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 at some point in C, then some Point C in ABF of C = 1 / B -, A, which equals the integral ABF of XDX. If we look at a picture, this graph here, if we thought about the base of the rectangle being B -, A and the height being FC, that being the average height, there's got to be some location somewhere where the area under the curve above the rectangle is the same as the area above the area under the curve that's not included. So if we thought about the area of the rectangle B -, A times the average height FC, that really equals the area under the curve. So ABF of XDX. If we then divided each side by b -, A, we'd get F of C equaling 1 / B -, a on the integral ABF of XDX. So that's the mean value theorem for a definite integral, the Fundamental theorem of Calculus Part 1. If F is continuous on the closed and bounded interval AB, then capital F of X = A to the XFTDT is continuous on AB and differentiable on the open interval AB, and its derivative is little F of X. Or the derivative of capital F of X equals DDX the integral A to XFTDT which is little F of X. So what we're really saying here is that a derivative and an anti derivative basically undo each other just like multiplication and division undo each other, or addition and subtraction undo each other. So if we start with our function little F of X, the derivative of that is capital F prime of X. So if we start with a proof, the derivative F prime X is by definition the limit as H goes to 0 of F of X + H -, F of X all over H. We can pull out the H so the limit as H goes to zero of one over HF of X + H -, F of X. Then going up to the original definition, that F capital F of X is A to the XF of TDT. So for F of X + H, we're going A to the X plus HF of TDT minus. For F of X we're going A to the XF of TDT. So if we change our bounds on the second one and make the minus into a plus, we then can see that the lower bound and the upper bound are the same variable. So we can rewrite it as the limit as H goes to 0 of 1 / H of the integral X to X + H of FTDT. So as H goes to 0, we know that X + H has to be approaching X. So now C has to be somewhere between X and X + H So if C is in between X and X + H, then C is also having to go to 0 if H is going to 0. So F of C is getting closer to F of X. Thus the limit as H goes to 0 of F of C equal F of X. The summary F prime of X is really equaling limit as H goes to 0 of 1 / H of the integral X to the X + H FTDT. Or the limit as H goes to 0 of F of C, where F of C is really just the limit as H goes to 0 of F of C is really equivalent to F of X. If we look at some examples, this first one, if we take the integral, we have five X -, X ^2 / 4 and we're going to go from the upper bound to the lower bound. So we're going to put in four 5 * 4 - 4 ^2 / 4 minus the lower bound 5 * 3 - -3 ^2 / 4. So we get 20 - 4 -, 15 -, 9 fourths. So we'd get 16 minus 15 would be 60 fourths -9 fourths. Actually, let's distribute. So we have 16 - 15 + 9 fourths, or we'd have 1 + 9 fourths or 13 fourths, the next one. So the area under the curve 5 -, X / 2, the area under the curve from -3 to 4 is going to be 13 fourths. So this next one, negative π thirds to negative π force, 4 secant squared T is going to give us 4 tangent t - π / t, and then we're just going to stick in our upper bound and our lower bound. So we're going to get 4 tangent negative π force, minus π / -π force. That's all the upper bound -4 tangent negative π thirds minus π / -π thirds. So 4 tangent negative π force would give us -4 +4. So the first part's going to go to 04 tangent negative π thirds is going to be -4 root 3 + 3. So we're going to get 4 root 3 - 3. So the integral negative π thirds to negative π force of four secant squared t + π / t ^2 DT is 4 root 3 - 3. And actually that may not be the area under the curve, that is the area between the X axis and the curve. The curve may have some above the axis and some below. The Fundamental theorem of calculus Part 2. If F is continuous at every point of the closed interval AB and if F is any anti derivative of F on AB, then the integral ABF of XDX is F of b -, F of A. Our proof here. Let's let some new function G of X equal the integral A to the A to X of FTDT. If F is any anti derivative of F on AB, then we know that capital F of X has to equal G of X plus some constant. Since F and G are both continuous on AB, we know that F of X equals the G of X + C has to also be continuous on AB. Because if we have something that's continuous and we add a constant to it, it's still continuous. So then if we have FB minus FA, that's really just G of B + C -, g of A + C. So the CS are going to cancel when we distribute the negative G of B -, G of A, G of B is going to be A to BF of TDT. G of A is just A to A of F of TDT, which is really 0. So we get A to B of FTDT. This is an extension. If we go from A constant to a function of FTDT and let that equal F of X from a constant to a function, what we do is we're actually going to take the derivative of the integral. The derivative of the integral is really just going to give us out the derivative of the function. So when we do this, what we're going to do is we're going to use the chain rule concept. We're going to put in the upper bound V of X into the function times the derivative of the upper bound function minus putting in the lower bound F of a times the derivative of the lower bound. Now what's going to happen whether it was on the upper bound or the lower bound, when we put in a constant, the derivative of a constant is always 0. So by using the chain rule concept, we literally stick in the upper bound times the derivative of the upper bound minus the lower bound times the derivative of the lower bound. So when we look at this, we get Y equaling the integral squared of X0 sine t ^2 DT is asking us to find the derivative of Y. So we take the derivative of one side, we take the derivative of the other. The derivative of an integral are going to cancel. However, the integral bounds are not both constants. So we're going to put in the upper bound sine of 0 ^2 times the derivative of the upper bound minus sine of the lower bound sqrt X ^2 times the derivative of the lower bound. So we get 0 minus sine X * 1 / 2 square roots of X or negative sine X2 square roots of X. This next one we're going to have Y equaling 0 to sine X of DT square root 1 -, t ^2 where the absolute value of X is less than π halves. So taking the derivative of one side, we're going to take the derivative of the other. The derivative and the integral are going to cancel each other out, but we have to take into account that we have functions as our bound. So we're going to put in the upper function 1 / sqrt 1 minus sine squared X times the derivative of sine X minus, putting in the lower bound one over the square root 1 -, 0 ^2 times the derivative of the lower bound. Well, the derivative of sine is cosine, and we know that one minus sine squared X is really cosine squared X. So we end up with cosine X over cosine X because the bottom one had to be positive due to the restriction that we were given in the original that the absolute value of X had to be less than π halves. So we know that cosine X over cosine X is just one. To find the area under a curve, we're going to subdivide the AB interval at the zeros of F, IE we're going to set the original function equal to 0 and figure out where it crosses the X axis. We're going to integrate F over each sub interval and then we're going to add the absolute values of the integrals. So in this first example, we get Y equal X to the one third if I set the original function equal to 0, so 0 equal X to the one third, we know X is 0. So looking at a graph of this from -1 to 8, this is absolutely not to scale. We're going to have from -1 to 0. That area is going to be underneath the X axis. If we didn't take the absolute value of that, that that area would be a negative area telling us where it's located. It's located under the X axis from zero to 8. It's above the X axis, so we're going to take the absolute value -1 to 0 of X to the one third DX plus the absolute value 0 to 8 of X to the one third DX. We're going to add 1 to the exponent and divide by the new exponent. We're going to add 1 to the exponent and divide by the new exponent for both of those pieces. So when we put in the upper bound 0 to the Four Thirds minus -1 to the Four Thirds, we're going to end up with 3/4. When we take the absolute value of that, when we have 3/4 of eight to the Four Thirds -0 to the Four Thirds, we're going to end up with 3/4 of 16. So the total area between the X axis and the curve is 12 3/4. Although by our picture we can see some of the area was under the curve and some of it was above the curve, some of it was below the X axis and some of it was above the X axis. Another example if we wanted to take the derivative of an integral, Remember once again the derivative of an integral are going to cancel each other out. It's kind of like a function inside of a function or composition function. So we're going to look at this two different ways. First, can we actually evaluate the integral of cosine TDT? And the answer is yes. What gives us out? Cosine T is a derivative sine. So we're going to have, we're going to ignore the derivative part for just a second. We're going to say sine T is square root of X to 0. Putting in the upper bound sine of square root of X minus sine of 0, sine of 0 is 0. So we have the derivative of sine square root of X. Now the derivative of sine square root of X derivative of sinus cosine times the derivative of the inside piece 1 / 2 square roots of X. Not too bad. So we took the integral. We knew the integral of cosine T. Once we had the integral and evaluated it, then we took the derivative. Well, the power of this formula or the power of this fundamental theorem is sometimes we don't know how to find the integral of the cosine T. So using the original function, this one up here, we're going to put in the upper bound cosine squared of X times the derivative of the upper bound minus the cosine of the lower bound zero times the derivative of the lower bound 0 derivative of any constant of 0. So we get cosine squared of X 1 / 2 sqrt X which is the same thing that we got when we took and evaluated the integral and then did the derivative. So when we look at the next one, the derivative of 0 to T to the fourth square to udu. Remember square root of U is really just U to the half. So they want us to actually evaluate the integral and then take the derivative. So if I add 1 to the exponent 3 halves divided by the new exponent 2/3, putting in the upper bound minus the lower bound and simplifying, we get we get the DDT of 2/3 T to the 6th. Now if we take the derivative of that, we bring down the exponent to 1 less power, so we get 4T to the fifth. Then it's wanting us to do it showing it without taking the integral, but to actually substitute the upper bound times the derivative of the upper bound minus the lower bound times derivative lower bound. So substituting in T to the 4th, we get square root of T to the fourth times the derivative of T to the fourth minus sqrt 0 times the derivative of 0 sqrt t to the 4th is just t ^2 and the derivative of T to the 4th is 4T cubed. So when we evaluate that, we get 4T to the fifth. And it is indeed the exact same thing whether I took the integral 1st and then the derivative or whether I substituted in the upper bound times derivative minus lower bound times derivative. Once again, the power comes from if we have a function that we don't know the integral of. Thank you and have a wonderful day.