Find the area of the region bounded by the graphs of the equations f(x)=-x^(2)+4x and y=0

We are asked to find the area of the region bounded by f(x)=-x^2+4x and y=0 .
We can factor the quadratic to y=-x(x-4); the x-intercepts are 0 and 4.
The equivalent question then is to find the definite integral int_0^4 (-x^2+4x)dx .
int_0^4 (-x^2+4x)dx=-x^3/3+2x^2 |_0^4
=(-64/3+32)-(0+0)=32/3
So the area bounded by the given curve and the x-axis is 32/3.
An alternative approach is to use a Riemann sum:
The area can be found by lim_{n->oo} sum_{i=1}^{n} f(c_{i})Delta x_{i}
If we use a regular partition and choose the right-hand endpoint in each subinterval we get:
A=lim_{n->oo} sum_{i=1}^{n} (-((4i)/n)^2+4((4i)/n))(4/n)
=lim_{n->oo}[(-64)/(n^2)sum_{i=1}^n(i^2/n-i)]
=lim_{n->oo}[(-64)/n^2(1/n((n(n+1)(2n+1))/6)-(n(n+1))/2)]
=lim_{n->oo}[(-64)/n^2(1/3n^2+1/2n+1/3-n^2/2-n/2)]
=lim_{n->oo}[(-64)/3-(32)/n-(64)/n^3+32+(32)/n]
=(-64)/3+32=32/3=10.bar(6)
The graph: