If you do in fact mean (as opposed to one of these being the derivative of
at some point), then integrating twice gives
From the initial conditions, we find
Eliminating , we get
Then
Variables are: a. Symbolic names made up by the programmer that represent locations in the computer's memory b. Operators that p
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The line y = x and y = -x + 4 intersect when at the point (2, 2).
Expresing y = -x + 4 in terms of x, we have x = 4 - y.
Thus, the area of the region bounded by the <span>graphs of y = x, y = −x + 4, and y = 0 is given by
![\int\limits^2_0 {(y-(4-y))} \, dy = \int\limits^2_0 {(y-4+y)} \, dy \\ \\ = \int\limits^2_0 {(2y-4)} \, dy= \left[y^2-4y\right]_0^2 =|(2)^2-4(2)| \\ \\ =|4-8|=|-4|=4](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_0%20%7B%28y-%284-y%29%29%7D%20%5C%2C%20dy%20%3D%20%5Cint%5Climits%5E2_0%20%7B%28y-4%2By%29%7D%20%5C%2C%20dy%20%5C%5C%20%20%5C%5C%20%3D%20%5Cint%5Climits%5E2_0%20%7B%282y-4%29%7D%20%5C%2C%20dy%3D%20%5Cleft%5By%5E2-4y%5Cright%5D_0%5E2%20%3D%7C%282%29%5E2-4%282%29%7C%20%5C%5C%20%20%5C%5C%20%3D%7C4-8%7C%3D%7C-4%7C%3D4)
Therefore, the area bounded by the lines is 4 square units.
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Expresing y = -x + 4 in terms of x, we have x = 4 - y.
Thus, the area of the region bounded by the <span>graphs of y = x, y = −x + 4, and y = 0 is given by
Therefore, the area bounded by the lines is 4 square units.
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