integration/areas question (1 Viewer)

[ezzy85]

how would you go about doing this q? its an extenstion from the cambridge book.

Find the value of k for which y = kx bisects the area enclosed by the curve 4y = 4x - x^2 and the x axis.

when finding the point of intersection i got [ 4(1-k), 4x(1-k) ]. i tried splitting the area into different parts and making simltaneous equations but can't work it out. I got the area of the curve to be 8/3.

THanks alot.

 

[wogboy]

Firstly the area enclosed between the curve y = x - (x^2)/4 and the x-axis is =

integral of (x - (x^2)/4 dx)

= (x^2)/2 - (x^3)/12 + C

Now substitute & subtract the limits 4 and 0:

= 16/2 - 64/12 - 0

= 8/3


Because y = kx meets the curve y = x - (x^2)/4 at x = 4(1-k), the area between these two lines/curves is given by:

integral of (x - (x^2)/4 - kx dx)

= (x^2)/2 - (x^3)/12 - k(x^2)/2 + C

= -(x^3)/12 + [(x^2)/2] * (1-k) + C

now substitute & subtract the limits 4(1-k) and 0:

= -64/12 * (1-k)^3 + 16/2 * (1-k)^2 * (1-k)

= -16/3 * (1-k)^3 + 8 * (1-k)^3

= 8/3 * (1-k)^3

So the area between the curve and the line for any value of k is:

Area = 8/3 * (1-k)^3

Since the area required is 4/3 units^2 (half of 8/3), then:

8/3 * (1-k)^3 = 4/3

(1-k)^3 = 1/2


k = 1 - cube_root(1/2)

you can also rationalise the denominator, and make it:

k = 1 - [cube_root(4)]/2

 

[ezzy85]

thanks alot