equal y=x and y=4-3x

x=4-3x or 4x=4 x=1. when x=1, y=1, that's where the two lines intersect, at the point (1,1).

draw a horizontal line from the intersection to the y axis. That creates 2 right triangles, one above the other

one is bordered by the lines x=0, y=1 and y=x. It's a triangle with height = 1 and base = 1. The area is 1/2 the base times the height or 1/2(1x1) = 1/2

The triangle just above it is bordered by x=0, y=1 and y=4-3x. It is a right triangle with height 3 and base 1

Its area is 1/2(3x1)=3/2 It helps to sketch the two lines and divide the area into the two triangles

Add the 2 triangle areas together, 1/2 + 3/2 = 4/2 = 2 the answer is 2

Answer 2

The answer is 2) 2

Graphing a quick sketch of y=x, y=4-3x, and x=0, we can see that the region bounded by the graphs of these equations is a triangle going from x=0 to x=1, with top boundary y=4-3x and bottom boundary y=x.

We set up an integral from 0 to 1: ∫(4-3x-x)dx. Integrating, we get 4x-2x² from 0 to 1. So ( 4·1 - 2·1² ) - ( 4·0 - 2·0² ) = 2