calculusintegration
I need to find the area of the region that is bounded by $y=x^2-4$ and $y=2x-1$
I think I solved it, but I don't know what the right answer is so I'm not sure!
I got:
$$da = wl$$
$$=(x^2-4)-(2x-1)dy$$
$$=(x^2-2x-3)dy$$
$$\int{}da = \int{(x^2-2x-3)dy}$$
$$a = \frac{x^3}{3}-x^2-3x+c$$
HINT They ask for the area of the yellow region:
The areas would be given by integrals $\int_{x_1}^{x_2} \left(y_\text{top}(x) - y_\text{bottom}(x)\right) \mathrm{d} x$ with appropriate choices of boundaries $x_1$ and $x_2$ and functions $y_\text{top}(x)$ and $y_\text{bottom}(x)$.
Solving for $y$ produces:
$$ y =\frac{1}{2} \sqrt{9x^2-36} $$
What we need to find is the area of $y$ from $x=2$ to $x=3$.
Thus we put:
$$A = \frac{1}{2}\int_2^3 \sqrt{9x^2-36} dx$$
Make an "hiperbolic" substitution:
$$3x = 6 \cosh u$$
$$3dx = 6 \sinh u du$$
We get the new limits are $0$ and $b = \cosh^{-1}\frac{3}{2} = \log({\frac{3}{2} +\sqrt{\frac{5}{4}}})$
$$A = 6\int_0^b {{{\sinh }^2}udu} $$
This integral is very similar to the integral of $\sin^2 x$,
$$A = 6\int_0^b {{{\sinh }^2}udu} =6 \left. {\frac{{\sinh \left( {2u} \right) - 2u}}{4}} \right|_0^b$$
Then you have
$$A = 3\frac{{\sinh 2b - 2b}}{2}$$
After a myriad of algebraic steps I end up with
$$A = \frac{9}{4}\sqrt 5 - 6\log \phi $$
Where $$\phi$$ is the golden ratio number.
Note that
$$\eqalign{ & \log 8 - 3\log \left( {3 + \sqrt 5 } \right) = 3\log 2 - 3\log \left( {3 + \sqrt 5 } \right) = \cr & - 3\log \left( {\frac{{3 + \sqrt 5 }}{2}} \right) = - 3\log \left( {1 + \frac{{1 + \sqrt 5 }}{2}} \right) = \cr & - 3\log \left( {1 + \phi } \right) = - 3\log {\phi ^2} = - 6\log \phi \cr} $$
EDIT: Remeber the value is just half the value of the total area.
Best Answer
I take it you mean $y=x^2-4$ and $y=2x-1$.
Draw a picture. We get a familiar parabola, and a straight line. The straight line $y=2x-1$ meets the parabola where $x^2-4=2x-1$. This can be rearranged to $x^2-2x-3=0$. The quadratic factors as $(x-3)(x+1)$, so the meeting points are at $x=-1$ and $x=3$.
Note that the finite region caught between the two has the line above the curve. Thus our area is $$\int_{-1}^3\left((2x-1)-(x^2-4)\right)\,dx.$$ Before integrating, simplify the integrand a bit.