Three points in the plane calc problem

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The discussion revolves around solving a calculus problem involving the integral of a piecewise linear function defined by three points in the plane: A(6,-3), B(11,8), and C(15,0). The user initially miscalculated the area under the curve by not properly accounting for areas above and below the x-axis. It was clarified that the area above the x-axis should be considered positive, while the area below should be negative. The correct approach involves calculating the areas of two triangles and one rectangle, then combining these areas appropriately. The user is encouraged to double-check their calculations and ensure they are applying the correct signs for the areas.
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We're doing a review this week and I'm already having problems. I don't remeber doing this problem and cannot find similar problems in the book. Here's the question:

You are given the three points in the plane A= (6-3), B = (11,8), and C=(15,0). The graph of the function f(x) consists of the two line segments AB and BC. Find the integral \int_6^{15} f(x)dx by interpreting the integral in terms of sums and/or differences of areas of elementary figures.

1.) \int_6^{15} f(x)dx = _______


Here's what i done:

the first thing i did was plugged in A,B, & C points into a graph. after that, i connected the dots and found out that there were two triangles. So i used the area of the triangle formula to find the area of each triangles and added it together. I got an answer of 43.5, but it's incorrect. I'm not good at math, so can someone point me to the right direction?
 
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Sorry,but i haven't made the calculations...But are u sure you added the areas with the right signs??The one above the Ox axis is positive,but the one below has a negative area...

Daniel.
 
What the integral does is add up all the area under the line ABC and "ABOVE THE X AXIS" .. you should have 2 triangles and 1 rectangle... you really need a graph to do this problem at your level...
 
wait... is that a A=(6,3) or A=(6,-3)... this makes a huge different...if it is a -3, you have to sum up all area above the x-axis and minis the below the x axis...
 
I bet it is A(6,-3).Else,he wouldn't have screwed up the calculations.

Daniel.
 
You may also wright analitically equation for straight line, which goes through 2 points.
If P1 (x1, y1) and P2 (x2, y2), then

(x-x1)/(x2-x1)=(y-y1)/(y2-y1) is the equation. And now You just have to integrate from x1 to x2.
Sorry for bad english.
 
yea point A is (6,-3), sorry about that typo. ok i think i see the two triangles and the rectange now. So what i did was found the total area of the two triangle above the x-axis using 1/2*b*h. after that, i found the area of the rectange under the x-axis using L*W. subtract the area of above x-axis with the below and found an answer of 21 which is incorrect. can someone show me what I'm doing incorrect? i think I'm doing the calculations correctly (double checked acouple of times).
 
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