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.
y=3+√X, y=3+1/5x What is the area?
Looks pretty straightforward, both equations say y = ...
let's find their intersection:
3 + √x = 3 + (1/5)x
√x = x/5
square both sides,
x = x^2/25
25x - x^2 = 0
x(25-x) = 0
x = 0 or x = 25
if x = 0, y = 3
if x = 25, y = 8
effective height of regions
= 3+√x - 3 - (1/5)x
= x^(1/2) - (1/5)x
area = ∫(x^(1/2) - (1/5)x) dx from x = 0 to 25
= [ (2/3)x^(3/2) - (1/10)x^2 ] from 0 to 25
= ( (2/3)(125) - (1/10)(625) - 0)
= 125/6
= appr 20.833
To find the region enclosed by the given curves, we need to determine the points of intersection first.
Setting the two equations equal to each other:
3 + √x = 3 + (1/5)x
Subtracting 3 from both sides:
√x = (1/5)x
Squaring both sides:
x = (1/25)x^2
Multiplying both sides by 25x to eliminate the fraction:
25x^2 = x
Now solving for x:
25x^2 - x = 0
Factoring out an x:
x(25x - 1) = 0
Setting each factor equal to zero:
x = 0 or 25x - 1 = 0
If x = 0, then y = 3 + √0 = 3+0 = 3
If 25x - 1 = 0, then x = 1/25, and y = 3 + (1/5)(1/25) = 3 + 1/125 = 3.008
Therefore, the points of intersection are (0, 3) and (1/25, 3.008).
Now let's plot the curves and shade the enclosed region.
The first curve, y = 3 + √x, is a curve that starts at (0, 3) and increases as x increases.
The second curve, y = 3 + (1/5)x, is a straight line with a positive slope of 1/5.
The enclosed region is bounded by these two curves and the x-axis.
Since the two curves intersect at a finite number of points, we will integrate with respect to x.
To find the area of the region, we need to integrate the difference between the two curve equations over the interval where they intersect.
Thus, the area of the region can be calculated as follows:
Area = ∫[a, b] (3 + (1/5)x - (3 + √x)) dx
Simplifying:
Area = ∫[a, b] ((1/5)x - √x) dx
To determine the bounds of integration, we set the two equations equal to each other:
3 + √x = 3 + (1/5)x
√x - (1/5)x = 0
Solving for x:
x(√x - (1/5)) = 0
x = 0 or √x = (1/5)
x = 0 or x = 1/25
So, the bounds of integration are from x = 0 to x = 1/25.
Now, we can integrate to find the area:
Area = ∫[0, 1/25] ((1/5)x - √x) dx
To find the antiderivative of ((1/5)x - √x), we can use the power rule and the rule for integrating √x:
Area = (1/10)x^2 - (2/3)x^(3/2) | [0, 1/25]
Evaluating the integral:
Area = [(1/10)(1/25)^2 - (2/3)(1/25)^(3/2)] - [(1/10)(0)^2 - (2/3)(0)^(3/2)]
Area = [(1/10)(1/625) - (2/3)(1/√625)] - [0]
Area = (1/6250) - (2/3)(1/25)
Area = 1/6250 - 2/75
Area = (15 - 500)/37500
Area = -485/37500
Therefore, the area of the region enclosed by the given curves is -485/37500.
To sketch the region enclosed by the given curves, start by plotting the two equations: y = 3 + √X and y = 3 + (1/5)x.
First, let's analyze the curves and understand the intersection points:
1. Set the two equations equal to each other to find the intersection point:
3 + √X = 3 + (1/5)x
√X = (1/5)x
2. Square both sides of the equation:
X = (1/25)x^2
3. Rearrange the equation and solve for x:
(25x^2) - x = 0
x(25x - 1) = 0
From this, we obtain two possible solutions:
1) x = 0
2) x = 1/25
Now, let's understand the nature of the curves:
The first curve, y = 3 + √X, is an upward-opening square root curve starting from the point (0,3).
The second curve, y = 3 + (1/5)x, is a linear equation with a positive slope of 1/5.
Next, we should determine which variable to integrate with respect to, x or y, to find the area of the region.
Since the curves are defined explicitly in terms of equations y = f(x), it is more convenient to integrate with respect to x. Moreover, as we can see from the equations, the region has a vertical boundary on the left at x = 0 and a vertical boundary on the right at x = 1/25.
So, the integral to find the area (A) of the region is given by:
A = ∫ [0, 1/25] (y2 - y1) dx
Now, let's calculate the area:
First, let's calculate the y-values of the curves at the intersection points, x = 0 and x = 1/25:
For y = 3 + √X:
At x = 0, y = 3 + √(0) = 3
At x = 1/25, y = 3 + √(1/25) = 3 + 1/5 = 16/5
For y = 3 + (1/5)x:
At x = 0, y = 3 + (1/5)(0) = 3
At x = 1/25, y = 3 + (1/5)(1/25) = 3 + 1/125 = 376/125
Now, we can set up the integral:
A = ∫[0, 1/25] [(3 + (1/5)x) - (3 + √X)] dx
A = ∫[0, 1/25] [((1/5)x) - (√X)] dx
A = ∫[0, 1/25] ((1/5)x - √X) dx
Evaluating this integral will give you the area of the region enclosed by the given curves.