# Cal Ii

1. Jan 20, 2005

We're doing a review this week and i'm already having problems. I dont 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?

2. Jan 20, 2005

### dextercioby

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.

3. Jan 20, 2005

### vincentchan

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....

4. Jan 20, 2005

### vincentchan

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.....

5. Jan 20, 2005

### dextercioby

I bet it is $$A(6,-3)$$.Else,he wouldn't have screwed up the calculations.

Daniel.

6. Jan 20, 2005

### Yegor

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.