 Welcome to the Grand Valley State University calculus screencasts. In this episode, we're going to learn how to use the second fundamental theorem of calculus to analyze a function that's defined by a definite integral. As a reminder, recall that the second fundamental theorem of calculus says that if we integrate little f of t dt on some interval from a constant c to a variable x, then we get an antiderivative of little f that passes through the point c0. Put another way, the second fundamental theorem of calculus tells us that if we differentiate an integral function, we get back the integrand. And this theorem is fundamental in that it tells us that differentiation and integration are closely related. More specifically, differentiation here undoes integration, and so differentiation is basically the inverse of integration. Now we're going to use the second fundamental theorem of calculus to obtain information about functions that are defined by definite integrals. So let's consider an example. Let's let big f be the function defined as the definite integral from 1 to x of the natural log of a sine squared of t plus a half dt. Now that integrand is pretty complicated, so it doesn't seem very likely that we're going to find an antiderivative for it, and that means the first fundamental theorem of calculus is going to be of much help to us in understanding this function. But we can use the second fundamental theorem of calculus to learn a lot of things about this function, and that's what we're going to do in the remainder of this video. Now the second fundamental theorem of calculus tells us that if we differentiate big f of x, that means we're differentiating this integral from 1 to x of a little f of t dt, we get back little f of x, which in this case is the natural log of the sine squared of x plus a half. Now that we know the derivative of this function big f, we can determine things like where big f is increasing and decreasing, where its critical points are. So pause the video for a moment and use this formula we just found for f prime to determine the critical points of big f on the interval from negative 2 to 3. Then resume the video when you're ready. Now f prime is the log of sine squared of x plus a half, and the sine squared of x plus a half is never less than a half, so f prime is defined everywhere. That means the only critical points of big f will occur when the log of the sine squared of x plus a half is 0. Now a log is 0 only when what's inside is 1, so we're looking for the values of x that make the sine squared of x plus a half equal to 1. And that's going to happen if we solve for sine of x when the sine of x is plus or minus root 2 over 2. Or on the interval from negative 2 to 3, that happens when x is negative pi over 4, x is pi over 4, and x is 3 pi over 4. So those are where the critical points of f are. Now we know the critical points. Let's determine the intervals where big f is increasing and decreasing. So pause the video for a moment, determine those intervals, and then classify the critical points we just found as local or global maximums or minimums or neither. Then resume the video when you're ready. Now to determine the intervals in which f is increasing or decreasing, we need to know the intervals in which f prime is positive and negative, and we just look at points that are around our critical points. So for example, to the left of negative pi over 4 is negative pi over 2, and big f prime of negative pi over 2 is positive, to the right is 0 and big f prime of 0 is negative, so f is increasing to the left of negative pi over 4 and decreasing to the right. That makes the point that x equals negative pi over 4 a relative maximum of f. We do the same thing for the other two critical points, testing points around pi over 4, testing 0 to the left, pi over 2 to the right, we see that f is decreasing to the left of pi over 4 and increasing to the right, and that makes the point that x equals pi over 4 a relative minimum of f. Finally, testing points to the left of 3 pi over 4 and to the right of 3 pi over 4, in f prime, we see that f is increasing to the left and decreasing to the right of 3 pi over 4, And that makes the point at 3pi over 4 a relative maximum of f. Now recall that f prime is defined everywhere. Pause the video for a moment and explain why that means that f has to attain an absolute maximum and an absolute minimum value on the interval negative 2 to 3. Then approximate the absolute maximum and minimum values to 3 decimal places. And resume the video when you're ready. The fact that big f prime exists everywhere means that big f is a continuous function. Call that the extreme value theorem tells us that every continuous function on a closed and bounded interval achieves its maximum and minimum value on that interval. To find the maximum and minimum values of big f on the interval, we just have to test the values of f at the critical points and compare those values to the values of f at the end points. Given the way big f is defined, we are unlikely to be able to find the exact values of f at these critical points and at the end points. However, we can approximate these values as close as we like by using, say, a middle sum. Let's approximate f at the left end point negative 2 using middle sums. f of negative 2, by definition, is the definite integral from 1 to negative 2 of the log of the sine squared of t plus a half dt. Now the middle sum using 100 subintervals is about 0.2515. Same using 200 subintervals, same using 300 subintervals. Since these three sums all agree to 3 decimal places, we can conclude that big f of negative 2 is about 0.251. If you had any difficulty calculating these middle sums, because the lower limit is actually bigger than the upper limit on the integral, remember that if we interchange the limits, that multiplies the integral by negative 1. We can use middle sums to approximate the values of f at the critical points and the other end point just as easily, and the values are as shown here. So the absolute maximum value of big f on the interval from negative 2 to 3 is about 0.619, and that occurs at x equals negative pi over 4. The absolute minimum value of f on this interval is about negative 0.021, and that occurs at the critical point pi over 4. Okay, we now have f prime, so we can calculate f double prime and learn things about big f based on what f double prime tells us. So pause the video, use this formula for f prime to calculate f double prime, and find all the inflection points of f on the interval from negative 2 to 3. Then resume the video when you're ready. We find the second derivative of f using the chain rule. We take the derivative of the log, get 1 over sine squared x plus a half times the derivative of sine squared x plus a half, that's 2 sine x cosine x. The inflection points of big f are going to occur when f double prime changes sign from positive to negative or negative to positive, and that's only going to happen when we have points where f double prime is 0. And f double prime is 0 when sine x cosine x is 0, and that's only going to happen when the sine of x is 0 or the cosine of x equals 0. Sine of x is 0 at x equals 0 in the interval from negative 2 to 3, and the cosine of x is 0 at negative pi over 2 and pi over 2 on the interval from negative 2 to 3. And f double prime changes sign at these three different points, so that means these three points are all inflection points at x equals negative pi over 2, x equals 0, and x equals pi over 2. Putting all this information together, we can draw a reasonably accurate graph of this very complicated function, big f, defined by the integral from 1 to x of ln sine squared t plus a half dt, just using the second fundamental theorem and the information we've already gleaned about what derivatives and second derivatives tell us about our function. That concludes this particular screencast, and we look forward to seeing you come back in the future.