Asked by Pat
Find the area of the region bounded by the graphs of the given equations:
y=x, y=2√x
y=x, y=2√x
Answers
Answered by
Jai
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
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
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