Eva is solving the equation 4(x+32)=8
1 answer:
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now, what's the slope of a line parallel to that one above? well, parallel lines have exactly the same slope.
We first obtain the equation of the lines bounding R.
For the line with points (0, 0) and (8, 1), the equation is given by:
For the line with points (0, 0) and (1, 8), the equation is given by:
For the line with points (8, 1) and (1, 8), the equation is given by:
The Jacobian determinant is given by
The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v
Therefore, the integration is given by:
![63 \int\limits^1_0 \int\limits^{1}_0 {(5u-23v)} \, dudv =63 \int\limits^1_0\left[\frac{5}{2}u^2-23uv\right]^{1}_0 \\ \\ =63\int\limits^1_0(\frac{5}{2}-23v)dv=63\left[\frac{5}{2}v-\frac{23}{2}v^2\right]^1_0=63\left(\frac{5}{2}-\frac{23}{2}\right) \\ \\ =63(-9)=|-576|=576](https://tex.z-dn.net/?f=63%20%5Cint%5Climits%5E1_0%20%5Cint%5Climits%5E%7B1%7D_0%20%7B%285u-23v%29%7D%20%5C%2C%20dudv%20%3D63%20%5Cint%5Climits%5E1_0%5Cleft%5B%5Cfrac%7B5%7D%7B2%7Du%5E2-23uv%5Cright%5D%5E%7B1%7D_0%20%5C%5C%20%20%5C%5C%20%3D63%5Cint%5Climits%5E1_0%28%5Cfrac%7B5%7D%7B2%7D-23v%29dv%3D63%5Cleft%5B%5Cfrac%7B5%7D%7B2%7Dv-%5Cfrac%7B23%7D%7B2%7Dv%5E2%5Cright%5D%5E1_0%3D63%5Cleft%28%5Cfrac%7B5%7D%7B2%7D-%5Cfrac%7B23%7D%7B2%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D63%28-9%29%3D%7C-576%7C%3D576)
For the line with points (0, 0) and (8, 1), the equation is given by:
For the line with points (0, 0) and (1, 8), the equation is given by:
For the line with points (8, 1) and (1, 8), the equation is given by:
The Jacobian determinant is given by
The integrand x - 3y is transformed as 8u + v - 3(u + 8v) = 8u + v - 3u - 24v = 5u - 23v
Therefore, the integration is given by: