I got the 2x +5 problem now im on

-x^2 + 4 from x = -2 to x = 2 and i'm off by .5

 

n=?
give me a few, let me make a quick sketch, it should help

 

n is still 4

 

aight, on it

 

not the best pic, but look at this
blue is left
red is right
green is mid

so what you are doing is finding the area of each one of those rectangles and adding them up to approximate the area created from the equation and the x-axis

since the equation is symmetric over the interval you can see that both the right side and left side approaches will give the same answer, and also one rectangle will have 0 area, since the point lies on the x-axis
the midpoint approach works better here because of this

 

which values do i use though... for 2x + 5 I used .5f(2) +.5f(2.5) + .5f(3) + .5f(3.5)

 

so for the left side you would use the first point in the interval, then use the left side of the next rectangle, so from the one i just drew you would use
-2, -1, 0, 1 as your four points to evaluate the height, and each one has a width of 1
for the right side you would use
-1, 0, 1, 2 to evaluate height and width is still 1
for midpoint use
-1.5, -.5, .5, 1.5

 

Oh ok...since the first problem had a difference of 2/4 = .5 and this one is 4/4 = 1 ...DUH it's all making sense now lol

 

yep, just braek the size of the interval into how ever many equal pieces they want

then use that as your width and find the height of each rectangle, which is just f(x) at the point, either left side, right side, or mid, then add up all the areas

this is just a chapter that helps visualize what integration means, this will never be done again, and isnt really important, but understanding what an integral is is very important

next you will find area between two equations, so instead of the x-axis as your bottom of the rectangle you use the point on another equation

its just as easy, but people get real confused at this point for some reason, just draw lots of pics and see what the hell is really happening and its cake

 

wtf is the point of this crap