Find the area of the region bounded by the graphs of the given equations:
y=x, y=2√x
Req'd: area bounded by y = x and y = 2*sqrt(x)
First thing to do here is to find their points of intersection, so we'll know the bounds. We can do it algebraically or graphically.
To find algebraically the points of intersection, we just use substitution. Since y = x,
y = 2*sqrt(x)
x = 2*sqrt(x)
x^2 = 4x
x(x - 4) = 0
x = 0 and x = 4
To set up the integral, we choose whether vertical strips (dx) or horizontal strips (dy) to be used. In here, let's just use vertical (dx). The bounds for this is from x = 0 to x = 4, and since it is dx, all expressions must be in terms of x.
Therefore,
Integral (2*sqrt(x) - x)dx from 0 to 4
= (2*(2/3)*(x^3/2) - (1/2)x^2) from 0 to 4
= ((4/3)x^3/2 - (1/2)x^2) from 0 to 4
= [ ((4/3)(4^(3/2)) - (1/2)*4^2 ] - [ ((4/3)(0^(3/2)) - (1/2)*0^2 ]
= (4/3)(8) - (1/2)(16) - 0
= 32/3 - 8
= 32/3 - 24/3
= 8/3 square units
Hope this helps~ :3
To find the area of the region bounded by the graphs of the given equations y=x and y=2√x, we need to determine the points of intersection.
Setting the equations equal to each other:
x = 2√x
Squaring both sides:
x^2 = 4x
Rearranging the equation:
x^2 - 4x = 0
Factoring out x:
x(x - 4) = 0
Setting each factor equal to zero and solving for x:
x = 0 or x - 4 = 0
If x = 0, then y = 0.
If x - 4 = 0, then x = 4, and substitute into the equation y = x to get y = 4.
So, the points of intersection are (0,0) and (4,4).
The region bounded by the graphs of the given equations is a triangle with a base of length 4 and a height of 4.
Therefore, to find the area of the region, we use the formula for the area of a triangle:
Area = (base * height) / 2
= (4 * 4) / 2
= 8.
Therefore, the area of the region bounded by the graphs of y=x and y=2√x is 8 square units.
To find the area of the region bounded by the given equations y = x and y = 2√x, we need to determine the points of intersection and then integrate the difference between the two curves.
Step 1: Find the points of intersection.
To find the points of intersection, we set the two equations equal to each other:
x = 2√x
To solve this equation, square both sides:
x^2 = (2√x)^2
x^2 = 4x
Rearrange the equation:
x^2 - 4x = 0
x(x - 4) = 0
From this, we can determine that there are two solutions: x = 0 and x = 4.
Step 2: Set up the integral.
Since the curves intersect at x = 0 and x = 4, we integrate the difference between the two equations from x = 0 to x = 4:
Area = ∫[0,4] (2√x - x) dx
Step 3: Evaluate the integral.
We can use the power rule of integration to evaluate the integral:
Area = [2/3 * (2x^(3/2)) - 1/2 * (x^2)]|[0,4]
Substitute the upper and lower limits:
Area = [2/3 * (2(4)^(3/2)) - 1/2 * (4^2)] - [2/3 * (2(0)^(3/2)) - 1/2 * (0^2)]
Simplify:
Area = [2/3 * (2*8) - 1/2 * 16] - [2/3 * (2*0) - 1/2 * 0]
Area = [16/3 - 8] - [0 - 0]
Area = (16/3 - 8)
Area = 16/3 - 8/1
Area = (16 - 24)/3
Area = -8/3
Thus, the area of the region bounded by the graphs of y = x and y = 2√x is -8/3 square units.