How do i get the area under the shaded region??

Excel is giving me the best fit curve equation in the form of Xo = A*vo^2 + B*vo +C,

if i get the equation in the form of vo = ____ then i can simply integrate from 0 to Xo.

I don't understand the graph. Is Xo and Vo your axes? If so what are the boundaries on the region?

quadratic forumula will get you vo

Looking at it again it is a real mess. You have the vertical axis labeled as Xo, but then refer to it as the y axis. Then you have the horizonal axis labeled as Vo. But then you are multiplying Xo * Vo which means they are points on the axis. But then you have function that draws the curve labeled as Xo(Vo) implying they are axes.

I'll proceed with the assumption the axes are X (vertical) and V (horizontal), and Vo and Xo are points on the axes, and that your function is miswritten and should be X = A*V^2 + B*V + C

If so the area under the curve is the definite integral from 0 to Vo of: (A*V^2 + B*V + C) * dV

yea Xo is veritcal and Vo is horzitonal, but I want the area that is shaded in the picture,

i was doin Xo times Vo cuz its a rectangle, then minus the area under the curve to get the shaded region

the boundaries are at Xo and Vo

This post was edited by FamilyGuyViewer on Mar 4 2016 01:57pm

You asked for the area under the curve (the title of the thread!). Just subtract it away like you planned. And fix your notation.

okj, oi was just wondering if u can switch the function around in terms of Xo, instead of Vo.

wat if i plot the Xo values on the X axis and V on the Y axis, then will I get the shaded I needed, if i integrate directly from 0 to xo

This post was edited by FamilyGuyViewer on Mar 4 2016 09:37pm

well you can but it is going to be a nightmare solving for V(x) because it is quadratic in V.

yea you can invert your function and integrate

messy AF though

how would i invert it lmao