2. find the area of the region bounded by the graphs of y= x y=4 -3x and x=0
I would start by visualizing what the bounded area looks like on a Cartesian plane.
No calculus need here. Since the two lines intersect at (1,1) the region is just a triangle with height=1 and base=4/3, so its area is 2/3
If you insist on integrating, you can do it two ways.
Split the region in two, changing lines at x=1
a = ∫[0,1] x dx + ∫[1,4/3] (4-3x) dx = 1/2 + 1.6 = 2/3
Or, if you use horizontal strips, each strip has width equal to the distance between the lines
a = ∫[0,1] (4-y)/3 - y dy = ∫[0,1] (4-4y)/3 dy = 2/3
To find the area of the region bounded by the graphs of y = x, y = 4 - 3x, and x = 0, you can use the concept of definite integrals from calculus. Here's how you can go about solving it step by step:
1. Draw the given graphs on a coordinate plane to visualize the region bounded by them. The graph of y = x is a straight line passing through the origin with a 1:1 slope. The graph of y = 4 - 3x is a line with a y-intercept of 4 and a slope of -3. The line x = 0 is simply the y-axis.
2. Find the points of intersection between the curves y = x and y = 4 - 3x. To do this, set the two equations equal to each other and solve for x:
x = 4 - 3x
4x = 4
x = 1
Therefore, the points of intersection are (1, 1).
3. Identify the limits of integration. Since the region is bounded by the vertical line x = 0 and the curves y = x and y = 4 - 3x, the limits of integration for x will be from 0 to 1.
4. Set up the integral to find the area. Since the region is bounded by the curves vertically, you need to integrate the difference between the top curve (y = 4 - 3x) and the bottom curve (y = x) with respect to x:
Area = ∫[0 to 1] (4 - 3x - x) dx
5. Evaluate the integral to find the area. Integrating (4 - 3x - x) gives you:
Area = ∫[0 to 1] (4 - 4x) dx
= [4x - 2x^2] evaluated from 0 to 1
= (4(1) - 2(1)^2) - (4(0) - 2(0)^2)
= (4 - 2) - (0 - 0)
= 2 square units
Therefore, the area of the region bounded by the graphs of y = x, y = 4 - 3x, and x = 0 is 2 square units.