Use calculus to find the area of the triangle with the given vertices: (0,0) (10,1) (-1,2)
This completely blindsided me. What's going on here?
First, find the equations of the three lines that go through these points.
L1 -> the line going through (0, 0) and (-1, 2)
L2 -> the line going through (0, 0) and (10, 1)
L3 -> the line going through (-1, 2) and (10, 1)
Put all equations in slope-intercept form (y = mx + b). I'm going to call these three functions respectively.
Draw the triangle. From x = -1 to x = 0, you can find the area of the part of the triangle to the left of the y-axis by finding the integral of the difference of the two functions L3(x) - L1(x):
Then, from x = 0 to x = 10, you find the area of the part of the triangle to the right of the y-axis by finding the integral of the difference of the two functions L3(x) - L2(x):
So the complete integral will be
I would break the problem up into two pieces: the area in the triangle between x = -1 and x = 0, and the area in the triangle between x = 0 and x = 10. For each of those triangles, you're going to have to compute the following:
Then you add the two together to find your total area. Make sense? I should say that this isn't the only way to solve this problem using calculus, but it's the way I would do it.