We're now discussing properties of definite integrals when an integrand is an even or an odd function, and the interval of which we're integrating is symmetric about the origin. That is the variable ranges between plus and minus some real number. Let's integrate the odd function sine x for x between plus and minus Pi on two. Here's the graph, and the areas under the curve have been colored green for positive and pink for negative. Clearly, the green and pink areas have the same magnitude as the regions exactly coincide under 180-degree rotation about the origin, hence, the overall area must become zero, as the positive and negative areas cancel out. We can check the answer directly using the Fundamental Theorem of Calculus. An antiderivative is cos x, which we evaluate between plus and minus Pi on two, and see that the answer quickly becomes zero, confirming the result. Consider another odd function, say y equals x cubed, and we integrate it between plus and minus one. Here's a sketch of the curve, and again the positive area is colored green and the negative area pink. Again, the regions match exactly under 180 degree rotation. So, the positive and negative areas cancel out, and the definite integral is zero. Again, we can check the answer using the fundamental theorem, quickly confirming that zero is the overall area. If we integrate Newton's serpentine curve mentioned earlier, also between plus and minus one, then, again, the positive and negative areas cancel out to give zero. These examples of special cases of a general phenomenon, no matter how complicated or not function might be, if we integrate between plus and minus a, where a is a positive number, then the positive and negative areas exactly cancel out regardless of where they appear. Any positive area to the right cancels out with a negative area to the left, obtained by rotating the curve 180 degrees about the origin. Similarly, any negative area to the right, cancels out with a positive area to the left. If the function's odd, you always get a net zero area. What happens if we integrate an even function over a symmetric interval? Consider the even function cos x integrated between plus and minus Pi on two. In this case, the curve sits above the x axis so there's no possibility of cancellation. However, the area to the left of the y axis is an exact mirror image of the area of the right because the cosine function is even. Taking the mirror image in the y axis doesn't change the sign of the areas, the entire area is just twice the area on the right. That is, two times the definite integral from zero to Pi on two. We can find this using the fundamental thermal calculus by evaluating the antiderivative sine x between zero and Pi on two, and we get a total area of two. Thus the area under the curve splits exactly into two areas, size one above to the right and to the left of the y axis. This example is a special case of a general phenomenon. To integrate an even function of interval between plus and minus a, for some real number a, we just integrate from zero to a and double our answer. This makes perfectly good sense, as any particular region under the curve to the right of zero with above or below the x axis is matched exactly by a region obtained by reflecting in the y axis since the function is even, and this mirroring effect doesn't alter the sign of the area. When the integrand is odd, the area disappears, but when the integrand is even it suffices just to integrate over the positive interval and double your answer. This fact is quite useful and often creates some economies in evaluating areas when even the functions are involved. For example, let's find the area under the witch of Maria Agnesi of the interval between plus and minus one over the square root of three. Because the function is even, it's enough to integrate from zero to one on root three and then multiply by two. An antiderivative of one over x squared plus one, you might remember is the inverse tan function, so we evaluate this between zero and one on root three. Giving Pi on six, which we multiply by two, gives a total area to be Pi on three. We finish with a very difficult example which uses just about every trick in the book and takes advantage of the knowledge we've built up regarding odd and even functions. This is not a routine example, and it's included more for interest and for those of you that enjoy a challenge. So, please don't worry if you don't follow all the details, and pause the video if I go too fast. We want to find this definite integral. Notice, the interval is between plus and minus Pi on two, which is a clue that we should be exploiting even and odd functions. The integrand, however, is very complicated. If only the strange denominator were not there, then the integrand will become the nice even function cos x, which we looked at before and found the total area to be two. Maybe we can exploit this if only we could handle the strange denominator. Well, if you separate out the denominator as a function in its own right, ignoring the cos x, you get y equals one over one plus e of minus x. It's actually not all that strange. It's an example of a curve we discuss in the last video called the logistic curve. If you apply your techniques of curve sketching, you can see it looks something like this. Where it crosses the y axis at y equals a half is an inflection, and the curve is sandwiched in between two horizontal asymptotes. The curve looks like a stretched out S-shape, almost like an integral sign falling over, an example of a sigmoid curve. It has full 180-degree rotational symmetry about the inflection point. You can spin it around 180 degrees about the y intercept, and the curve will reproduce itself. Now, 180-degree rotational symmetry characterizes the property of being odd, provided the point of rotation is the origin. This gives us an idea. If we can shift the curve down a half of a unit, then the point of rotation, the y intercept becomes the origin and will have an odd function. This phenomenon is illustrated here. Just by subtracting a half from the rule for the function in the top diagram, the function in the bottom diagram is odd. You can check the algebra if you wish to confirm that you get a new rule which is indeed odd. We don't actually need to know any details about the rule for this new function; we only need to know it's odd. Here is our original problem. We write the integrand as cos x times one over one plus e to the minus x. Then, do a trick, you've seen many times before in different contexts. We subtract and add a half to the second factor in integrand, which doesn't change the overall value. Then, we separate the interval into two pieces. The first piece has an integrand cos x times a complicated bit. The second piece is integrand that just cos x over two. Now, in the first piece, we have two factors, cos x is an even function. The second piece is just the rule for that odd function we discussed a little while ago by shifting a sigmoid curve down half a unit. Multiplying the rules for an even function by an odd function, in fact, gives an odd function. So, the entire integrand in the first piece is odd. So, the definite integral must be zero as we're integrating between plus and minus Pi on two. Hence, the first piece is zero, and in the second piece we can bring a half of the front of that familiar integral involving just cos x. So, the answer becomes a half two, which is one. Thuss the final answer to our original problem is the very pleasant number one. In today's video, we discussed even functions, whose graphs have reflectional symmetry in the y axis, and odd functions, which graphs have 180-degree rotational symmetry about the origin. The terminology arises from functions built using even and odd powers of the input variable x, and leads to surprising connections between functions and generalizations of polynomials. We saw that one builds definite integrals of intervals of the symmetric about the origin than the area under an odd function evaluates automatically to zero, and the area under the even function is twice the area of the positive interval. 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.
