Finding x of line bisecting parabola

[TheFallen018]

Hey, I have this question I've been trying to figure out in an integration textbook. The part of the question that I'm having trouble understanding is basically this.

With the parabola below, find the x coordinate of A, if the line OA divides the shaded area into two equal parts.
View attachment 7519

The area of the parabola is 4 units^2.

The equation of the function is -3(x^2-2x))Would someone be so kind as to help me out with this one. I'm even quite sure how to approach this. Thanks.

 

[HallsofIvy]

The area under a curve, y= f(x), and above the x-axis between x= a and x= b is given by $\int_a^b f(x) dx$. Whoever gave you this problem expected you to know that! Taking $x_0$ to be the x-coordinate of point A, the area of the region to the left is $-3 \int_0^{x_0} x^2- 2x dx$. The area of the region to the right is $-3\int_{x_0}^2 x^2- 2x dx$. Do those two integrals, set them equal and solve for $x_0$.

(Now that I have said all that, I notice this can be done without any Calculus at all! A parabola is symmetric about its axis. The x-coordinate that divides the parabola into two equal parts is the x-coordinate of the vertex. you can get that by "completing the square". Calculate it both ways and see if you get the same answer.)

 

[MarkFL]

Hey, I have this question I've been trying to figure out in an integration textbook. The part of the question that I'm having trouble understanding is basically this.

With the parabola below, find the x coordinate of A, if the line OA divides the shaded area into two equal parts.The area of the parabola is 4 units^2.

The equation of the function is -3(x^2-2x))Would someone be so kind as to help me out with this one. I'm even quite sure how to approach this. Thanks.

You might find this thread helpful:

http://mathhelpboards.com/questions-other-sites-52/phyllis-question-yahoo-answers-regarding-finding-line-divides-area-into-equal-parts-7197.html

 

[Greg]

Taking $x_0$ to be the x-coordinate of point A, the area of the region to the left is $-3 \int_0^{x_0} x^2- 2x dx$. The area of the region to the right is $-3\int_{x_0}^2 x^2- 2x dx$. Do those two integrals, set them equal and solve for $x_0$

The problem states that $\overline{OA}$ divides the parabola into two equal parts, not that $x=A$ divides the parabola into two equal parts.

 

[HallsofIvy]

Yes, I misread the problem. Thanks. Taking A to have coordinates $(x_0, 6x_0- 3x_0^2)$, the line OA has equation $$y= (6- 3x_0)x$$. The area under the parabola and above that line is given by $\int_0^{x_0} 6x- 3x^2- (6- 3x_0)x dx= \int_0^{x_0} 3x_0x- 3x^2 dx$. Since you already know that the area of the parabola is 4, set that integral equal to 2 and solve for $x_0$.