From the figure shown, the interval is divided into 5 equal parts making each subinterval to be 0.2.
Part A:

The approximate the area of the region shown in the figure using the lower sums is given by:
![Area= [y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2]+[y(0.8)\times0.2] \\ +[y(1)\times0.2] \\ \\ =[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2]+[\sqrt{1-(0.6)^2}\times0.2] \\ +[\sqrt{1-(0.8)^2}\times0.2]+[\sqrt{1-(1)^2}\times0.2] \\ \\ =(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2)+(0\times0.2) \\ \\ =0.196+0.183+0.16+0.12=0.659](https://tex.z-dn.net/?f=Area%3D%20%5By%280.2%29%5Ctimes0.2%5D%2B%5By%280.4%29%5Ctimes0.2%5D%2B%5By%280.6%29%5Ctimes0.2%5D%2B%5By%280.8%29%5Ctimes0.2%5D%20%5C%5C%20%2B%5By%281%29%5Ctimes0.2%5D%20%5C%5C%20%20%5C%5C%20%3D%5B%5Csqrt%7B1-%280.2%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%280.4%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%280.6%29%5E2%7D%5Ctimes0.2%5D%20%5C%5C%20%2B%5B%5Csqrt%7B1-%280.8%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%281%29%5E2%7D%5Ctimes0.2%5D%20%5C%5C%20%20%5C%5C%20%3D%280.9798%5Ctimes0.2%29%2B%280.9165%5Ctimes0.2%29%2B%280.8%5Ctimes0.2%29%2B%280.6%5Ctimes0.2%29%2B%280%5Ctimes0.2%29%20%5C%5C%20%20%5C%5C%20%3D0.196%2B0.183%2B0.16%2B0.12%3D0.659)
Part B:
The approximate the area of the region shown in the figure using the lower sums is given by:
![Area= [y(0)\times0.2]+[y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2] \\ +[y(0.8)\times0.2] \\ \\ =[\sqrt{1-(0)^2}\times0.2]+[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2] \\ +[\sqrt{1-(0.6)^2}\times0.2] +[\sqrt{1-(0.8)^2}\times0.2] \\ \\ =(1\times0.2)+(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2) \\ \\ =0.2+0.196+0.183+0.16+0.12=0.859](https://tex.z-dn.net/?f=Area%3D%20%5By%280%29%5Ctimes0.2%5D%2B%5By%280.2%29%5Ctimes0.2%5D%2B%5By%280.4%29%5Ctimes0.2%5D%2B%5By%280.6%29%5Ctimes0.2%5D%20%5C%5C%20%2B%5By%280.8%29%5Ctimes0.2%5D%20%5C%5C%20%5C%5C%20%3D%5B%5Csqrt%7B1-%280%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%280.2%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%280.4%29%5E2%7D%5Ctimes0.2%5D%20%5C%5C%20%2B%5B%5Csqrt%7B1-%280.6%29%5E2%7D%5Ctimes0.2%5D%20%2B%5B%5Csqrt%7B1-%280.8%29%5E2%7D%5Ctimes0.2%5D%20%5C%5C%20%5C%5C%20%3D%281%5Ctimes0.2%29%2B%280.9798%5Ctimes0.2%29%2B%280.9165%5Ctimes0.2%29%2B%280.8%5Ctimes0.2%29%2B%280.6%5Ctimes0.2%29%20%5C%5C%20%5C%5C%20%3D0.2%2B0.196%2B0.183%2B0.16%2B0.12%3D0.859)
Part C:
The approximate area of the given region is given by
The question asks for the value of

where

.
First let's look at what that surface looks like.
Letting

yields

<span>Letting

yields

</span><span>Letting

yields

</span>
Therefore

is the area of the triangle defined by the three points

.
We can thus reformulate the integral as

.
By definition on the plane

thus <span>

</span>
![I=\int_{z=0}^6\left[2x+\frac{x^2}6-\frac{zx}3\right]_{x=0}^{6-z}dz=\int_{z=0}^62(6-z)+\frac{(6-z)^2}6-\frac{z(6-z)}3\right]dz](https://tex.z-dn.net/?f=I%3D%5Cint_%7Bz%3D0%7D%5E6%5Cleft%5B2x%2B%5Cfrac%7Bx%5E2%7D6-%5Cfrac%7Bzx%7D3%5Cright%5D_%7Bx%3D0%7D%5E%7B6-z%7Ddz%3D%5Cint_%7Bz%3D0%7D%5E62%286-z%29%2B%5Cfrac%7B%286-z%29%5E2%7D6-%5Cfrac%7Bz%286-z%29%7D3%5Cright%5Ddz)
<span>
![I=\int_{z=0}^6\frac{z^2}2-6z+18=\left[\frac{z^ 3}6-3z^2+18z\right]_{z=0}^6=36-108+108](https://tex.z-dn.net/?f=I%3D%5Cint_%7Bz%3D0%7D%5E6%5Cfrac%7Bz%5E2%7D2-6z%2B18%3D%5Cleft%5B%5Cfrac%7Bz%5E%203%7D6-3z%5E2%2B18z%5Cright%5D_%7Bz%3D0%7D%5E6%3D36-108%2B108)
</span>
Hence

<span>
</span>
This is classic Pythagorean solving:
Side^2 + Side^2 = Hypotenuse^2
The hypotenuse is the longest side of a triangle, and it connects the leg and the base usually. In this triangle, 18 is the hypotenuse -- so since we are solving for the unknown side, the equation is:
a^2 + 5.7^2 = 18^2 -- now solve
a^2 = 18^2 - 5.7^2
a^2 = 291.51
a = sqrt(291.51)
a = 17.07 -- because height is the square root of the difference of the squares of 18 and 5.7
The answer is choice B.
Answer:
Step-by-step explanation:
yes