Answers for Ch. 6 Review: Applications of the Integral

Transcription

1 Answers for Ch. 6 Review: Applications of the Integral. The formula for the average value of a function, which you must have stored in your magical mathematical brain, is b b a f d. a d / / ( 8 ) 5 5. ( ). d cos d.8 [ sin ] ( sin sin ). Ooh! A calculator won t do you much good here, huh? a a ( a / / ) a / a a a d a a 5. First, the average value: + d + a d Did you catch that? You can t find the antiderivative of the fraction as it s written, but if you divide each of the numerator terms by the denominator, it s a much simpler epression to deal with. ( ) d + 6 Now, determine what value of c makes f (c) equal to. c + c c + c c c ± However, the interval in the problem is [½, ], so c is the only such value. 6. The average value: π sin d π [ cos ] π π ( cosπ ( cos ) ) π ( ( ) ( ) ) ( + ) π π π Then determine when f (c) takes on that value. sin c π c arcsin π I m starting to think I put too many of this sort of question on the review. Ugh. First the average value: 5 ( 6) d 5 5 ( 6) d. Page of

2 For this antiderivative, I m going to use u- substitution, letting u 6, so that du d. This gives 5 / 6 d u du. Notice that the limits of integration have changed; when, u () 6, and when 5, u (5) 6. / / / ( ) u 8 And now, solve the equation f (c). (c 6) / Then I ll take the square root and solve for c. c + 6 ± ± 6 c 6 ± 6 c ± 6 While both of these results are negative, only one of them is in the interval: c 6 9. In general, I would advise that you start with a graph on these. That s what I m going to do. (c 6) / Squaring both sides gives c c c The average value: ( ) u + 6, so that du d. 5 d. I ll let 6 ( + 6) 6 limits have changed: 6 gives u ( 6) + 6 6, and 5 gives u ( 5) + 6. d du, where the u u du u And, finally, solve f (c) 6. The way to find the area of a region like this is to integrate the top curve minus the b bottom curve: ( top ) case, it s a bottom d. In this ( 8 + ) d. And at this point, I m officially tired of working out antiderivatives by hand. I ll be going with a calculator for all of these power-rule sorts of things. This one comes out to. 6 ( c + 6) Cross-multiplying, ( c + 6). Page of

3 . Here s the graph:. This time the functions are in y, not, which means that the graphs are turned sideways. Here they are. Those intersection points came from the graph, but they could also be found by setting the two functions equal to each other and solving the resulting quadratic. 7 d 8 If you rotate the diagram 9, the blue one goes on top, and the red on the bottom.. A graph: As there are two separate regions here, each with a different curve on top, so it s going to take two integrals to find the area. For brevity, I will name the parabola p() and the cubic c(). The integrals ( c p ) d required are p c d + +. (Why not the red on top? Because the - coordinates of the blue are positive, and of the red are negative; we want the difference when we subtract to be a positive height on the skinny little rectangles.) ( blue ) red dy y ( y + y + 9) y dy 9 6 Page of

4 . Another cubic/parabola combination When, y ( ( )) 8, so the point is (, 8). The slope comes from the derivative. dy ( ) ( ) d, and dy ( ( ) ). d Then the equation of the line is y 8, or, equivalently, ( ) y. On to the graph. GeoGebra makes really nice looking graphs, right? Once again, I will cleverly name the parabola p() and the cubic c(). ( ) + p c d c p d Trig! The cosine is on top and the secant on the bottom. ( cos ( sec )) π d.. + Take a look at that window. If I had not known that the tangent was going to intersect the cubic a second time, I wouldn t have been able to tell from the initial graph I saw. I got it to show up on my calculator by doing a zoom fit, initially for [, ], and then going out to 6. Clearly, the top curve is the cubic and the bottom is the line. 5 (( ) ( ) ) d With volumes of revolution, it s also a good idea to start out with a graph of the two-dimensional region. The basic idea on all volume integrals is to integrate the area of a cross-section. 5. The first task is to find the tangent. That requires a point and a slope. First the point. Page of

5 When that region is revolved about the - ais, the cross-sections are circles, and the integral will look like π R d, where the radius here is +. 6 π ( + ) d π.5. 5 Just for good measure, here s the D version. y z ( π ( + ) π ( + ) ) d.59 Here s a D version. z y 8 5 π 7. The D graph: 8. For this set of problems, the ais of revolution has moved away from the - or y-ais. However, the same (πr πr ) integrand idea still applies. This is more interesting than the last one. When this is revolved about the -ais, there will be a gap. This is something that uses the so-called washer method. The cross sections are circles with holes in them, and the basic set-up is ( π π ) R r d, where R is the outer radius, in this case from the -ais to the line, and r is the inner radius, from the - ais to the parabola. The outer radius goes from y to the square root (in red below), and the inner radius from y to the quadratic (in blue). Page 5 of

6 9. These are turned sideways again, functions of y rather than. Since the distance from the -ais to a point on the graph of each function is its y-coordinate, the radii here are greater than that by one unit. That means each radius is + y. For the red radius, R + ( + ) +, and for the blue, r ( π ( + ) π ( + ) ) 9 π 5.. In D: d y The radii are + in each case. Substituting, the outer radius becomes + y + y, in red, and the inner radius is + y, in blue. The integral is therefore (( ) ( + ) ) π y y dy 8 5 π You ll notice that I factored out the π this time. D: Page 6 of

7 . Here s a graph. D: y This time the radii are a little trickier to figure out. Well, the inner radius isn t it s zero, since there s going to be no gap between the shaded region and the ais of revolution. Here s a color-coded version to help with the outer radius.. More formulas In each case, the green segment has length y, since it goes from the -ais to the curve. Sometimes y is negative, and sometimes it s positive. The red segment is a radius with length + y, since the ais of revolution is two units from the -ais and y, being negative, makes it shorter. The blue radius is also + y: in this case, y is positive, and increases that two-unit distance from the ais of revolution to the -ais by y. π ( + ) π + ( ) y d d 5 π.6. This one is a bit more challenging. The inner radius always goes from to the cubic curve, and is y + y +. The outer radius changes, though. While it s also, sometimes (so R ), and sometimes y + (so R y + ). In the diagram below, the blue segments both show the inner radius to the cubic. The red shows the outer radius to the slanted line, and the green shows the outer radius to the vertical line. Page 7 of

8 like b side d a, where a and b are the places that starts and ends at and side is the length of that cross-section perpendicular to the -ais. This is going to take two integrals, both in terms of y. π (( + ) ( + ) ) ( ( y ) ) dy.5 y y dy π If this problem seems really obnoious, I agree. It s harder than what you d get on a test (although certainly you should be able to follow what s going on). The difficulty is in the ugliness of the epressions more than anything else. Parentheses can get oppressive. + The red segment with length y in the circle is the same red segment as in the square. Thus the integral becomes y d y d. Of course, we 9 need the variable in the integrand to match the d, and the equation for the circle is how we make that happen: if + y 8, then y 8. 9 ( 8 ) d This net set of three questions are all based on the same region: a circle with radius 9 centered at the origin. I ll be needing the equation for that: + y 8. If you don t understand where that came from, think about the definition of a circle.. Cross sections perpendicular to the -ais are squares. Since we integrate the area of a cross section, it s going to be something First footnote, and I m on page 8! I keep going back and forth on whether or not to hyphenate cross-section (or, you know, cross section ). I ve decided not to worry about consistency. On the question sheet, I think I consistently decided against it, but not here. Page 8 of

9 . With equilateral triangle cross-sections, the circle is the same as before, but this time the cross-sectional shape has the area of a triangle. I ve decided to use the formula A absin C. bh d y y d y d ( ) d The integral is y y sin 6 d 9 y d ( 8 ) d Finally! These last three questions are in the classic AP free-response style. a) The area is just a straightforward integral, and the limits of integration are given. 7 d b) Revolving about y will leave a gap. I ll need washers. The outer radius in the diagram below is red and the inner is blue.. This time we have isosceles right triangles and put the hypotenuse in the plane. The area is easiest to do as A bh. I ve already run off the copies; I ll fi the question before it goes online, so if that s where you found the questions, you should only see the unambiguous version of the question. So R and r y. π ( ) R r d π d π 6 d Note that the instructions here said write but do not evaluate a Page 9 of

10 simplified integral. This is what it is asking for. c) The area of a rectangle is base times height, obviously, and this time the height is specified as four times the length of the base. The base is the green segment below, also known as y, so the height is y. y y d y d d 6. Weird, right? Two different regions. a) The tangent line is at, so first I shall find the other coordinate of the point of tangency. f () 6() + 6() 7. The point is (, 7). Then I ll get the slope, which requires a derivative. f () + 6 f () () () + 6 And then the tangent line is y 7 ( ). I could certainly leave it in this form, but I anticipate wanting slope-intercept in a moment, so that will be y + 8. b) The region S is bounded on the top by the line everywhere, but the bottom boundary is sometimes the - ais and sometimes the cubic. It s clear from the graph that the - intercepts of the cubic are at the origin and at 8 (and you can check that by calculating f () and f (8)). The -intercept of the tangent line is at, which is again something you can check algebraically. The area is then broken into two regions one for [, ], and the other for [, ], as seen in the graph below. y d 96 blue: ( + 8) red: ( + 8 ( )) 9 d Page of

11 9 8 Then the total area is c) And now for the other region. Cross sections perpendicular to the -ais are semicircles. The red bar here is a diameter. 7. This time the region is split up when the top curve changes. a) d + 8 ( ) 6 d b) Here s a graph with the radius of the region indicated. That s a horizontal radius, so we start with, but is from the y-ais to the right. Here s another graph. Since that y value is the diameter, the radius must be y ( 6 + 6) 8 +. The area of a semicircle is π r, so the required integral is 8 8 π + d. Since + r, r. That gives the integral 8 8 π r dy π dy. It has to be dy because the skinny side of the red rectangle cuts off a narrow piece of the y-ais. In order to turn the there into a formula containing y, I ll rearrange the formula for f (). y y / y Thus the integral I need is 8 / π ( y ) dy. Page of

12 Done. c) Um, the height is h(y)? I don t see a function called h in the problem! That actually makes it a simpler question. When I get to the place where I need the height, I put in h(y), and that s it. I do need to consider what the base of the rectangle is, though. The cross-sections perpendicular to the y-ais are horizontal, like the red segment. The base is therefore the right-hand -value minus the lefthand -value. We already have that the left y /. y 6 y 6 (y 6). This gives a base of / ( 6) y y, and an integral of 8 base height dy 8 / ( 6) y y h y dy. Without a function for h, there s no way to know what the rectangles look like, so no D graph. Page of