Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Mean value theorem for definite integrals. If F is continuous on a closed and bounded interval AB, then there's some point in C within that interval where F of C = 1 / b -, A of the integral A to BF of XDX. The Fundamental Theorem of Calculus Part 1. If F is continuous on the closed and bounded interval AB, then capital F of X equals the integral ETA X of FTDT is continuous on AB and differentiable on the open interval A/B i.e. We're not saying it's differentiable at the end points, but at every point in between, and its derivative is little F of X. That is, F prime of X equals the derivative of the integral AXFTDT, which is just little F of X. So capital F prime of X equal little F of X. Proof. F prime of X by definition of a derivative, is the limit as H goes to 0 of capital F of X + H -, F of X all over H. If we factored out the 1 / H, we'd have the limit as H goes to zero of one over HF of X + H -, F of X. Now the definition of F of X + H is A to X + H of F of TDT minus A to X of F of TDT. Now remember we had a property that says we could change these bounds, but if we do, it also changes the sign. So instead of a minus, now it's going to turn into a plus and we're going to go from X to a. Now, if our A on the top and the A on the bottom of two integrals that we're adding are the same, we can rewrite it as a single integral from the base to the other base, or from the lower to the upper X to the X + H of F of TDT. So from there, as H approaches 0, then X + H has to approach X. Now, since there's some C between X and X + H, we know that C has to also be approaching X. So F of C approaches F of X. Thus the limit is H approaches 0 of F of C has to equal F of X. In summary, the derivative of capital F of X equals the limit as H approaches 0 of 1 / H from X to X + H of FTDT, the limit is H goes to 0 of F of C is just really F of X. We have an area of a rectangle B -, a times FSC equaling the integral ABF of XDX because the integral is the area under a curve. So what this picture is showing is that there's got to be some height C where the amount above the rectangle and the amount below the rectangle that weren't in the curve had to be the same. So the base distance b -, a times the height F of C has to equal the area under the curve. If we solve for F of C, we see that F of C equal 1 / b -, A to the integral ABF of XDX. If we did some examples, we have -3 to the four of 5 -, X ^2 X / 2 DX. If we figure out this integral, we would realize that that's 5 from -3 to four of five DX minus from -3 to four of X ^2 DX. So 5 is going to turn into 5X, and we're going to go from the upper bound to the lower bound minus X / 2. We're going to add 1 to the exponent, and we're going to divide by the new exponent, and that's going to also go from -3 to 4:00. So when we stick the bounds in, we have 5 * 4 - 5 * -3 - 4 ^2 / 4 - -3 ^2 over 4 so -3 That was awful to read, sorry -3. So 5 * 420 + 15 - 4 ^2 / 4 is going to give us 4 - 9 fourths. So that's going to give us 35 - 16 - 9 seven fourths. And so 35 -, 1 3/4 or 33 and a fourth. If we look at another example using some trig, what gives us out secant squared T as a derivative and that would be tangent. So for tangent T and remember π / t ^2 we could think of this π * t to the -2. So if we add 1 to the exponent and divide by the new exponent, we get negative π / t and that whole integral is going to go from negative π thirds to negative π force. So then we're just going to stick in the bounds 4 tangent negative π force minus π / -π force, IE the upper bounds that whole thing minus sticking in the lower bounds. So 4 tangent negative π thirds minus π over negative π thirds. So tangent of negative pipe force is -1 negative divided by negative is going to make that a positive. The Pi's are going to cancel, leaving us a four. So -4 + 4 is going to give us 0 for our first term here we're going to get 4 tangent of negative π thirds is going to be a negative root 3 +3. So when we distribute our negative through, we're going to get plus 4 ^2 roots of 3 -, 3. So the area under the curve 4 secant squared t + π / t ^2 from negative π thirds to negative π force would be four 3 - 3 Fundamental Theorem of calculus Part 2. If F is continuous at every point of AB on the closed and bounded interval and capital F is any anti derivative of F on AB, then A to B of F of XDX equals F of b -, F of A. The proof. Let's give some new function G of XA to the XF of TDT. If F is any anti derivative of F on AB then we know that F of X has to equal G of X plus some constant C because these functions are just being different by a constant because the derivative of a constant zero. Since F&G are both continuous on the closed bounded interval AB, then F of X has to equal G of X + C has to be continuous on A&B FB minus F of A then is just G of B + C -, G of A + C What happens is those two constants then cancel. So we get G of B -, G of A, which by our original definition, A to BFTD, T -, A to A of FTDT, but A to A is 0. So we get a BFTDT. This is an extension that's not in our book that our book uses a lot that I wanted to talk about. If we're talking about going from a constant to a new function, so A to some function V of XF of TDT would equal F of X or F prime of X is just the chain rule, the function of the inside times the derivative of the inside. So if we thought about undoing an integral, a derivative and an integral undo each other. So if we have DDX of A to some function V of XF of TDT, that's really just DDX of F of X. So if we thought about this, we would take FV of X * v prime of X minus FA of V prime of A. Basically I'm putting in my top portion for my F of T and then I'm taking the derivative of the top minus, plugging in the bottom, taking the derivative of the bottom. The derivative of the bottom though the derivative of a constant is always 0. So we end up with just the function times the derivative of the function. So we put in V of X into the F equation that was part of the integral and then take it times the derivative. So if we look at an example of this, if we have Y equaling square root of X to 0 of sine t ^2 DT. So we're going to take the derivative of each side and a derivative and an integral basically undo each other. So we're going to have DDX of Y which is just going to be DYDX on the right hand side. We're going to put in the upper limit the zero times the derivative of the upper limit minus sticking in the lower limit times the derivative of the lower. So the derivative of 0 is just zero. So that first term is going to go to 0. We know that sqrt X ^2 is going to give us X and the derivative of square root of X is 1 / 2 square roots of X. So this is all going to evaluate to negative sine X / 2 square roots of X. The derivative and the integral basically are undoing each other, but because one's a function, we have to actually evaluate it, putting in the upper bound times the derivative of the upper bound minus the lower bound times the derivative of the lower bound. If we looked at Y equaling y = 0 to sine X of DT square root over sqrt 1 -, t ^2 where X absolute value of X has to be less than π halves. We're doing that so that the sine of the absolute value of X less than π halves would have to be a positive value. So if we take the derivative of each side, we're going to get DYDX is going to equal putting in the upper bound. So one over square root 1 minus sine squared X times the derivative of the upper bound minus sticking in the lower bound, one over square root 1 -, 0 ^2 times derivative of lower bound derivative of 0 is 0. So this last term is really going to go to 0. The derivative of sine is cosine. We know by our Pythagorean identity 1 minus sine squared X is cosine squared X. The square root of cosine squared X is going to be cosine X. Cosine X over cosine X is one. We don't have to worry about whether it's positive or negative due to that original restriction. That said, the absolute value of X had to be less than π halves. To find an area under a curve, we're going to subdivide A to B at the zeros of F. We're going to integrate F over each sub interval, and then we're going to add the absolute values of the integral. So if we look at an example, we have Y equal X to the one third from -1 less than or equal to X less than or equal to 8. Y equal X to the one third is going to be a curve, and we're going to realize that it's going to cross the X axis when X is 0. So we're going to say let y = 0 and solve for X. So we're going to do an integral from -1 to 0 of X to the one third DX plus the integral zero to 8 of X to the one third DX. We're going to take the absolute value of each of these. If we think about this first piece here, if we weren't taking the absolute value, we'd get a negative value because that area is underneath the X axis. So when we evaluate this, we're going to have add 1 to the exponent, divide by the new exponent, still need the absolute value here, add 1 to the exponent, divide by the new exponent, still having the absolute value. So when we evaluate this, we're going to get 3/4 from the first piece plus 3/4 * 16 / 16 -, 0 for the second piece. Literally just sticking the upper bound minus the lower bound and evaluating. So 3/4 + 12 is going to give us 12 3/4. So the total area that's enclosed by the curve in the X axis, taking into only account the positives is going to be 12 3/4. We wanted the derivative of X from zero to square root of X of cosine TDT. Well, we can actually evaluate the integral of cosine TDT. The integral of cosine is going to be sine and then we're going to put in the upper bound minus the lower bound. So sine of square root of X minus sine of 0, sine of square root of X is going to be sine. Square root of X sine of 0 is just 0. Now we're going to take the derivative of sine square root of X. So the derivative of sine is cosine. And by the chain rule we then have to take the derivative of the square root of X which is 1 / 2 sqrt X. So we get cosine squared of X / 2 sqrt X. The other method to do this is to go immediately to sticking in the upper bound times the derivative of the upper bound minus the lower bound times the derivative of the lower bound. So cosine square root of X, the derivative of square root of X is 1 / 2 sqrt X cosine of 0 is one, but the derivative of 0 is 0. So what we see here is that we get the same answer whether we actually find the integral and then take the derivative or if we stick in the upper bound times the derivative of the upper bound minus the lower bound times the derivative of the lower bound. The power of the sticking in the functions. The second portion that we just did is that sometimes we can't evaluate the integral, the given integral. So this next one DDT from zero to T to the 4th of square root of Udu. Remember square root of U is really U to the half. So adding one to the exponent, dividing by the new exponent of T to the fourth, zero to T to the fourth, putting in the upper bound minus the lower bound, we would get T 2/3 T to the 6th. Now if we took the derivative of that, we bring down the exponent to 1 less power. So we get 4T to the fifth. Now if we do it the other method where we just put in the upper bound times the derivative of the upper bound minus sticking in the lower bound derivative lower bound. The square root of T to the fourth is t ^2. The derivative of T to the 4th is 4T cubed square root of 00 derivative of 00. So we get 4T to the fifth. Either method that we use, we got the same. Answer out. Thank you and have a wonderful day.