find the area of region R bounded by the curves y=3x , x=2y and 2x+y=5

You need to divide the region into two parts at x=1, where 3x and 5-2x intersect.

a = ∫[0,1] (3x - x/2) dx + ∫[1,2] (5-2x - x/2) dx

See the region at

http://www.wolframalpha.com/input/?i=plot+y%3D3x%2Cy%3Dx%2F2%2Cy%3D5-2x

To find the area of the region R bounded by those curves, we need to first determine the points of intersection of the curves. Then we can use the integral to calculate the area between those points.

1. Find the intersection points:
- The first curve is y = 3x.
- The second curve is x = 2y.
- The third curve is 2x + y = 5.

We can find the intersection points by solving these equations simultaneously.

Substituting x = 2y into the equation 2x + y = 5:

2(2y) + y = 5
4y + y = 5
5y = 5
y = 1

Substituting y = 1 into the equation x = 2y:

x = 2(1)
x = 2

So the point of intersection of the second and third curves is (2, 1).

Now, substitute y = 3x into the equation x = 2y:

x = 2(3x)
x = 6x
-5x = 0
x = 0

Substituting x = 0 into the equation y = 3x:

y = 3(0)
y = 0

So the point of intersection of the first and second curves is (0, 0).

2. Calculate the area using integrals:

To find the area of the region R bounded by the curves, we consider the horizontal and vertical strips of infinitesimal width within the region and integrate accordingly.

Since the region is symmetric over the y-axis, we calculate the area from y = 0 to y = 1.

The area can be calculated using the integral ∫(upper limit)(lower limit)(f(x) - g(x)) dx or ∫(upper limit)(lower limit)(g(y) - f(y)) dy, where f(x) and g(x) are the equations of the curves.

In this case, we will use the integral: A = ∫(y = 0 to y = 1)(2y - 3x) dx.

Since we've obtained x in terms of y in the equation 2y - 3x = 0, we can substitute x = (2/3)y into the integral:

A = ∫(y = 0 to y = 1)(2y - 3(2/3)y) dy
= ∫(y = 0 to y = 1)(2y - 2y) dy
= ∫(y = 0 to y = 1) 0 dy
= 0

Therefore, the area of region R bounded by the given curves is 0.