In today's video, we introduce and apply the Fundamental Theorem of Calculus, which provides an elegant formula that uses antiderivatives to find exact values for definite integrals, that is, areas under curves. We discuss common antiderivatives and how they are represented using indefinite integrals, which use the same symbolism as definite integrals except the terminals are missing. The definite integral is a number representing the area under a curve over some given interval. It's represented symbolically by putting together an integration symbol, and integrand f of x representing the rule of the function, a differential dx telling us what input variable is being used throughout and terminals a and b that are endpoints of the interval. In principle, we can find the value of that definite integral by taking a certain limit with an expression known as a Riemann, sum discussed and illustrated last time. This looks like a very involved and complicated way to find areas under curves. Under certain circumstances, it seems that a miracle occurs and all this complexity can be avoided. The fundamental theorem of calculus states, that the definite integral can be evaluated as a difference of two expressions, capital F of b minus capital F of a. Where capital F of x is any antiderivative of little f of x. The terminology needs explaining which we'll do in a moment. This theorem holds, that whenever the function 'little f', whose rule appears in the integrand of the integral, is continuous over the interval from a to b, that is, you can imagine drawing the curve without lifting your pen off the paper. Before illustrating applications to this theorem, we need first to spend time clarifying and exploring carefully the notion of antiderivative. To say capital F of x is an antiderivative of little f of x, means that the derivative with respect to x is little f of x. In previous videos, you're used to taking the derivative in all sorts of settings, but here, little f of x is not being differentiated instead, it appears as a derivative itself. To say that the derivative of capital F of x is entirely equivalent to saying the capital F of x is an antiderivative of little f of x. These two statements convey the same information. There is a subtlety however, notice how little f of x is the derivative of capital F of x is an antiderivative. It turns out that antiderivatives are not unique, in fact, all antiderivatives of f of x differ by addition of a constant. I'll explain why later in the video. You'll see constants appearing a lot in integration formulae represented typically by a capital C called a constant of integration, but more about that later. The whole point of the fundamental theorem is to find areas under curves as simply as possible and the technique relies on finding antiderivatives of functions. If we're able to do this, then we get exact answers, no need for approximations. There's a twist to all of this. The process of differentiation is relatively straightforward once you've learned the basic rules like the Chain, Product and Quotient rules, which turn differentiation essentially into a mechanical procedure. Going the other way however, forming antiderivatives is far from mechanical, and requires great skill and ingenuity. Let's start to build a table with some common useful functions, their derivatives and antiderivatives. For example, here are some simple decreasing powers of x. You're familiar with their derivatives. The derivative of x is one, so the x becomes an antiderivative of one. Antiderivatives can differ by a constant. So, let's add capital C to cover all possibilities. This is the C mentioned before, the constant of integration. The derivative of x squared is 2x. So, the derivative of half x squared will be just x. So, that x squared on two becomes an antiderivative of x. Again, we had a constant C to cover all possible antiderivatives. The derivative of x cubed is three x squared. So, the derivative of one-third of x cubed, it becomes an antiderivative of x squared and we add C. These are special cases of the pattern. In general, the derivative of x to the n is nx to the n minus one and an antiderivative of x to the n is the result of adding one to the exponent, which becomes n plus one and placing n plus one in the denominator, and don't forget to add C. Now, there's an important exception, we can only divide by n plus one if it's non-zero. So, we have to stipulate that n is not equal to minus one. Let's focus then on some negative parts of x including the problematic case one on x equal to x to the minus one. Again, the derivatives are straightforward. The antiderivatives of x to the minus two and x to the minus three are taken care of by the pattern we observed earlier of the antidifferentiating x to the n. What about an antiderivative of x to the minus one. It's a non-trivial fact that the derivative of ln of x is one over x. So, ln of x is an antiderivative and as usual, we add C. But, we're not yet out of the woods, this only makes sense if x is positive. You can't take the log of a negative number, yet of course x to the minus one makes sense for negative numbers. There's a device for covering the negative case which I'll mention later. Let's consider the two circular functions. The derivatives you know, but be careful to include the minus sine when differentiating cos x. Since the derivative of sin x is cos x and sin x is an anti-derivative of cos x and don't forget the plus C. Now, the derivative of cos x is minus sin x. So, to get an antiderivative sin x, we have to get rid of the minus sign, which we do by adding a minus sign to cos x. And, sounding like a broken record, don't forget the plus C. Let's also add rows to the natural exponential and logarithm functions whose derivatives we know well. Antidifferentiating e to the x is immediate. Now, for a really tough one. An antiderivative of ln of x is xln of x minus x. You have to be very clever to guess that. There is a systematic way to find this antiderivative using a technique called integration by parts, which you'll learn about if you take courses in calculus that follow on from this course. But it's very easy to check that it actually works. If you differentiate the expression, splitting it up and using the product rule on the first piece, you will see in a few steps that the answer really does simplify to ln of x confirming that we really have found an antiderivative. We've spent time discussing antiderivatives, but we haven't used the fundamental theorem yet to find any areas under curves. The theorem tells us that the area is capital F of b minus capital F of a. Because this expression is so useful, it gets its own square bracket notation. Write f of x between the brackets with a the subscript and b the superscript, as an abbreviation for F of b minus F of a. When people see this, they just often say, evaluate capital F between a and b, which means, technically speaking to form F of b minus F of a. The theorem now has this compact form, and we say, the area under the curve little f of x for x between a and b is capital F of x evaluated between a and b, while capital F of x is any antiderivative of little f of x. Let's try it out on an example. Let's find the area under the parabola y equals x squared for x between zero and one. We already know the answer's one-third from an earlier video, so this is a good test case for this new method. The area is an anti-derivative of x squared evaluated between zero and one. But, we know an anti-derivative namely x cubed on three. So, we just evaluate this by subbing in one for x and zero for x and taking the difference, which quickly becomes one-third as expected. Notice how effortless this becomes; no complicated telescoping sums, Gauss trick or any of the other paraphernalia we've used in an earlier video. Let's look at another example from an earlier video and evaluate it by direct methods. The area under the curve y equals 12x squared minus five, again for x between zero and one. The area is just an antiderivative of 12x squared minus five evaluated at zero and one. How do we know how to put four x cubed minus five x inside the square brackets? Just as we can differentiate piece by piece and bring constants out of the front, we can perform similar manipulations to find antiderivatives. We discussed before that an antiderivative of x squared is x cubed on three, so an antiderivative of 12x squared should be 12 times this, which is four x cubed. An antiderivative of one is x, so an antiderivative of minus five should be minus five times x. Adding these two antiderivatives together gives four x cubed minus five x. Then subbing in one for x and zero for x and taking the difference, the answer quickly simplifies to negative one, which happily coincides with the answer we found in an earlier video. Let's find the area under the branch of the hyperbola y equals one on x in the first quadrant for x between one and two. The area is given by this definite integral. It's a common and harmless abuse of notation to put the integrand and the differential together as a single fraction making everything more compact. By the fundamental theorem, all we have to do is evaluate some antiderivative of one on x between one and two. But ln of x is an antiderivative, so evaluating between one and two gives log of two minus log of one, which becomes just log of two. Hence, the area we're looking for is the natural logarithm of two. Notice that, there's nothing special about two in this calculation. We can replace two by any positive real number k and the area turns out to be ln of k. This diagram implicitly assumes k is bigger than one. Suppose instead that k is less than one. The definite integral is still ln of k as before, but now integrating from one to k in fact moves from right to left, instead of left to right, so that the green area drawn here calculated in a backward direction becomes negative. This matches nicely the fact that ln of k is negative. If we really want to regard the green area as positive, we have to integrate forward from k to one, and we can swap terminals by multiplying by minus one to get negative ln of k, which is indeed positive and all is well. Let's find the area under the sine curve between Pi on two and Pi. An antiderivative is negative cos x. Evaluating between Pi on two and Pi quickly produces the answer one. Amazing. Who would have expected the area to be such a nice number? If we integrate from Pi on two all the way to two Pi, then the curve slips below the x-axis and you expect by the symmetry to add two lots of the green area, which is called pink in the diagram, to be counted negatively, so the total answer should be negative one. Let's check. We evaluate the same antiderivative, but now, between Pi on two and two Pi and quickly see that the answer simplifies to negative one as expected. Here's the fundamental theorem again. We have a natural notation to describe the antiderivative. We write capital F of x using the same expression as the definite integral but without the terminals. This is called an indefinite integral. The indefinite integral describes an antiderivative of the integrand little f of x, so is a function of x. By contrast, the definite integral is an area under the curve, so it's a real number. We mentioned earlier the fact that antiderivatives of f of x differ by a constant, this assumes all functions are continuous on a particular interval of interest. I'll briefly explain the reason. Suppose, that capital F of x and capital G of x are both antiderivatives of little f of x, so that their derivatives are both equal to f of x. We want to show that their rules are the same, but differ by a constant. The trick is to differentiate the difference capital F of x minus capital G of x, which is the difference in the derivatives, which is f of x minus f of x which is zero. If the derivative is zero, then the tangent lines to the curve with rule capital F of x minus capital G of x must be horizontal everywhere on the interval, which means, the curve must be itself just a horizontal line. So, F of x minus G of x equals some constant C. So, that f of x equals g of x plus C. Indeed, we've shown the two antiderivatives differ by a constant. This explains the origin of the constant of integration C that you will see all the time in so-called integration formula, which is just equations linking functions to the antiderivatives and use the indefinite integral notation. For example, this one just says that the general antiderivative of x is x squared on two plus C. This one says that the general antiderivative of x squared is x cubed on three plus C. These are special cases of the formula that says the general antiderivative of x to the n is x to the n plus one or the n plus one plus C, provided n does not equal minus one. It's common to use the same capital C for the constant of integration. The exceptional case to integrate x to the minus one, also written as the integral of dx on x is ln of x plus C, provided x is greater than zero. You'll also see the same thing with magnitude signs, which is a clever device for covering the case also when x is negative. Why this works is explained carefully in the notes. We also have the integral of e to the x, e to the x plus C, and two integration formulae for the circular functions. Since the derivative of tan x is sec squared x, we get that the integral of sec squared x is tan x plus C. There are many more formulae which you'll see if you open the covers of almost any calculus textbook. There are also some general principles of manipulating indefinite integrals which are made explicit in the accompanying notes. In today's video, we introduced and applied the fundamental theorem of calculus. Which provides a simple and elegant formula using antiderivatives to find exact values for definite integrals. We found antiderivatives for a number of common functions and then used some of them to try out this new method to calculate areas under a variety of curves. You also expressed antiderivatives using indefinite integrals, which uses the same symbolism as definite integrals except the terminals are missing. Then listed a number of common integration formulae. These all employ a constant of integration typically denoted by adding capital C, which arises because of the fact that all antiderivatives of a given function differ by a constant. Please read the notes and when you're ready, please attempt the exercises. Thank you very much for watching and I look forward to seeing you again soon.