What does "integrate" mean (in basic terms)?

I'm not sure how good this explanation will be, but here goes:

Integration is a form of calculus. Basically, it is a method of adding up lots of tiny pieces and finding what the overall sum is. Integration can be used to show that the area of a circle is pi r² (but you will need to be comfortable with it), and other similar uses like volumes of spheres. Also, in a similar way, it is usd to calculate moments of inertia for various objects. But I expect you mean, what is meant by integration, say of x² between a and b?

Integration of functions like this is basically finding the area under the x² curve between the points x=a and x=b. If you draw this graph and shade the area, you can approximate the area by using lots of rectangles or trapeziums. But these are only estimates, with the width of each rectangle taken as a very small number. The smaller the number, the more accurate the answer. These methods are still sums however.


So now we come to integration. An integration is theoretically a sum of rectangles with the step width (usually called dx) tending to zero (ie: getting closer and closer to zero). When you 'do' an integration to evaluate the actual answer, you must use the standard rules. I won't go in to them here as there are hundreds and I don't know most of them myself. But the most basic is that if you integrate xⁿ you get xⁿ⁺¹ /(n+1) + C where C is a constant number. So if you integrate x² you get x³ /3 + C. This is integration without limits (or with the lower limit zero). If you use the limits we had of a and b, we are only wanting a small part of the area under the graph. So we have to use our answer above with x=b, but also take away the bit we don't want, before x=a. So the answer is:

b³ /3 + C - [a³ /3 + C]
As you can see the constants will cancel out.

I'm not sure if this has helped at all!

Neil M

The basic part of integration, as said above is to find areas under curves. The proccess of integration though is really changing differential equations back into regular equations. However as Neil said, it can also be used to find the area under a curve. If you know calculus notation, I can show why this is true(try drawing a curve so that you can see why this relation is)

As dx tends to zero, it becomes dx; likewise, dA becomes dA and dy we can regard as 0. Thus,

dA/dx=y
We can write a function statement to solve this for the amount of area at a given x. Write back if you understand this much...

Hope this helps,

Brad