Since the straight line is always above the parabola, you can just integrate between the two x values. If you were to integrate with respect to y, you would have to change integrands at the lower point where the two graphs intersect.
No, because (1) this is not a "double integral", (2) that would be integrating over a rectangle, not the region given, and (3) those numbers are wrong.Now, should I integrate from -11/50 to 0 and then from 0 to 91/50?
Well, the wouldn't "use Calculus", would it? You can "use Calculus" by treating this as an "area between graphs" problem. The straight line from (0, 0) to (-1, 11) is given by y= -11x and from (-1, 11) to (4, 1) by y= -2x+ 9. For x between -1 and 0, the triangle area is between those two graphs. The line from (0, 0) to (4, 1) is given by y= x/4. For x between 0 and 4, the triangle area is between y= -2x+ 9 (above) and y= x/4 (below).Question #2: Use Calculus to find the area of the triangle with the given vertices:
Here, I have no clue. I thought I could just find the midpoint between one of the segments and then the distance to the opposite vertex in order to get my height...?
There was something about this type of exercise in the "Introductory Tutorial to Multiple Integration" thread, but since we haven't done that in class, I'm not sure I can do in the test.
Sorry for the long read! I just need someone to point me in the right direction![/QUOTE]