Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
2y=5x , y=3 and 2y+1x=6
It is easier to integrate with respect to the variable
Area =
Help!!!!
you can see the region here:
http://www.wolframalpha.com/input/?i=plot+2y%3D5x+%2C+y%3D3+and+2y%2B1x%3D6+for+0%3C%3Dx%3C%3D2
Integration over y will be easier, since the horizontal strips are all parallel to the upper edge.
If you integrate over x, then you have to break the region into two parts, because the lower boundary changes at x=1.
To sketch the region enclosed by the given curves, let's first plot the individual curves and see how they intersect.
1. Start with the equation 2y = 5x. We can rewrite it as y = (5/2)x.
- This equation represents a straight line passing through the origin (0,0) with a slope of 5/2.
2. Next, consider the equation y = 3.
- This equation represents a horizontal line parallel to the x-axis, passing through the y-coordinate 3.
3. Lastly, we have the equation 2y + x = 6. To isolate y, we subtract x and divide by 2.
- The equation becomes y = (6 - x)/2.
- This equation represents a straight line passing through the points (6, 0) and (0, 3).
Now, let's plot these curves on a coordinate plane:
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The first curve, y = (5/2)x, starts at the origin (0,0) and goes up diagonally to the right.
The second curve, y = 3, is a horizontal line parallel to the x-axis, passing through the point (0,3).
The third curve, y = (6 - x)/2, starts at the point (6,0) and goes diagonally down to the left, intersecting with the other two curves.
The enclosed region is the triangle formed by these three curves. To calculate the area, we need to find the boundaries of integration: the x-values where the curves intersect.
Setting the equations y = (5/2)x and y = 3 equal to each other, we can find the x-coordinate of the point of intersection:
(5/2)x = 3
x = (2/5) * 3
x = 6/5
Substituting this x-value into either equation, let's find the y-coordinate:
y = (5/2) * (6/5)
y = 6/2
y = 3
So, the point of intersection is (x, y) = (6/5, 3).
Next, setting y = 3 and y = (6 - x)/2 equal to each other, we can find another point of intersection:
3 = (6 - x)/2
6 = 6 - x
x = 0
Substituting this x-value back into either equation, let's find the y-coordinate:
y = (6 - 0)/2
y = 3
So, the point of intersection is (x, y) = (0, 3).
Now, we integrate with respect to x, since the intersection points are defined by x-values.
The boundaries of integration are x = 0 to x = 6/5.
The area is given by the integral:
Area = ∫[0, 6/5] [(6 - x)/2 - 3] dx
Simplifying this integral will give us the area of the region enclosed by the given curves.
To further evaluate the integral, if needed, you can use techniques such as integration by substitution or integration by parts.
I hope this explanation helps!