Find the area in the plane bounded by the two curves y = sqrt{x} and y = 4 x.?

Answer 1

get the intersection of the two curves..

√x = 4x .. squaring...
x = 16x²
thus x = 0 or x = 1/16

within that interval , √x > 4x ... check with 0.01
√0.01 = 0.1
while 4(0.01) = 0.04

thus the area is

∫ [√x - 4x] dx .... x from 0 to 1/16

= 2/3 x^(3/2) - 2x^2 from 0 to 1/16
= (2/3) (1/64) - 2(1/256)
= 1/96 - 1/128


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Answer 2

The curve and the line intersect at x=0 and x=0.0625. Graph the functions.
The area bounded = sqrt(x)-4x
Integrate between 0 and 0.0625
integ x^(1/2)-4x dx (0,0.0625)
x^(1/2+1)/(1/2+1) - 4 x^2/2
=(2/3)x^(3/2)-2x^2
at 0, the integral vanishes
at 0.0625
(2/3)(0.0625)^(3/2)-2(0.0625)^2
0.0104167-0.0078125
0.0026042

Answer 3

These two curves intersect at x = 0 and x = 1/4. Therefore, the area is Integ[0,1/4](4x - sqrt(x))dx. You can finish it.