Square:
Consider the area between the graphs and . This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals
where , , and
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First off, to find a, b and c I believe we have to find the intercepts, and to do that we need to make the equations equal to each other and solve for x. But I'm having problems because of the y values when trying to isolate them.
I'm stuck there. The square root poses a problem for me.
How can I isolate and solve for x?
By your integral setups, you are integrating over x; therefore, you need the x coordinates. Since you need xs, you need to know the the vertex.
Or you could integrate over like skeeter suggested over y. Then you will only need the y coordinates.
Did you ever get the foiling process right?
Your first answer was off as Mr. Fantastic pointed out?
Oh indeed I corrected that, I got 60 and 21 as 2 of my values, which are correct.
To find the vertex it's -b/2a. Our function is ... How would I proceed to find the vertex? Is there a different formula?
And yes I know I could integrate w/r to y, but the first part of the question asks me to do it the way portrayed in the topic title, and the 2nd part asks me to find the area using integration w/r to y.
I'm not sure that what you have above is correct. In fact the area between 2 functions is the larger function - the smaller function over some bound (either x or y)
So I would write
Let us re-arrange your equations such that
Find the intersection of these points, i.e. when y satisfies
I hope I expanded that right of Mr. Fantastic is going to jump on me!
Then simply integrate from the bounds that you find.
A little tip, if you notice the above is a nice quadratic function! We can then find the roots by factoring.