Finding x coordinates for a point dividing shaded area into equal parts

[kais]

Hey guys,
Having trouble with my very last question, it goes

the curve in the picture shown has equation y=bx(x-2)

(a) find b given that the shaded area is 4units^2
(b) find the x-coordinate of the point A if the line OA divides the shaded area into equal parts picture link is (https://s.yimg.com/hd/answers/i/05eb8de185c44299b48201c0d62b3fbf_A.png?a=answers&mr=0&x=1449657862&)

For (a) i took the definite integral between 0,2 of bx(x-2) dx and eventually ended up with b=-3
so now i have y=-3x(x-2) and i have to find the x-coordinate of the point of a line that divides the shaded area into equal parts.

Im pretty sure from here i have to find my limits to make a definite integral using

My problem is im not sure how i can find it since i only have the once function y=-3x(x-2). How do i get the function of the line? is it just y=mx?

Cheers in advance

​

 

[stapel]

the curve in the picture shown has equation y=bx(x-2)

(a) find b given that the shaded area is 4units^2
(b) find the x-coordinate of the point A if the line OA divides the shaded area into equal parts
picture link is (https://s.yimg.com/hd/answers/i/05eb8de185c44299b48201c0d62b3fbf_A.png?a=answers&mr=0&x=1449657862&)

The link returns a page containing only the word "Unauthorized".

Please reply with a clear description of how "A", "O", and "b" relate to the given curve.

For (a) i took the definite integral between 0,2 of bx(x-2) dx and eventually ended up with b=-3

so now i have y=-3x(x-2) and i have to find the x-coordinate of the point of a line that divides the shaded area into equal parts.

What is "the shaded region"? Thank you!

 

[Ishuda]

Hey guys,
Having trouble with my very last question, it goes

the curve in the picture shown has equation y=bx(x-2)

(a) find b given that the shaded area is 4units^2
(b) find the x-coordinate of the point A if the line OA divides the shaded area into equal parts picture link is (https://s.yimg.com/hd/answers/i/05eb8de185c44299b48201c0d62b3fbf_A.png?a=answers&mr=0&x=1449657862&)

For (a) i took the definite integral between 0,2 of bx(x-2) dx and eventually ended up with b=-3
so now i have y=-3x(x-2) and i have to find the x-coordinate of the point of a line that divides the shaded area into equal parts.

Im pretty sure from here i have to find my limits to make a definite integral using

My problem is im not sure how i can find it since i only have the once function y=-3x(x-2). How do i get the function of the line? is it just y=mx?

Cheers in advance

​

First, I can't see the image. I get an "Unauthorized". Next, if the shaded region is the region between x=0 and x=2, than the -3 is also what I get. Finally, the x which divides the region into two equal parts. That would be the x such that, in your integral, the a is zero and the b is such that the area is 2 units².

However, an easier, IMO, way is to note that the function is a parabola, that a parabola is symmetric about the vertex, and the vertex occurs at an x halfway between the two zeros (if it has two real zeros as this one does). So what is the x value of the vertex?

 

[kais]

Sorry guys the link worked for me so i thought it was all good Heres the picture

 

[Ishuda]

Sorry guys the link worked for me so i thought it was all good Heres the picture

Good. For the second part, I misunderstood the question and you are correct, as I now understand the question, in that you need the integral of a line f(x)=mx from x=0 where f(0)=0 to x=b where f(b)=A. So, what is m in terms of A and b. Also note that A=y(b)=-3b(b-2) since (b, A) is a point on the parabola. So now you should have all the information needed to find b (since the integral of mx from 0 to b is 2 units²). Let us know if you run into difficulties.

 

[kais]

Alright, so far ive got

y=mx
mx= -3b(b-2)
mx+3b^2-6b=0
x(m+3b-6)=0 here got rid of x because parabola and line go through origin and root x=0
m+3b-6=0
b= 6-m/3
If thats right that will give me my boundary for my integral right?

So now i need to do an integral which will look something like which will have 2, 6-m/3 boundary ∫ -3x(x-2) dx - ∫ 6-m/3 dx which has a 0 , 6-m/3 boundary
am i on the right track? sorry for all the questions i just really want to get my hear around this

 

[Ishuda]

Alright, so far ive got

y=mx
mx= -3b(b-2)
mx+3b^2-6b=0
x(m+3b-6)=0 here got rid of x because parabola and line go through origin and root x=0
m+3b-6=0
b= 6-m/3
If thats right that will give me my boundary for my integral right?

So now i need to do an integral which will look something like which will have 2, 6-m/3 boundary ∫ -3x(x-2) dx - ∫ 6-m/3 dx which has a 0 , 6-m/3 boundary
am i on the right track? sorry for all the questions i just really want to get my hear around this

Since what is wanted is the x-coordinate of A, the integration is from 0 to b, and we have labeled that x coordinate b, I think it might be easier to keep things in terms of b.

You are on the right track but you've made some mistakes: First see above in red: Although you have treated the x as b, you should actually use the x=b to get
mb= -3b(b-2)
or
m = -3(b-2)
which would give
b = 2 - m/3
and not your equation of b = 6 - m/3.

Since the point (b, A) is also on the line f(x)=mx, we must have
f(x) = A x /b,
that is a line which goes though vertical values 0 (at x=0) and A (at x=b). This gives another equation for m
m = A/b
or
A/b = -3(b-2)
or
A = -3b(b-2)
which you should expect since the point A is on both the parabola and line. Now integrate
$\displaystyle \int_0^b -3x(x-2)\, dx\, - \int_0^b\, mx\, =\, 2\, units^2$

 

[kais]

Im sorry to be annoying mate, but im just not making any progress. I know what the answer is but i cant make my integral actually reach it

so far ive got
∫-3x(x-2) dx
cancels down into
∫ -3x^2 - 6x dx
which the anti derivative is
[-x^3 +3x^2]

but im not sure how to integrate the mx. Sorry again and i really appreciate all the help so far

 

[Ishuda]

Im sorry to be annoying mate, but im just not making any progress. I know what the answer is but i cant make my integral actually reach it

so far ive got
∫-3x(x-2) dx
cancels down into
∫ -3x^2 - 6x dx
which the anti derivative is
[-x^3 +3x^2]

but im not sure how to integrate the mx. Sorry again and i really appreciate all the help so far

The m is just a constant so you can treat it as m times the integral of x. Like you would have $\displaystyle \int 3x\, dx\, =\, 3\, \int\, x\, dx$, you would have $\displaystyle \int mx\, dx\, =\, m \, \int\, x\, dx$