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Minh Flip Bottle
Jun 07, 2021

A calculus (-ish) kind of problem

in Math Problems

"Given the function y = x^4 - (m+1)x^2 + m. Find the value(s) of m such that the area defined by the curve y and the x-axis that is above the x-axis is equal to the total area defined by the curve y and the x-axis that is below the x-axis." (Sorry if this is a bit wordy).



4 comments
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Ian Fowler
Jun 07, 2021

I still a bit confused. Do you mean the area between the roots?

Since this a "w shaped" function for all m>-1 and the bump in the middle flattens as m ----> -1. Fort m = -1 there is no more bump and the curve has a loosely phrased "parabola-like shape" and at that point, this question doesn't make sense to me.


So I have to assume for this question you do mean the area between the roots, which are always 1, -1, sqrt(m), -sqrt(m). Also m>0 in order to have any area above the x-axis.


When m > 1 the +/-root m roots are on the outside of the 1,-1 roots

When 0 < m < 1 the +/- root m roots are between the 1,-1 roots.


So 2 answers:

m > 1: m = 5

0 < m < 1: m = 1/5

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Minh Flip Bottle
Jun 08, 2021  ·  Edited: Jun 08, 2021

Yes, the area between the roots. And by the way, m > 1 is clearly the same as m = 5. So which one? And it seems like 0 < m < 1 does not work, I checked on Desmos.

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Ian Fowler
Jun 08, 2021  ·  Edited: Jun 08, 2021

You have to break it into 2 cases:

1) 0 < m < 1 where the sqrt(m) root < the 1 root, then the bounds of integration are from 0---->1

The answer is m = 1/5 and the right side of the photo clearly proves it.


2) m > 1 where the sqrt(m) root > the 1 root, then bounds of integration are from 0 ----> sqrt(m)

The answer is m = 5. Solution is on the lower right. I couldn't be bothered busting it up into 2 integrals, 0---->1 and 1------> sqrt(5), to verify that the areas are negatives of each other. I'll leave that to you.


And by the way, m > 1 is clearly the same as m = 5. So which one?

The case restriction is m > 1 which leads to the solution m = 5.

So the question "which one?" doesn't make any sense.


This is reason #1 why doing it yourself is superior than relying on Desmos - which is clearly wrong.


At any rate, it is a good problem. Thanks.



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Minh Flip Bottle
Jun 09, 2021

Oh yeah, sorry for the stupid question. And thanks for the good solution.

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4 comments