find the area between the two curves y=1/2x and y=x square root of 1-x^2

first we have to know where they intersect, so

(1/2)x = x√(1-x^2)
times 2
x - 2x√(1-x^2) = 0
x (1 - 2√(1-x^2) = 0
x = 0 or x = ±√3/2

I made a quick sketch and found two symmetrical regions, so let's just take it from 0 to √3/2 and double that answer.

Area = ∫x(1-x^2)^.5 - x dx from 0 to √3/2
= [ -(1/3)(1-x^2)^(3/2) - (1/2)x^2 ] from 0 to √3/2
= ...

I will let you do the messy arithmetic

i don't understand how to integrate x(1-x^2)^.5

First, I think Reiny dropped a factor of 1/2 in his integral (-x/2, not -x), but that's not a big worry (except that the answer is wrong!) :-)

To integrate x√(1-x^2), substitute u=1-x^2

Then you have du = 2x dx

x√(1-x^2) = √u du/2 = 1/2 u^(1/2) du
integrate that to get 1/3 u^(3/2)

so, ∫x(1-x^2)^.5 = 1/3 (1-x^2)^(3/2)

After all the messy arithmetic, you should wind up with 5/48

oops. dropped the minus sign, which Reiny slyly retained. . .

i don't get 5/48 when i plug in root 3/2

To find the area between the two curves y = (1/2)x and y = x√(1 - x^2), you need to find the points of intersection of these two curves and then calculate the definite integral of their difference over that interval.

Step 1: Finding the points of intersection:
Set the two equations equal to each other and solve for x:
(1/2)x = x√(1 - x^2)

Square both sides of the equation to eliminate the square root:
(1/4)x^2 = x^2(1 - x^2)

Now simplify the equation:
(1/4)x^2 = x^2 - x^4

Rearrange the equation:
0 = 3/4x^2 - x^4

Factor out an x^2:
0 = x^2(3/4 - x^2)

Set each factor equal to zero and solve for x:
x^2 = 0 --> x = 0
3/4 - x^2 = 0 --> x^2 = 3/4 --> x = ±√(3/4) = ±√3/2

So, the points of intersection are (0, 0) and (√3/2, 1/2).

Step 2: Calculate the area:
The area between the two curves can be found by taking the integral of their difference over the interval of x between the points of intersection.

Integrate y = (1/2)x - x√(1 - x^2) with respect to x, from x = 0 to x = √3/2:

A = ∫[(1/2)x - x√(1 - x^2)] dx, evaluated from 0 to √3/2

Perform the integration using the antiderivative rules:
A = [(1/4)x^2 - (1/3)(1 - x^2)^(3/2)] evaluated from 0 to √3/2

Substituting the limits into the antiderivative:
A = [(1/4)(√3/2)^2 - (1/3)(1 - (√3/2)^2)^(3/2)] - [(1/4)(0)^2 - (1/3)(1 - (0)^2)^(3/2)]

Simplifying the expression:
A = [(3/16) - (1/3)(1 - 3/4)^(3/2)] - 0

A = (3/16) - (1/3)(1/4)^(3/2)

A = (3/16) - (1/3)(1/8)

A = 3/16 - 1/24

A = (9 - 2) / 48

A = 7 / 48

So, the area between the two curves y = (1/2)x and y = x√(1 - x^2) is 7/48 square units.