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# Area under a curve watch

1. Sketch the curve with equation y=x(2-x) for 0 <_x<_3

A straight line L is passing trough the origin (0,0), intersects this curve at a point with x-coordinate k, where 0<k<2.

Show that the area of the region bounded by the curve y=x(2-x) and the line L is k^3/6

Find the value of k such that this area is exactly one eight of the area under the curve y=x(2-x) from x=0 to x=2.

2. (Original post by Teatime)
Sketch the curve with equation y=x(2-x) for 0 <_x<_3

A straight line L is passing trough the origin (0,0), intersects this curve at a point with x-coordinate k, where 0<k<2.

Show that the area of the region bounded by the curve y=x(2-x) and the line L is k^3/6

Find the value of k such that this area is exactly one eight of the area under the curve y=x(2-x) from x=0 to x=2.

y=x(2-x) therefore y=2x-x^2

You already know one co-ordinate for the stright line 'k' sub that into the quadratic equation. Then solve simultaneously.
3. (Original post by Teatime)
Sketch the curve with equation y=x(2-x) for 0 <_x<_3

A straight line L is passing trough the origin (0,0), intersects this curve at a point with x-coordinate k, where 0<k<2.

Show that the area of the region bounded by the curve y=x(2-x) and the line L is k^3/6

Find the value of k such that this area is exactly one eight of the area under the curve y=x(2-x) from x=0 to x=2.

Sorry, I meant find the equation of the line.

Equation of the line is as follows, we know L intersects at X=K therefire to find the corresponding Y co-ordinate we sub that into the equation to get 2K-K^2. The gradient therefore we know as we now have two points on the line, as it passess the origin (0,0) and (k, 2K-K^2). Therefore the gradient of the line is: (2-K).

Now integrate (2x-x^2)-(2-K).
4. (Original post by Bhaal85)
Sorry, I meant find the equation of the line.

Equation of the line is as follows, we know L intersects at X=K therefire to find the corresponding Y co-ordinate we sub that into the equation to get 2K-K^2. The gradient therefore we know as we now have two points on the line, as it passess the origin (0,0) and (k, 2K-K^2). Therefore the gradient of the line is: (2-K).

Now integrate (2x-x^2)-(2-K).
Why integrate the gradient and not the whole line?
5. (Original post by XTinaA)
Why integrate the gradient and not the whole line?
That is the equation of the line.
6. (Original post by Teatime)
Sketch the curve with equation y=x(2-x) for 0 <_x<_3

A straight line L is passing trough the origin (0,0), intersects this curve at a point with x-coordinate k, where 0<k<2.

Show that the area of the region bounded by the curve y=x(2-x) and the line L is k^3/6

Find the value of k such that this area is exactly one eight of the area under the curve y=x(2-x) from x=0 to x=2.

the sketch is attached.

L: y = mx
m = (k^2-2k)/k = k-2
=> y = (k-2)x

A = integral[0,k] of f(x)
f(x) = 2x - x² - kx + 2x = (4-k)x - x²
A = [(4-k)/2 x² - 1/3 x³][0,k]
= -1/3 k³ - 1/2 k³ +4k²= 4k² -1/6 k³
7. (Original post by Bhaal85)
That is the equation of the line.
Doubt it. If expressing as y=mx+c, then y=(2-k)x

So integrate y=x(2-x)=2x-x², then y=2x-kx, chuck in the values of x (namely, 0 and k), there should be your answer I think...
8. (Original post by XTinaA)
Doubt it. If expressing as y=mx+c, then y=(2-k)x

So integrate y=x(2-x)=2x-x², then y=2x-kx, chuck in the values of x (namely, 0 and k), there should be your answer I think...
That's what I meant, I forgot the X, and the server was just soooooooooooooooooooooooooooo busy, so I couldn't amend it.
9. (Original post by GeneralGrievous)
the sketch is attached.

L: y = mx
m = (k^2-2k)/k = k-2
=> y = (k-2)x

A = integral[0,k] of f(x)
f(x) = 2x - x² - kx + 2x = (4-k)x - x²
A = [(4-k)/2 x² - 1/3 x³][0,k]
= -1/3 k³ - 1/2 k³ +4k²= 4k² -1/6 k³
Isn't it the integral of y=(2x-x^2)-(2x-kx) therefore integrate of:

Y=-x^2+kx giving

(-X^3)/3+(KX^2)/2
10. (Original post by Teatime)
Sketch the curve with equation y=x(2-x) for 0 <_x<_3

A straight line L is passing trough the origin (0,0), intersects this curve at a point with x-coordinate k, where 0<k<2.

Show that the area of the region bounded by the curve y=x(2-x) and the line L is k^3/6

Find the value of k such that this area is exactly one eight of the area under the curve y=x(2-x) from x=0 to x=2.

The graph itself is an upside down parabola with points:

(0,0) (1,1) (2,0) sketch accordingly.
11. (Original post by Bhaal85)
Isn't it the integral of y=(2x-x^2)-(2x-kx) therefore integrate of:

Y=-x^2+kx giving

(-X^3)/3+(KX^2)/2
However by integrating between k and 0 it yeilds an answer of (K^3)/6.
12. (Original post by Bhaal85)
However by integrating between k and 0 it yeilds an answer of (K^3)/6.
Which is what is asked for.
13. (Original post by XTinaA)
Which is what is asked for.
The way the original poster wrote the answer was somewhat ambiguos.
14. (Original post by Bhaal85)
The way the original poster wrote the answer was somewhat ambiguos.
Well would it not have simplified to x^1/2 or sqrtX?
15. (Original post by XTinaA)
Well would it not have simplified to x^1/2 or sqrtX?
No, its K raised to the power of 3 then divided by 6.
16. (Original post by Bhaal85)
No, its K raised to the power of 3 then divided by 6.
I know, I'm saying it is not then ambiguous.
17. (Original post by XTinaA)
Well would it not have simplified to x^1/2 or sqrtX?
k^3/6 and (k^3)/6 are two completely different things.

K^3/6 can be written like you said, K^1/2.

However (K^3)/6 can be thought of as 1/6 * K^3.

Like I said, two completely different meanings.
18. (Original post by Bhaal85)
k^3/6 and (k^3)/6 are two completely different things.

K^3/6 can be written like you said, K^1/2.

However (K^3)/6 can be thought of as 1/6 * K^3.

Like I said, two completely different meanings.
Yeah but if the former was meant it wouldn't be written as it was.
19. (Original post by XTinaA)
Yeah but if the former was meant it wouldn't be written as it was.
Yeah, well considering I help most people on this section, you tend to see a lot of typo errors due to the arkwardness of keyboards, or perhaps they are not mathematically minded, and not many people like loading up special characters so that proper math characters can be displayed.
20. (Original post by Bhaal85)
Yeah, well considering I help most people on this section, you tend to see a lot of typo errors due to the arkwardness of keyboards, or perhaps they are not mathematically minded, and not many people like loading up special characters so that proper math characters can be displayed.
I think this isn't worth wasting energy over

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