 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# The Definite Integral - Concept

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The definite integral is an important operation in Calculus, which can be used to find the exact area under a curve. The definite integral takes the estimating of approximate areas of rectangles to its limit by using smaller and smaller rectangles, down to an infinitely small size.

I want to talk about what the definite integral is. It's a really tough concept and maybe one of the toughes- toughest concepts in all of Calculus, but it's really important. And it answers the question how do you find the exact area under the curve? Well, let's start with the area under the curve problem.
Let f of x be a non-negative and continuous function on the interval ab, that's what we have drawn here. Find the area of the region lying beneath the curve y=f of x. So below this red curve and above the x axis. So above here. From x=a to x=b. So between these 2 vertical lines. We're looking for the area, the exact area of this region.
Now before we were talking about approximating the area with rectangular sums. The idea is to take the interval form a to b and chop it up into a certain number of sub-intervals and in this case we have 4. n=4. n is the number of sub-intervals, and we use 4 rectangles to approximate the area under this curve. Now each rectangle is the same width and that width delta x is b-a the total width of the interval divided by 4, the number of rectangles. So that's what delta x is. This is a left hand sum. The left hand sum is the sum of areas of a bunch of rectangles. Each of these areas is a height times a width. Delta x is the width in this first rectangle, f of x sub 0 is the height. In the second one delta x is the width and f of x sub 1 is the height, right? We're getting the heights from the left hand x values for each rectangle. So that the height of this rectangle will be f of this value, right? That's where the rectangle touches the curve. So we sum from 0 to 3. There are 4 rectangles here but we're summing from 0 to 3 and that's a little, little confusing. This is the way summation is written in shorthand. This is called sigma notation and when the sums have a lot of terms, it's much more convenient to use this notation even though it's a little confusing. But notice every term in this sum has the form f of x sub something times delta x. This something is the index of the sum and we're going from 0 to 3, right? That represents these numbers here. Okay, this is the left hand sum and it's a an approximation of the area under the curve and as you can see this left hand sum gives me an underestimate.
Now, as you might imagine, this sum gets more accurate the more rectangles we use. So as n gets larger, we get more and more accuracy. Now let's see how they make this exact.
Let's say that we have n rectangles. So I'm not going to specify the number. Just n. Delta x, the width of each rectangle will be b-a over n. So each of these widths is b-a over n. b-a again is this entire width divided by n the number of rectangles gives me the individual width here. The left hand sum is now the sum from i=0 to n-1. Notice we don't go all the way up to n. x of n would be this number here b, and we don't use that height to find the last the the the height of the rectangle. Right? The last rectangle's f of x sub n-1 times delta x. So anyway, we sum from 0 to n-1. This is n rectangles and it's still an approximation no matter how big n is, it's still an approximation. It just gets better and better as n approaches infinity. And here's the idea. Here's where the Calculus comes in.
To get the exact area we take the limit as n approaches infinity of this left hand sum. Now notice n the number that we're taking to infinity is this value. It also happens to correspond to the number of rectangles so we're making, we're using more and more rectangles. And if the value of the left hand sum gets closer and closer to some limit, then that will be the value of my area.
So just a quick recap. We've been talking about approximating the area under a curve with a left hand sum, that's a rectangular sum that uses rectangles that reach up to the left then touch the curve at the left hand corner. But we cannot see the same thing with right hand sums. Take a look at this.
The area equals the limit as n approaches infinity of this sum. This is the left hand sum, right? The sum from i=0 to n-1 is a left hand sum. But we can also sum from i=1 to n. The sum looks exactly the same but the indices are different. We're going from 1 to n this makes it a right hand sum. It turns out that when n approaches infinity this gives me the exact same limit and that's the exact same area. So you can calculate the area either using a left or right hand sum. What we did before with the left hand sum wasn't special, we could do it for the right hand sum as well.
Now this idea of the limit that gives me the exact area, it's so important that we give it a name and it's called the definite integral. So here's my definition of definite integral.
If a function f is continuous, we define the definite integral to be and it's red. The integral from a to b of f of x with respect to x is this limit as n approaches infinity of the left hand sum or this limit as n approaches infinity of the right hand sum. Either of these limits will give you the value of this definite integral.
Now note here I didn't specify whether or not f of x was non-negative. It just has to be continuous for this definition to work. But if f of x is not negative, then this value will exactly equal the area under the curve. So let's go back to our original picture here.
Remember this problem, the area under the curve. This area is exactly given by the definite integral if f of x is non-negative and that's why it's so important. We don't have to approximate any more. We can get the exact value.