What is the area bounded by $y=x^2-1 $ & $ y=kx$?

Question

I am having trouble figuring out how to solve this with $k$ as a constant instead of a given value. How would I find the numbers to put into the integral? this is what i have so far.

$\int kx-(x^2-1)\;dx$

I know how to integrate and solve, it's what the integral is bound by is the part I'm struggling with.

Answer 1

Here's a simple graph for visualization:

Notice that the area is bounded by two $x$ coordinates, which incidentally happen to be the solutions to the intersection of the parabola and the line.

They are the solutions of the equation $kx=x^2-1$.

Can you figure out the bounds now?

Answer 2

HINT

1. Find the x coordinates $x_1$ and $x_2$ of the intersection points of $y=x^2-1$ and $y=kx$
2. Evaluate

$$A=\int_{x_1}^{x_2} dx \int_{x^2-1}^{kx} dy$$

Answer 3

Solve $$x^2-1=kx$$ and get the solutions $$ x_1 = \frac {k+ \sqrt {k^2+4}}{2} $$ and $$ x_2 = \frac {k- \sqrt {k^2+4}}{2} $$

The desired area is then $$ \int _{x_2}^{x_1} (kx - x^2+1)dx = \sqrt {k^2+4}( \frac {k^2}{6} + \frac {2}{3})$$