From a geometric perspective, an integral works by using rectangles to approximate the area under the curve you are taking the integral over.
By making the intervals on the x axis smaller and smaller, you in turn make the rectangles thinner and thinner, which in turn gives a better approximation of the area because there is less “error” in the approximation. p. 354 in Stewart gives a good picture of this. The areas of the rectangles in the upper left grossly overapproximate the area under the curve, whereas the rectangles in the bottom right do a much better job. They are smaller, and thus so are their uniform areas. Compare the “overlap” of the pic in the upper left to that of the lower right. The rectangles do a much better job of minimizing the error.
Of course, this intuition can only get you so far. Analysis prioritizes rigor, and as such, you need to consider algebraic inequalities to get a more air-tight view of what is occurring. Look at p. 366. under the definition where it says "the precise meaning of the limit … "
Integrals are limits. Based on the definition of a Riemann sum, they are limits as n goes to infinity in particular. Now look at the inequality you see there. There are several things you should notice:
-
http://www.texify.com/img/\normalsize\!%20\int_a^b%20\!%20f(x)%20\%2C%20\mathrm{d}%20x%20.gif
and
http://www.texify.com/img/\LARGE\!\normalsize%20\sum_{i%3D1}^n%20f(x_i)%20\Delta%20x.gif
are both NUMBERS. If the symbols bother you, use algebra and let p stand for the integral part, and q stand for the Riemann sum.
-
http://www.texify.com/img/\LARGE\!\normalsize%20\sum_{i%3D1}^n%20f(x_i)%20\Delta%20x.gif
is an APPROXIMATION. Specifically, it is contingent on n, where n is the number of rectangles you decide to use to approximate the area under the curve.
Perhaps you are wondering “why rectangles?” The answer is because if you use the distance of the intervals on the x-axis, this distance is a length. Lengths have numbers … so do sides of rectangles. So we have a side of a rectangle. The other side of the rectangle is the value of
http://www.texify.com/img/\normalsize\!f(x)%20%20.gif
on the interval, which of course is a height of the curve … but it is also a height, so it too can be the height of a side of a rectangle.
But how do we make this idea of “as close to the conjectured limit as we please”? The answer lies in the delta-epsilon definition of a limit. Epsilon is some positive number representing that “as close as we please.”
Now, recall the definition of a function: I have a set of numbers called the DOMAIN and another set of numbers called the RANGE. Numbers in the range are contingent on numbers in the domain in this way: If I have a number in a range, this implies that there is some number in the domain that makes it so. So, suppose
http://www.texify.com/img/\normalsize\!f(x)%20%3D%20x%20%2B%206.gif
. Let the domain be all real numbers EXCEPT for -1. Then 5 cannot be in the range of f(x). Why? Because if it was, this would contradict the assumption that -1 is not in the domain.
Now, we take this one step further as we tie this into the definition of a limit at infinity. Making the function as “close” to a conjectured limit as we like means that there will always be some number of rectangles N with this property:
for all numbers LARGER than the number of rectangles I use, and for any choice of numbers in an interval in the domain (which of course lies on the x-axis), the amount of error (i.e.
http://www.texify.com/img/\normalsize\!\epsilon.gif
) that the number of rectangles that I chose to use to approximate the conjectured limit is as small as I want it to be.
Since
http://www.texify.com/img/\normalsize\!\epsilon.gif
is ANY positive number, it represents ANY amount of error. When you used the definition of a limit to solve limits in Calc I, your value of
http://www.texify.com/img/\normalsize\!\epsilon.gif
depended on
http://www.texify.com/img/\normalsize\!\delta%20%20.gif
By satisfying this set of inequalities, you find the limit. Since the Riemann sum is a limit at infinity, evaluating an integral is evaluating a limit.
Real analysis is “advanced” calculus, so it is assumed that you know trig, algebra, geometry, elementary calculus, and of course arithmetic. As such, there are no such books as you requested. However, I would highly recommend Euler’s “Elements of Algebra.” The way the English is written is somewhat awkward, but if you can get past that, Euler explains arithmetic/precalc concepts better than any textbook out there.
For geometry, there isn’t really any I can recommend. I would say Euclid’s “Elements,” but translations can be phrased extremely awkwardly. Still, it’s worth checking out
http://aleph0.clarku.edu/~djoyce/java/elements/elements.html
For calculus, I would recommend Howard Anton’s text. I found his explanations to be a lot better personally.
btw, If you’re in Calc III, you need to get a handle on 3D shapes, ESPECIALLY EVERYTHING IN CHAPTER 10 AND 12. You will be in trouble if you don’t, especially when you have to set up multiple integrals on 3D shapes, but can’t because you can’t visualize them. There are also certain integrals that have domains that are MUCH easier to evaluate using polar coordinates.
Actually, you need EVERYTHING in order to understand Chapter 16. You need to understand vector sums in order to get line integrals; you need to understand how the norm of a cross product determines the area of a parallelogram so you can approximate surface areas, you need to understand derivatives of vector valued functions because the tangent vectors of said function are used to make said parallelogram. Make every effort to understand each concept because it will come back to haunt you if you do not.
As usual, keep asking questions on stuff you don’t get.