Calculus problem?

Question

Find the area of the region bonded by the curves.

y = 8x^2 and y= square root of x


step by step please! thanks :)

Answer 1

First you need to find where the curves intersect by setting the 2 functions equal to each other:
8x^2 = sqrt(x). Square both sides: 64x^4 = x
Set one side = 0: 64x^4 - x = 0
Factor: x(64x^3 - 1) = 0
Set each factor = 0 and solve: x = 0, x^3 = (1/64), x = (1/4)
So the limits of integration are x=0 to x=1/4.
Next you have to figure out which function is the top function. You can do this by graphing, or plugging in a value in the interval like x = 1/8.
Sqrt(x) is the top function, 8x^2 is the bottom.

So the integral is Integral(Sqrt(x) - 8x^2) with limits from 0 to 1/4.
The antiderivative is (2/3)x^(3/2) - (8/3)x^3 evaluated from x=0 to x= 1/4

The answer is (2/3)(1/4)^(3/2) -(8/3)(1/4)^3
=(2/3)(1/8) - (8/3(1/64)
=1/12 - 1/24 = 1/24

Answer 2

y = 8x^2
y= sqrt(x)

These curves intersect at x = 0, and at x such that 8x^2 = sqrt(x).
We want x such that 64x^4 = x (squaring both sides), i.e 64x^3 = 1
This means x^3 = 1/64, so x = 1/4.

The area between the two curves is the absolute value of the
difference of the areas for the interval [x=0, x=1/4]

Let S be the area between the two curves. Since sqrt(x) dominates
8x^2 in the interval [0, 1/4] the area will be given by the
formula

S = int(0, 1/4, sqrt(x)- 8x^2) dx

So integrating,

S = (2/3)x^(3/2) - (8/3)x^3 | [0, 1/4]

The integral is 0 at 0 so we just need to evaluate at x=1/4

S = (2/3)(1/4)^(3/2) - (8/3)(1/4)^3
= (2/3)(1/4)(1/2) - (8/3)(1/64) = 1/12 - 1/24 = 1/24

The area between the two curves is 1/24.

Answer 3

first find out where they both intersect so you can set up your upper and lower bounds, they intersect and o and .25. So your intergral lower bound is 0 and your upper bound is .25. Now you just simply do the top curve minuse the lower curve. So your ingeral is sqrt(x)-8x^2between 0 and .25. The you inegrate. The integral of sqrt of x is 2/3*x^(3/2) and the other one is 8/3x^3. so you get 1/12-(1/24)=1/24

Answer 4

Curves intersect when 8x² = √x
64x^(4) = x
x(64x³- 1) = 0
64x³ - 1 = 0, x = 0
x = 1/4, x = 0
A = ∫ (√x - 8x² ) dx between lims. 0 and 1/4
A = (2/3)(x^3/2) - (8/3) x³ between stated lims.
A = (2/3)(1/4)^(3/2) - (8/3).(1/4)³
A = (2/3).(1/8) - (8/3).(1/64)
A = 2/24 - 4 / 96
A = 8/96 - 4/96
A = 4/96
A = 1/24