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.