00:01
In this problem, we are really using the definition of an integral, which is finding the area under a curve or between two curves.
00:09
So in this case, we are given a function, y equals x over x squared plus 1.
00:15
And we need to find the values of m such that we enclose the region of the area.
00:21
So something that we can do pretty easily is just find the derivative of this function.
00:27
And we can use the quotient rule to do that.
00:29
We would get y prime equals 1 minus x squared over x squared plus 1 squared.
00:36
Now the reason why you could do that is you could graph y and y prime, and to visualize this function, we could have some idea of what area we're enclosing.
00:47
But in this case, we don't need the derivative because we are integrating.
00:52
So we are also given the equation of a general line, y equals mx.
00:58
And so we can set mx equal to our y function.
01:02
Mx will be equal to x over x squared plus 1.
01:06
And now from here on out, it's basically just some algebra.
01:10
We can multiply to get x times the quantity, mx squared plus m equal to x.
01:17
Now we can do some factoring.
01:19
We can have x times mx squared plus m minus 1 all equal to 0.
01:26
And why did we do this factoring? well, we can have x equal to zero as one's solution.
01:34
Now, since we know x is zero, we can take the our inner function, if you will, or our inner quadratic and set it equal to zero.
01:44
Mx squared plus m minus 1 equal to 0.
01:48
Now if we do that, we can get x squared by itself by manipulating our ms.
01:55
So we would have x squared equals 1 minus m all over m.
01:59
And now if we take the square root of each side, we can get x...