Hhhhhhhhhhhhhhhhhhh the answer is 8
7) Vertex = V(2,3)
9) Vertex = V(2,1)
17) The smaller the absolute value of the coefficient of x², the larger is the opening of the parabola. (to compare, just take the absolute value of a)
Coefficients are (5), (-3) and (1):
Widest: f(x) =x², followed by f(x) = -3x² and the narrowest is f(x) =5x².
<span>Volume = 3534.29
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They are listed very weirdly so i will go according to your numbering system
2- 1, 20, 2, 10, 4, &5
1- 1, 14, 2, 7
4- 3,6,9,12,15
3- 2,4,6,8,10
6-5,10,15,20,25
5- 6,12,18,24,30
8- 7,14,21,28,35
7- 8,16,24,32,40
Looks like the integral is
where ![R=[0,1]\times[0,3]](https://tex.z-dn.net/?f=R%3D%5B0%2C1%5D%5Ctimes%5B0%2C3%5D)
. (The inclusion of 
will have no effect on the value of the integral.)
Let's split up 
into 
equally-sized rectangular subintervals, and use the bottom-left vertices of each rectangle to approximate the integral. The intervals will be partitioned as
![[0,1]=\left[0,\dfrac1m\right]\cup\left[\dfrac1m,\dfrac2m\right]\cup\cdots\cup\left[\dfrac{m-1}m,1\right]](https://tex.z-dn.net/?f=%5B0%2C1%5D%3D%5Cleft%5B0%2C%5Cdfrac1m%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac1m%2C%5Cdfrac2m%5Cright%5D%5Ccup%5Ccdots%5Ccup%5Cleft%5B%5Cdfrac%7Bm-1%7Dm%2C1%5Cright%5D)
and
![[0,3]=\left[0,\dfrac3n\right]\cup\left[\dfrac3n,\dfrac6n\right]\cup\cdots\cup\left[\dfrac{3(n-1)}n,3\right]](https://tex.z-dn.net/?f=%5B0%2C3%5D%3D%5Cleft%5B0%2C%5Cdfrac3n%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac3n%2C%5Cdfrac6n%5Cright%5D%5Ccup%5Ccdots%5Ccup%5Cleft%5B%5Cdfrac%7B3%28n-1%29%7Dn%2C3%5Cright%5D)
where the bottom-left vertices of each rectangle are given by the sequence
with 
and 
. Then the Riemann sum is
