Integration and antiderivatives

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1 Integration and antiderivatives 1. Evaluate the following:. True or false: d dx lim x / x et dt x cos( 1 4 t ) dt sin x dx = sin(π/) 3. True or false: the following function F (x) is an antiderivative of f(x) = 1 : x { 1 F (x) = x x < 1 1 x x > 4. True or false: Suppose f and g are continous functions on the same domain and g(x) is never zero. Then an antiderivative of f(x) g(x) is an antiderivative of f divided by an antiderivative of g. 5. Sketch the region between y = sin(x) and y = e x on the interval [, π]. Compute its exact area. What is the average value of e x over the interval [, π]? x 1. Note that for any integrable function f(x), lim x f(t) dt =. Thus we can apply l Hôpital s Rule (check all the details!) to the limit and obtain, via the fundamental theorem of calculus: x lim et dt x x cos( 1 4 t ) dt = lim e x x cos(x ) = e Where the factor of in the denominator comes from the chain rule and the final equality is obtained by continuity of the functions e x, x and cos x.. False. The definite integral is simply a number, and the derivative of a number is. 3. True. The two pieces of F differentiate to f (check this!). 4. This is false. Suppose F (x) and G(x) are antiderivatives of f and g respectively. Then by the quotient rule: ( ) F (x) f(x)g(x) F (x)g(x) = G(x) G (x) and this is not necessarily equal to f(x g(x). For example, if f(x) = x and g(x) = ex, then an antiderivative for f is of the form F (x) = x + C for C a constant and for g is G(x) = e x + D for D a constant. Then F (x) G(x) = x + C e x + D so ( ) F (x) = (x x C)e x G(x) e x + De x + D x e x = f(x) g(x) for any choice of C and D. 1

2 5. The area is given by A = (e x sin x) dx = The average value of e x is given by: e x dx sin x = [e x ] π [ cos x] π = e π = e π 3 Ave = 1 π e x dx = eπ 1 π Different methods for integrating a function Consider the function f(x) = 4 x 1. What is the domain of f?. Sketch the graph of f. What geometric shape do you notice? 3. Using basic geometry, compute f(x) dx 4. Subdivide the domain of f into 8 subintervals of equal length. Compute the upper and lower Riemann sums for this subdivision. Is this a good approximation? 1. The domain of f is [, ].. The graph of f is: This is a semicircle(!) 3. The integral will be the negative of the area between the curve and the x-axis, which is half the area of a circle of radius. Therefore f(x)dx = π.

3 4. The subintervals will have length.5. By symmetry, I can just compute the Riemann sums over four subintervals, say [,.5], [.5, 1], [1, 1.5], [1.5, ]. The function is increasing on these intervals, so the lower Riemann sum is 15 L =.5 (f() + f(.5) + f(1) + f(1.5)) = ( ) 6.79 Similarly, the upper Riemann sum is: U =.5 (f(.5) + f(1) + f(1.5) + f()) = ( ) 4.79 Given that π 6.83 the lower Riemann sum is not too bad an approximation, while the upper Riemann sum is quite bad. Area versus integral Consider the function f(x) = sin x on the interval [, π]. 1. Using the fundamental theorem of calculus, compute the value of this integral. Is this surprising? Why or why not?. Suppose instead that we want to compute the area between the graph of f and the x-axis. Will this value be different from what you obtained in part 1? How should you go about computing it? 1. By the fundamental theorem of calculus sin(x) dx = [ cos x] π = = Thinking about this in terms of areas it may seem surprising that this integral is zero, as we know this area is non-zero. However, remembering that the integral is the signed area, and noting that sin is odd and has period π in fact this result is not surprising.. We compute the area by considering the intervals where f is positive or negative separately. Where f is positive, we count the integral of f. Where f is negative, we count the opposite of the integral: Area = sin x dx sin x dx = [ cos x] π + [cos x] π π = ( 1) + (1 ) = π 3

4 Multiple Choice - Final exam Spring 17 Select the best answer to the following questions. Justify your response. 1. Suppose f is a differentiable function. Then d dx sin x f(t)dt =? (a) f (sin(x)) cos(x) (c) cos(f(x)) f (x) (b) f(sin(x)) cos(x) (d) sin(f(x)) f (x). A species of plant has growth rate moodeled by the continuous function r(t). What does r(t) dt represent? t1 t (a) The total numbers of plants at t and t 1. (b) The number of plants at time t 1. (c) The change in growth rate between t and t 1. (d) The change in the number of plants between t and t Which of the following statements is true? (a) 1 sin(x ) dx < (b) 1 sin(x ) dx = (c) 1 sin(x ) dx > (d) None of the above, since 1 sin(x ) dx does not exist. 4. Which of the following is a good approximation for 5 3 sin(x)dx? (a) 3 sin(3) sin(3.5) + 4 sin(4) sin(4.5) (b) (c) sin(5) sin(3) 5 3 (d) (sin(3.5) sin(3)) + (sin(4) sin(3.5)) + (sin(4.5) sin(4)) + (sin(5) sin(4.5)) 5. If a function q(x) is not defined at x = 1. (a) lim x 1 q(x) cannot exist. 1 sin(3) + 1 sin(3.5) + 1 sin(4) + 1 sin(4.5) 4

5 (b) lim x 1 q(x) could be. (c) lim x 1 q(x) must approach ±. (d) None of the above. 6. Suppose k(x) is a continuous function. Which of the following is equal to 1 x k(x )dx? (a) 4 1 k(u) du (b) k(u) du (c) 4 1 k(u) du (d) k(u) 1 du 7. Which of the following most clearly illustrates the Mean Value Theorem applied to the situation for a car driving from Ithaca to Watkin s Glen? (a) At some point, the acceleration of the car is equal to the rate of change in its speed. (b) At some point, the car is exactly halfway between Ithaca and Watkin s Glen. (c) At some point, the car s instantaneous speed is equal to its average speed for the trip. (d) At some point, the car s distance from Ithaca is an absolute maximum. 8. You decide to estimate e by squaring longer and longer decimal approximations of e = (a) This will work because e is a rational number. (b) This will work because y = x is a continous function. (c) This will not work because e is an irrational number. (d) This will work because y = e x is a continous function. 9. Which of the following is the best linear approximation to e x for x-values near? (a) e x 1 + x (b) e x 1 + x (c) e x 1 x (d) e x 1 ex. 1. (b) by FTC and chain rule. (d) (rate means derivative!) 3. (a) by max-min inequality 4. (d) by definition of Riemann sums 5. (b) (Think of f(x) = x 1 x 1 ) 6. (a) By substitution u = x. 1 5

6 7. (c) 8. (b) The more digits you add in, the closer you get to the value of e, and by the continuity of x, the closer you will get to e by squaring the decimal expansions. 9. (b) by definition of the linearisation of e x at. Optimisation 1. A triangle is formed by a line through the point (, 4) and the coordinate axes, as shown below. Our goal is to determine the maximum and minimum possible areas of this triangle. Don t forget to label the figure with the variables you intend to use. (,4) (a) Give an equation (in one variable!) describing the area of such a triangle. (b) Is there a minimum area for such a triangle? If so, what is it? If not, why not? What about a maximum? (a) I assume we only want triangles where x and y are positive. Let b be the base of the triangle and h be the height. Let y = ax + h be the line through (, 4) (think about why the constant is h!). Then 4 = a + h so a = 4 h. By definition, b is the value of x such that y = (why?), so = 4 h b + h and hence b = h h 4. Therefore, letting A be the area of the triangle: A = b h = h h 4 (b) The domain of A is (4, ) (think about why!) and to find minima or maxima we want to find critical points: da h(h 4) h h(h 8) = dh (h 4) = (h 4) Therefore there is a critical point at h = 8. A is negative on (4, 8) and positive on (8, ), so by the first derivative test, the area reaches a minimum at h = 8, where A(8) = 16. Since lim h 4 A(h) = and lim h A(h) = so A does not have a maximum value. 6

7 . The 8-room Mega Motel is filled to capacity when the room charge is $5 dollars per night. For each $1 increase in room charge, 4 fewer rooms are filled each night. What charge per room will result in the maximum revenue per night? Let R be the revenue for one night, let N be the number of rooms filled that night and let P be the price per room that night. Then R = N P. From the text, we have the following relationship between N and P : N = 8 4 P 5 = 1 4P 1 So R = 4P (5 P ), and the domain of R is [5, 5]. Following the usual process (see above, for example), the maximum revenue is $65, obtained for P = $ Consider the function f(x) = cos x + cos x. You want to determine if f is ever negative. (a) Explain why you need only consider values of x in the interval [, π]. (b) Is f ever negative? Justify your answer. (a) f(x) is a sum of a constant, a π-periodic function, and a π-periodic function, so it is π-periodic, and we need only study it on [, π]. (b) f (x) = 4 sin x sin(x) = 4 sin x(1 + cos(x)) f (x) = for x =, π, π. f is decreasing on (, π) and increasing on (π, π) so by the first derivative test it has a minimum at π. f(π) =, so f is never negative on [, π] and therefore never negative on (, ). 4. A rectangular sheet of 8.5 inch by 11 inch paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. (a) Write L as a function of x. (b) What value of x minimises L? (c) What is the minimum value of L? (a) Using the triangles as drawn in the figure, using Pythagoras theorem, we have the following equations: (8.5 x) + y = x z = L x and furthermore, z + y = L x. Squaring both sides and rearranging, we find: L = 7 x3 x 8.5

8 (b) The domain of L is [, 4.5) (4.5, 8.5]. Taking the derivative of L with respect to x, we get a critical point at x = = Checking the sign of the derivative, L is decreasing on (, 4.5) and (4.5, 51/8) and increasing on (51/8, 8.5). Noting that the function is negative on (, 4.5), which is not possible for L, we see by the first derivative test that L has a minimum at x = (c) Evaluating at x = 51 8 and taking square roots, we find that the minimum value of L is approximately 11 inches. In the figure, w = L x. 8

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c) xy 3 = cos(7x +5y), y 0 = y3 + 7 sin(7x +5y) 3xy sin(7x +5y) d) xe y = sin(xy), y 0 = ey + y cos(xy) x(e y cos(xy)) e) y = x ln(3x + 5), y 0

c) xy 3 = cos(7x +5y), y 0 = y3 + 7 sin(7x +5y) 3xy sin(7x +5y) d) xe y = sin(xy), y 0 = ey + y cos(xy) x(e y cos(xy)) e) y = x ln(3x + 5), y 0 Some Math 35 review problems With answers 2/6/2005 The following problems are based heavily on problems written by Professor Stephen Greenfield for his Math 35 class in spring 2005. His willingness to