7. [12 marks] a. (6 marks) Use part 1 of the Fundamental Theorem of Calculus and the chain rule to differentiate the following integral. You must show the step where you use part 1 of the FTC and the step where you use the chain rule. cosXx f tIn(4 + 3t) dt 1 (6 marks) Use the comparison properties of integrals from section 5.2 of the textbook to show 4 0. Since f is continuous on [x, x + h], the Extreme all alue Theorem says that there are numbers u and v in [x, x + h] such that f(u) = m nd f(v) = M, where m and M are the absolute minimum and maximum values of f on x + h]. (See Figure 6.) By Property 8 of integrals, we have mh = f (t) dt = Mh tis, flu)h= \\""f (t) dt 0, we caDifferentiation Rules Observe that F is a composite function. In fact, new OF " - gx) - x' + 1. then we can write y = F(x) = /(g(x)). that is, 109. We know how to differentiate both f and a, so it would be useful to have a file that tells us hog," find the derivative of F e fog in terms of the derivatives of f and g. al turns out that the derivative of the composite function fog is the product of the cillalives of Fando. This fact is one of the most important of the differentiation rules any is called the Chain Rule. It seems plausible if we interpret derivatives as rates of change regard du/de as the rate of change of is with respect to x. dy/di as the rate of change nie .With respect to w. and foris as the rate of change of y with respect to x. If a changes twice as fast as a and y changes three times as fast as it. then it seems reasonable that's changes six times as fu . and so we expect that dy dy du dx du dx The Chain Rule If g is differentiable at x and f is differentiable at g(x), then the composite function F - fog defined by F(x) = f(g(x)) is differentiable at x and F' is given by the product F(x) =f'(g(x)) . g'(x) then In Leibniz notation, if y = f(u) and u = g(x) are both differentiable functions, dy dy du dx du dx ing to a change of Ax in r, that is, COMMENTS ON THE PROOF OF THE CHAIN RULE Let Au be the change in u correspond- Au = gur + Ax) - gl) Then the corresponding change in y is It is tempting to write Ay = flu + Au) - f(u)Is known that I f(x) dx = 17 and ( f(x)dx - 12, find , f(x)dx. SOLUTION By Property 5, we have [f()dx + (" f(oldx = [ f() dx 1, f(x)dx = ["f(x) dx - [f(x) dx = 17 - 12=5 Properties 1-5 are true whether a b. The following properties, in which we compare sizes of functions and sizes of integrals, are true only if a = b. Comparison Properties of the Integral 6. If f(x) = 0 for a s x = b, then (f(x) dx = 0. 7. If f(x) = g(x) for a s x = b, then (f(x) dx = ['g(x) dx. 8. If m s f(x) = M for a s x = b, then m(b - a) = [f (x)dx = M(b - a ) If f(x) = 0, then [" f(x) dx represents the area under the graph of f, so the geometric interpretation of Property 6 is simply that areas are positive. (It also follows directly from the definition because all the quantities involved are positive.) Property 7 says that a big- ger function has a bigger integral. It follows from Properties 6 and 4 because f - g = 0. Property 8 is illustrated by Figure 16 for the case where f(x) 2 0. If f is continuous, we could take m and M to be the absolute minimum and maximum values of f on the interval [a, b]. In this case Property 8 says that the area under the graph of f is greater than the area of the rectangle with height m and less than the area of the rectangle with height M. PROOF OF PROPERTY 8 Since m = f(x) = M, Property 7 gives S'max = [' f(x)dx = S'Max Using Property 1 to evaluate the integrals on the left and right sides, we obtain m(b - a) = f (x) dx = M(b - a ) Property 8 is useful when all we want is a rough estimate of the size of an integr he bother of using the Midpoint Rule