Pls help how to solve (x-2)+(2x+4)-x
1 answer:
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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>
First let's look at what that surface looks like.
Letting
<span>Letting
</span><span>Letting
</span>
Therefore
We can thus reformulate the integral as
By definition on the plane
</span>
<span>
Hence
<span>
</span>