to find out an area between curves using integration Find the area between $y^2=5x+1$ and $x^2=4y$. I have tried to find out the point of intersection, but it was a 4th power equation. How can I solve this?
 A: The vertex of the parabola $y^2 = 5x + 1$ which opens towards the +ve x-axis is at $\left( -\frac15, 0 \right)$ and the vertex of the parabola $x^2 = 4y$ which opens towards the +ve y-axis is the the origin $(0,0)$.
At the points of intersection of the two parabolae,
$$
\sqrt{5x + 1} - \frac{x^2}{4} = 0 \\
\implies f(x) = x^4 - 80x - 16 = 0.
$$
We observe that $f(-1) > 0$ and $f(0) < 0$ which means that one of the roots lies between $-1$ and $0$. It is pretty reasonable now to say that the root is somewhere in the close vicinity and to the right of the vertex of the parabola $y^2 = 5x + 1$, i.e. very close to $x=-\frac15$. Let's assume that the root lies exactly at this point.
Also observe that $f(4) < 0$ and $f(5) > 0$ which means the other root lies between $4$ and $5$.
If we define $g(x) = \sqrt{5x+1} - \frac{x^2}{4}$ and $A$ as our required area, then $A$ lies between 
$$
\int_{-\frac15}^4 g(x) \ dx
$$
and 
$$
\int_{-\frac15}^5 g(x) \ dx.
$$
This gives us a pretty tight bound: 
$$
7.2606 \text{ sq. units} < A < 7.4985 \text{ sq. units}.
$$
However, this bound can be improved pretty easily.
A: Hint: Graph, in which the vertical green line is at about 4.3735.

x = seq(-.5, 5, by=.0001)
y = x^4/16 -5*x - 1
max(x[y<0])
## 4.3735

A: There is a theorem called Greens identiy. It basically says:
$$
\int_{B}(\partial_{x}V_{1}(x,y)-\partial_{y}V_{2}(x,y)dxdy= \int_{\partial B}V_1(x,y)dy+V_2(x,y)dx=\int_{\partial B} V \cdot (dy,dx)^T
$$ 
Using $ V=(x,-y)$, you obtian $ 2\int_{B}1dxdy=2Area(B)$ and the right-hande side is just a contour integral over the boundary $\partial B$. Now you graphs form the boundary, so you know over what you have to integrate.
