Given that the parabola $y=x^2+1$ has a tangent line at point $P$, find the area, $S$, of the region enclosed by the parabola $y=x^2+1$, the tangent line, and the parabola $y=x^2$.
The part which I do not understand is how to express the part where it asks to find the region enclosed, as I find no way to express that given region in an integral.
Thanks in advance.
On
NOTE: In this method for finding the area enclosed, you will not need to find the actual points of intersection, see my method here.
Point $P$ can be expressed with coordinates $(a, a^2+1)$
Let $f(x)=x^2+a$
$f'(x)=2x$
$f'(a)=2a$
The equation of the tangent line is $y=2a(x-a)+a^2+1=2ax-a^2+1$
Setting this equal to $y=x^2$ to find the coordinates of intersection,
$x^2=2ax-a^2+1$
Here is the part you have gone up to, use this to confirm.
$x^2-2ax+(a^2-1)=0$
By the root coefficient relationship, where $a$ and $b$ are the roots ($a<b$),
$a+b=2a$
$ab=a^2-1$
$S=\int_a^b\left(2ax-a^2+1-x^2\right)dx=-\int_a^b\left(x^2-2ax+a^2-1\right)dx=\frac 16(b-a)^3=\frac 16 \left(\sqrt{(a+b)^2-4ab}\right)^3$
You know the values of $a+b$ and $ab$, can you solve the area enclosed from here?
You can also find the points of intersection, unlike what I did in my other answer.
Since the tangent line is $2ax-a^2+1$, you can find the intersection(s) between the tangent line and the parabola $x^2$.
Thus, solving for the intersection(s):
$x^2-2ax+a^2-1=0$
$x=a\pm 1$
You can solve your integral like this, but I prefer my other method.
$$\int_{a-1}^{a+1}\left(2ax-a^2+1-x^2\right)dx$$