hey need help with number 1

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- anonymous

hey need help with number 1

- schrodinger

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a and b

- anonymous

The integral is the area underneath the graph, and it is done by calculating the area of small rectangles. The area of those small rectangles can be calculated like this: \[\Delta xf(c)\] Where c is a point between the interval of x thats being considered. When you take the number of squares in an interval to infinity, the area gets exact.
Try to see what represents delta x and what represents f in these examples, and think that all the interval should be divided.
Try that first and tell me if you were able to do it.

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- anonymous

is funny,i don't even know the basics for this topic,am sorry about that :(

- anonymous

Ok, I'll explain it in detail.
The integral is the area, and we can aproximate it as follows:
|dw:1349809163005:dw|
Its obvious that the sum of the areas of these rectangles does not gives us the exact area, but if we make them smaller and smaller, the closer it will get to the area. To calculate these areas we need to multiply the base of the rectangle by its hight. Since we will make them smaller and smaller, it does not matter the size or how much of the real function we are missing because the error will be so small that it disapears when we take the limmit.
The hight of the rectangle is the value of the function inside of that rectangle at any point, its base is the x from the end of the rectangle minus the x of the begginig of the rectangle, and we get \[\Delta xf(c)\].
Do you understand it so far?