1 Super-Brief Calculus I Review.

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1 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES Remark 0.. Much of this set of lecture notes is adapted from a combination of Rogawski s Calculus Early Transcendentals and Briggs and Cochran s Calculus, and many of the examples used appear previously in these texts. For the purposes of this class, we will regard calculus as the study of limits and limit processes. Without yet formally recalling the definition of a limit, let s recall the main ways in which one applies the concept in an introductory Calculus I course (which the student should have recently taken). Super-Brief Calculus I Review. The Derivative. Given a function f and a point x, we wish to compute the instantaneous rate of change of f at x. Geometrically, we may identify this instantaneous rate of change with the slope of the line tangent to the graph of f at the point x. How can we compute such a thing? The process is achieved in just two major steps: a geometric approximation step and then a limit step. () Use the slope formula y 2 y to compute the average rate of change between x and any x 2 x other nearby point, which we denote x + h. Since the function f determines the y-values at x and x + h, the formula we obtain is the familiar difference quotient below: f(x + h) f(x) h If we take the distance h to be a very small number, then this slope should be a very close approximation to the slope we hope to compute. Indeed, in general a smaller choice of h should lead to a closer approximation. (2) Since step () gives us an infinite family of arbitrarily close approximations to the slope of the tangent line, the natural final step is to let h shrink to 0, i.e. compute the limit as h 0 of our approximations. So we obtain the formula f (x) = lim h 0 f(x + h) f(x) h When it exists, we call this value f (x) the derivative of f at x, and it is precisely the slope of the tangent line we seek. The Definite Integral. Now, given a function f and two points a and b with a < b, we wish to compute the area under the curve of f, bounded by the lines x = a and x = b. Again we proceed by a geometric approximation step, and then a limit step. The student should observe the analogy between this process and the one above for computing the derivative! () First we consider any positive integer n, and we subdivide the interval [a, b] into n equal pieces, indexed by the endpoints a = x 0 < x < x 2 <... < x k <... < x n = b. The width of each piece [x k, x k ] is x = x k x k = b a n. From each piece [x k, x k ], we select any test point x k, and we imagine drawing a rectangle with base [x k, x k ], and height determined by the test point f(x k ). The area of this rectangle is obviously (height) (width)= f(x k ) x. Adding together the area of each of the n rectangles, we obtain the Riemann sum:

2 2 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES n f(x k ) x k= This Riemann sum gives the approximate area of the region bounded by the curve of f. It is easy to see that, in most cases, choosing a finer mesh, i.e. subdividing [a, b] into more, smaller pieces, will yield a closer approximation to the actual area. In other words, choosing larger values of n (equivalently, smaller values of x) should lead to a closer approximation to the area we hope to compute. (2) Now, again, step () yields for us an infinite family of arbitrarily close approximations to the true area under the curve. So again we take a limit, this time as n, or equivalently, as x 0. We get the formula for the definite integral of f from a to b: b a f(x)dx = lim n k= n f(x k ) x This integral, when it exists, is the area under the curve of f bounded by x = a and x = b, as desired. The two processes above should have been the main focus of the reader s Calculus I course, together with the following remarkable theorem: Theorem. (The Fundamental Theorem of Calculus). Let f be any continuous function and let a be any point. Define the area function A of f (centered at a) by the rule A(x) = b a f(t)dt Stated informally, A is the function which takes a real number x for input, and for output computes the definite integral of f between the (non-variable) point a and the input x, i.e. gives the area under the curve of f between a and x. Then the following two statements hold: () A (x) = f(x); and (2) b f(x)dx = F (b) F (a), where F is any antiderivative of f. a So, again informally, the FTC (Fundamental Theorem of Calculus) may be read as: () The definite integral function of f is an antiderivative of f; and (2) any antiderivative of f also computes the definite integral of f. In other words, integration and antidifferentiation are essentially the same thing. So the (perhaps) seemingly unrelated geometric processes of taking the derivative and computing the definite integral are inverse processes of one another. Overall, there are two major goals for this course. The first is to develop a toolbox of more advanced integration skills, and use them for both purely mathematical and applied problems. The second major goal of the course is to develop new fundamental applications of the limit process, especially the development of infinite sequences and series. We begin with developing our integration techniques. 2 Integration by parts. At this point in our development of the calculus, the student should recognize that our differentiation techniques (chain rule, product rule, quotient rule, etc.) give us a tremendous amount of power to compute derivatives, but that our integration techniques are comparatively lacking. So far the only major integration tool we have developed is the substitution rule, which one should think of as a reverse chain rule. We now wish to expand the number of integration tools at our disposal by introducing a reverse

3 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 3 product rule. So consider a pair of differentiable functions u(x) and v(x). The product rule tells us that and hence d dx u(x)v(x) = u (x)v(x) + u(x)v (x) u(x)v (x) = d dx u(x)v(x) v(x)u (x) Now we antidifferentiate both sides (and invoke the Fundamental Theorem of Calculus) to obtain an integration rule, which we call integration by parts: u(x)v (x)dx = u(x)v(x) = v(x)u (x)dx Note that in the above expression we need not include any +C term, since there is an indefinite integral on both sides of the equation. We may also write the rule for integration by parts in the following memorable compact fashion, by using just a slight abuse of notation: udv = uv vdu Example 2.. Compute the following antiderivatives using integration by parts. () xe x dx (2) x sin xdx Solution to part (a). We see that the integrand xe x is a product of two functions (x and e x ). We need to choose an appropriate u and dv to apply integration by parts. There is no hard and fast rule for how to choose this, but a general guideline is that we should choose u to be whichever function becomes simpler when we derive it. Since x derives to (a simpler function ), we choose u = x and dv = e x dx. Deriving u and integrating dv dx, we get du = dx and v = e x. Plugging in u and v into the integration by parts formula, we get xe x dx = xe x e x dx = xe x e x. It may be necessary to use integration by parts more than one time, as the following examples illustrate. Example 2.2. Compute the following antiderivatives. () x 2 e x dx (2) e 2x sin xdx

4 4 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES The next example shows a trick for evaluating tough integration by parts integrals, that comes up fairly frequently in practice. Example 2.3. Compute e x cos xdx. Solution. First the student should notice why the problem is tougher than usual. Normally when we apply integration by parts, we want to choose u to be whichever function becomes simpler when we derive it. But in this case the derivative of e x is e x, which is not any simpler, and the derivative of cos x is sin x, which is also not any simpler! So the usual strategy for integration by parts actually will not be enough in and of itself. To compute this integral, we use integration by parts twice, and then a clever trick at the end to finish the problem. To start, we choose u = e x and dv = cos xdx. This gives du = e x dx and v = sin xdx. So applying integration by parts, we get e x cos xdx = e x sin x e x sin xdx. Now we apply integration by parts to the integral on the right-hand side above. Taking u = e x and dv = sin x, we get du = e x dx and v = cos x, so e x cos xdx = e x sin x e x sin xdx = e x sin x [ e x cos x + e x cos xdx] = e x sin x + e x cos x e x cos xdx. Now we have applied integration by parts twice, and at first glance it seems we have not made any progress: we started out wanting to compute e x cos xdx, and on the right-hand side above we have gotten an expression involving e x cos xdx, the very thing we are wanting to compute! While this initially seems frustrating, in fact it is good news that we have e x cos xdx appearing on both sides of the equation. Now we can be clever and just use algebra to finish the problem. Take the equality above, and add e x cos xdx to both sides to get: e x cos xdx + e x cos xdx = e x sin x + e x cos x. Collecting like terms, we get 2 e x cos xdx = e x sin x + e x cos x. Now we simply divide by 2 to find the value of e x cos xdx! e x cos xdx = 2 (ex sin x + e x cos x). General Strategy for Integrals of the Form e ax sin(bx)dx or e ax cos(bx)dx: () First use integration by parts twice in a row. This should give you an expression which has the integral you started with appearing on both the left- and the right-hand sides of the equality. Then (2) use algebra to solve

5 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 5 for the integral. The Fundamental Theorem of Calculus also implies that we may use integration by parts to compute definite integrals, using the following rule. b a u(x)v (x)dx = [u(x)v(x)] b a b a v(x)u (x)dx Example 2.4. Compute the following definite integrals. () 2 ln xdx (2) π/2 x cos 2xdx 0 3 Trigonometric Integrals. In the next few examples we will make use of the following well-known trigonometric identities, called respectively the Pythagorean identity and the half-angle formulas. sin 2 x + cos 2 x = sin 2 x = 2 ( cos 2x) cos 2 x = 2 ( + cos 2x) Example 3.. Evaluate the following integrals. () cos 5 xdx (2) sin 3 x cos 4 xdx. (3) sin 4 xdx General Strategy for Evaluating Integrals of the Form sin n cos m xdx: If n is odd use substitution rule with u = cos x and if m is odd use substitution rule with u = sin x. If both n and m are even, resort to the half-angle formulas (this is usually a bit longer). Example 3.2. Evaluate the following integrals. () sin 4 x cos 2 xdx (2) sin 3 x cos 2 xdx The student can see from the above examples that evaluating integrals involving powers of trigonometric functions often involves using identities to reduce the exponents involved to lower, more manageable levels. Of course, this process can tend to become tedious when dealing with large exponents. For this purpose, we introduce the following reduction formulas which the student may use at will from now on: sin n xdx = sinn x cos x n cos n xdx = cosn x sin x n + n n + n n sin n 2 xdx cos n 2 xdx tan n xdx = tann x n tan n 2 xdx sec n xdx = secn 2 x tan x n + n 2 sec n 2 xdx n

6 6 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES The above will more than suffice for our purposes. The student can consult Rogawski Section 7.2 for a much more complete list of reduction formulas. Deriving the Reduction Formula For Sine. Suppose n is a positive integer greater than or equal to 2, and we want to compute sin n xdx. Our strategy is to split off one copy of sine to get sin n xdx = sin n x sin xdx, and use integration by parts. Taking u = sin n x and dv = sin xdx, we get du = (n ) sin n 2 x cos x (by the chain rule) and v = cos xdx. So integration by parts yields sin n xdx = sin n x cos x + (n ) sin n 2 x cos 2 xdx. Next we note that cos 2 x = sin 2 x, and substituting above, we have sin n xdx = sin n x cos x + (n ) = sin n x cos x + (n ) = sin n x cos x + (n ) sin n 2 x( sin 2 x)dx sin n 2 xdx (n ) sin n 2 xdx (n ) sin n 2 x sin 2 xdx sin n xdx. Now since we have our desired integral sin n dx appearing on both sides of the equality above, we can use our trick from before and solve for it algebraically. We have sin n xdx + (n ) sin n xdx = n sin n xdx = sin n x cos x + (n ) sin n 2 xdx and therefore sin n xdx = n sinn x cos x + n n sin n 2 xdx which is exactly the reduction formula we sought. Example 3.3. Evaluate sec 6 xdx. Our last two problems are substitution rule problems. Example 3.4. Find tan xdx. Solution. Rewrite tan xdx = sin x cos xdx, and use the substitution rule with u = cos x. du = sin xdx, and so We have sin x tan xdx = cos x dx = u dx = ln u + C.

7 Unsubstituting u, we get CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 7 tan xdx = ln cos x + C = ln sec x + C. Example 3.5. Find sec xdx. sec x+tan x sec x+tan x, Solution. There is a clever trick to computing this integral. First we multiply the integrand by which is a form of and thus doesn t change the value. Then we use substitution rule with u = sec x + tan x, so du = (sec x tan x + sec 2 x)dx. Writing everything out: sec x + tan x sec xdx = sec x( sec x + tan x )dx sec x tan x + sec 2 x = dx sec x + tan x = u du = ln u + C. Unsubstituting u yields sec xdx = ln sec x + tan x + C. 4 Trigonometric Substitutions The main trick we will develop in this section is based on the following idea: Suppose we are trying to evaluate an integral which involves a term of the form a 2 x 2, where a is some positive real number. For this term to make sense, we must have x 2 a 2 and hence a x a. So x a, and hence we may set θ = arcsin x a. In other words, we may write x = a sin θ for some angle θ. Then if we make a change of variable, we may write a 2 x 2 = a 2 a 2 sin 2 θ = a sin 2 θ = a cos θ. In other words, by substituting a trigonometric function into some expression, it may become possible to simplify the original expression using our familiar trigonometric identities. We ll use this idea for the next few examples. Example 4.. Evaluate (6 x 2 ) 3/2 dx. Solution. We follow the basic idea outlined in the proof above, and we introduce a change of variable by letting x = 4 sin θ. Taking derivatives, we have dx = 4 cos θdθ. Now we apply the substitution rule:

8 8 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES dx = 4 cos θdθ (6 x 2 ) 3/2 (6 (4 sin θ) 2 ) 3/2 = (6( sin 2 4 cos θdθ θ)) 3/2 = 4 cos θdθ 6 3/2 (cos 2 θ) 3/2 = 64 cos 3 4 cos θdθ θ = 6 cos 2 θ dθ = 6 sec2 θdθ = tan θ + C. 6 Now to finish the problem all we need to do is unsubstitute θ, but this takes a moment s thought. Remember we initially introduced the variable θ by substituting x = 4 sin θ. Rewrite this as sin θ = x 4. Sketch a picture of a triangle with angle θ: you can draw one that has opposite side length x and hypotenuse 4 (because sine measures opposite over hypotenuse). On this triangle, the adjacent side length must be 6 x 2, by the Pythagorean theorem. So from the picture, one easily computes tan θ =. From here, we can unsubstitute θ to solve the problem: x 6 x 2 dx = (6 x 2 ) 3/2 6 tan θ + C = x 6 + C. 6 x 2 Example 4.2. Use integration to compute the area of a perfect circle of radius r. Fact 4.3. The following are all equivalent forms of the Pythagorean trigonometric identity (Why?): sin 2 x = cos 2 x cos 2 x = sin 2 x sec 2 x = tan 2 x + tan 2 x = sec 2 x General Strategy for Evaluating Integrals Using a Trigonometric Substitution: First identify whether or not a straightforward u-substitution is a better idea. If not, then () if you see a term of the form a 2 x 2 for some number a, try x = a sin θ. (2) if you see a term of the form x 2 + a 2 for some number a, try x = a tan θ. (3) if you see a term of the form a 2 x 2 for some number a, try x = a sec θ. Example 4.4. Evaluate x 2 (4 x) 5/2 dx. Example 4.5. Evaluate (+x 2 ) 2 dx. Example 4.6. Evaluate x 2 x 2 9 dx.

9 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 9 5 Integration of Rational Functions by the Method of Partial Fractions. Remark 5.. Expand motivating remarks. 3x Example 5.2. Evaluate x 2 + 2x 8 dx. Solution. The key to evaluating this integral is knowing that every rational function may be rewritten as a partial fraction expansion. Note that the denominator x 2 +2x 8 in the integrand above factors into (x + 4)(x 2). Then it is possible to find two real numbers A and B for which 3x x 2 + 2x 8 = 3x (x + 4)(x 2) = A x B x 2. The only trick is to find just what the values of A and B should be. To find the values of A and B, multiply both sides of the equation above by (x + 4)(x 2) to clear all denominators. Then combine like terms: 3x = A(x 2) + B(x + 4) = Ax 2A + Bx + 4B = (A + B)x + ( 2A + 4B) Now we have obtained the equality 3x = (A + B)x + ( 2A + 4B); it follows that we must have 3 = A + B and 0 = 2A + 4B. So we have a system of two equations in two variables; now we may resort to our algebra techniques! If we carefully solve the given system, we should get A = 2 and B =. Thus we may convert our original integral into the following, much easier problem: 3x 2 + 7x 2 Example 5.3. Evaluate x 3 x 2 2x dx. 5x 2 3x + 2 Example 5.4. Evaluate x 3 2x 2 dx. 3x x 2 + 2x 8 dx = 2 x + 4 dx + x 2 dx = 2 ln x ln x 2 + C. Note: The same techniques we have used above will work for the last example above, if we keep track of one important detail: Notice that the denominator x 3 2x 2 factors into x 2 (x 2). Since x appears as a factor with multiplicity 2; we must represent x twice in the partial fraction expansion: once as an x term, and once as an x 2 term. In other words, we will be able to find a partial fraction expansion of the form 5x 2 3x + 2 x 2 (x 2) = A x + B x 2 + C x 2, for some real numbers A, B, and C. Otherwise we may proceed as usual. 7x 2 3x + 3 Example 5.5. Evaluate (x 2)(x 2 2x + 3) dx.

10 0 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES Note: To solve the above, we must again keep track of an important detail. Notice that the polynomial x 2 2x + 3 which appears in the denominator is an irreducible quadratic, i.e. it cannot be factored into binomials with real coefficients. We may still find a partial fraction expansion for the integrand, but it will take the form 7x 2 3x + 3 (x 2)(x 2 2x + 3) = A x 2 + Bx + C x 2 2x + 3, where A, B, and C are some real numbers. A similar principle holds whenever an irreducible quadratic appears in the denominator of a rational expression. These integrals can in general be challenging to compute, and may involve both the method of completing the square (from the student s previous algebra course) and the use of the arctangent function. The following fact from a previous course will be helpful to remember: Fact 5.6. Example 5.7. Evaluate u 2 + a 2 du = a arctan u a + C z + z(z 2 + 4) dz. 4 x Example 5.8. Evaluate x(x 2 + 2) 2 dx. x 4 5x 3 + 9x 2 3x + 5 Example 5.9. Evaluate x 2 dx. 5x + 6 Remark 5.0. Expand examples: completing the square, long division needed. 6 Improper Integrals. Example 6.. Let b be any real number greater than. Compute the following integrals. () b dx (2) b x 2 dx Solution. We have b dx = [x] b = b and b x 2 dx = [ x ]b = b. Let us consider the geometric meanings of the expressions we have computed above. If we sketch the picture, the first integral b dx may be interpreted as the area of the box with the x-axis for a base, the line x = for a left side, the line y = for a top side, and the line x = b for a right side. If we allow b to grow large, i.e. we allow the right side of the box to slide further right, the integral confirms our intuition that the area of the box becomes larger and larger. It grows unboundedly, i.e. we have

11 lim b b dx = lim (b ) =. b CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES On the other hand, the second integral b x dx = 2 b behaves remarkably differently. If we compute the area bounded above by the graph of x between, say, and 2, we get 2 2 = 2. If we compute from to 3 we get 2 3 ; if we compute from to 4 we get 3 4, and so on. In general the area is increasing, which confirms our intuition. But what happens as b becomes very large? Notice that no matter how large b is, the area b will never exceed, i.e. the area does not grow unboundedly as in the previous example. Thus the region bounded by the curve x, the x-axis, and the line x = is infinitely long but has 2 b finite area. Since lim dx = lim b x2 =, we say that the region has area. b b This is our first glimpse at a phenomenon that we will consider extensively during our sequences/series unit later in the semester: that of an infinite process which yields a finite result. Definition 6.2. Let f be a continuous function. We define improper integrals with infinite bounds of integration as below, provided the given limits exist: a b f(x)dx = f(x)dx = lim b f(x)dx = lim a c a lim a b a b a f(x)dx + lim b f(x)dx for any real number a; f(x)dx for any real number b; b c f(x)dx for any real number c. If the limit exists and is finite, then we say the integral converges. If the limit does not exist, or is equal to or, then we say the integral diverges. Example 6.3. Evaluate each integral. () 0 e 3x dx (2) 0 +x 2 dx (3) x dx Example 6.4. Compute 0 xe x dx. We may also define improper integrals for functions with vertical asymptotes, as we see in the next example. Example 6.5. What should the value of 0 x dx be? Definition 6.6. Let f be a function continuous at x p and with a vertical asymptote at x = p. Define the improper integrals (with unbounded integrand) as follows: b a p a b p f(x)dx = lim c p f(x)dx = lim c p f(x)dx = lim d p + c a c a b d f(x)dx + lim d p + f(x)dx for a < p; f(x)dx for p < b; a d f(x)dx for a < p < b. Again, we say the integral converges if the limit exists and is finite; otherwise we say the integral diverges.

12 2 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES Example 6.7. Compute x 2 dx. Example 6.8. State whether the following integrals converge or diverge. If they converge, give the value of the integral. (Be careful!) () dx x /3 (2) x dx 3 Example 6.9. (2) Determine for what values of p the integral 0 Theorem 6.0. Let p > 0. Then () if p >, (2) if p <, (3) if p =, both () Determine for what values of p the integral x dx = p p but 0 x dx diverges but p 0 x dx and p 0 7 Applications: Probability x p dx diverges. x p dx = p p. x p dx diverge. x p dx converges or diverges. x p dx converges or diverges. Definition 7.. A probability density function p(x) is a non-negative function (i.e. p(x) 0 for all x) with the property that p(x)dx =. A random variable is an unknown quantity X that takes numerical values, subject to some probability distribution. For real numbers a < b, the probability that X is between a and b is denoted P (a X b). We say that X is a continuous random variable if there exists a continuous probability density function p(x) with the property that P (a X b) = b p(x)dx, for all a < b. a Example 7.2. Let p(x) = C x 2 +. () Assume p(x) is a probability density function. Find C. (2) Suppose X is a continuous random variable with density function p. Find the probability that X is between and 4. Example 7.3. Let p(x) be any probability density function, and X an associated random variable. Find the probability that X = a, where a is any real number. Definition 7.4. An exponential probability density is a function of the form p(t) = r e t/r, where r is some positive real number. The number r in the above expression is the mean or average value of a random variable X with density function p. Example 7.5. The waiting time T between customer arrivals in a drive-through fast food restaurant is a random variable with exponential probability density. If the average waiting time is 60 seconds, what is the probability that a customer will arrive within 30 to 45 seconds after another customer?

13 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 3 Definition 7.6. The standard normal density is the function p(x) = 2π e x2 /2. The above has average value 0 and standard deviation. More generally, the normal density with mean µ and standard deviation σ is the function p(x) = σ 2π e (x µ)2 /(2σ 2). If a random variable has a normal density function then we say it has a normal distribution or a Gaussian distribution. 8 Applications: Area Between Two Curves Example 8.. Find the area of the region between the functions f(x) = x 2 4x+0 and g(x) = 4x x 2, for x 3. Example 8.2. Find the area of the region bounded by the graphs of y = 8 x 2, y = 8x, and y = x. Example 8.3. Find the area enclosed by y = ln x and y = (ln x) 2. 9 Applications: Volume of a Three-Dimensional Solid Consider a 3-dimensional solid that extends in the x-direction from x = a to x = b. To estimate the volume of this solid, we could take a 3-dimensional Riemann sum: first we subdivide the interval [a, b] into n equal subintervals, thereby slicing the solid into n different 3-dimensional chunks, each of width x = b a n. Then we estimate the volume of each of the n chunks by taking the area of one cross-section (call it A(x k ), for k n), and multiplying it by the width x. The sum of these volume estimates gives an estimate for the volume of the entire solid, i.e. V n k= A(x k) x. Then, as in the case with integration, to find an exact volume all we need to do is take the limit of the above sum as n runs to infinity. So we get V = lim n k= n A(x k ) x = b a A(x)dx where A(x) is a function which takes a point x in [a, b] for input and returns the area of the cross section of the solid at x. Example 9.. Compute the volume of the solid whose base is bounded by the curve x = 4 y 2 and the y-axis in the xy-plane, and whose vertical cross sections (perpendicular to the x-axis) are half-circles. Example 9.2. Calculate the volume V of a pyramid of height 2 m whose base is a square of side 4 m. Example 9.3. Let R be the region bounded by the curve f(x) = (x + ) 2, the x-axis, and the lines x = 0 and x = 2. Find the volume of the solid of revolution obtained by revolving R about the x-axis. The previous example suggests a general method for computing the volumes of solids of revolution. Disk Method about the x-axis. Let f be continuous with f(x) 0 on the interval [a, b]. If the region R bounded by the graph of f, the x-axis, and the lines x = a and x = b is revolved about the x-axis, the volume of the resulting solid of revolution is V = b a π(f(x))2 dx.

14 4 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES Example 9.4. The region R is bounded by the graphs of f(x) = x and g(x) = x 2 between x = 0 and x =. What is the volume of the solid that results when R is revolved about the x-axis? Washer Method about the x-axis. Let f and g be continuous functions with f(x) g(x) 0 on [a, b]. Let R by the region bounded by the curves y = f(x) and y = g(x), and the lines x = a and x = b. When R is revolved about the x-axis, the volume of the resulting solid of revolution is V = b a π(f(x)2 g(x) 2 )dx. Example 9.5. The region R is bounded by the graphs of f(x) = x and g(x) = 4 x 2. Find the volume of the solid that results when R is revolved about the line y = 3. Now let f be a continuous function, and let R be the region bounded by the graph of f, the x-axis, and the lines x = a and x = b. The washer method gave us an easy way to compute the volume of the solid generated when R is revolved around the x-axis, but what about when R is revolved about the y-axis instead? We wish to have a convenient method for measuring this kind of solid of revolution. Let s consider again a type of 3-dimensional Riemann sum. Divide the interval [a, b] up into n equal subintervals, which divides R up into n chunks. When one of these chunks is revolved about the y-axis, instead of a prism, we get a a cylindrical shell. To estimate the volume of this shell, imagine cutting it in one place and stretching it out to look like a rectangular prism. Then its volume will be given by its width, which is the length of the subinterval X, times its height f(x k ) (where x k is some point in the subinterval), times its length, which is the circumference of the circle of radius 2πx k. So the volume of each chunk is approximately 2πx k f(x k ) x. When we add up the volumes of each of the n chunks, we get an estimate for the volume of the solid, given by V n k= 2πx kf(x k ) x. Then to obtain the exact volume, we again take the limit is n, to find that V = b a 2πxf(x)dx. The above gives a computation of the simplest case, where R is bounded by just one function f, and the x-axis. The shell method given below will be true in the more general case, where R is bounded above by a function f and bounded below by a function g. Shell Method about the y-axis. Let f and g be continuous functions with f(x) g(x) on [a, b]. If R is the region bounded by the curves y = f(x) and y = g(x) between the lines x = a and x = b, the volume of the solid generated when R is revolved about the y-axis is V = b 2πx(f(x) g(x))dx. a Example 9.6. Let R be the region bounded by the graph of f(x) = sin x 2, the x-axis, and the vertical line x = π 2. Find the volume of the solid generated when R is revolved about the y-axis. Example 9.7. Let R be the region bounded by the graphs of f(x) = x(5 x) and g(x) = 8 x(5 x). Find the volume of the solid obtained by rotating R about the y-axis. Example 9.8 (Gabriel s Horn). Let R be the region bounded by the graph of y = x and the x-axis, to the right of the line x =. () What is the volume of the solid generated when R is revolved about the x-axis? (2) What is the volume of the solid generated when R is revolved about the y-axis? Example 9.9 (Washer Method about the y-axis). Let R be the region bounded by the graphs of y = x, y = 4 x, and the x-axis. Compute the volume of the solid generated by rotating R about the y-axis.

15 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 5 Example 9.0 (Shell Method about the x-axis). Let R be the region bounded by the graph of y = x 2, the y-axis, and the lines y =, y = 4. Compute the volume of the solid generated by rotating R about the x-axis. Example 9. (Volume of a Torus). Assume a > 0 and 0 < b < a are constants. Consider the circle (x a) 2 + y 2 = b 2 in the plane, which has center (a, 0) and radius b. The solid generated by revolving the interior of the circle about the y-axis is called a torus (or donut). Compute the volume of this torus. 0 Applications: Arc Length and Surface Area Fact 0.. Let f(x) be a differentiable function, and assume f (x) is continuous on [a, b]. The the arc length s of the graph of f(x) from x = a to x = b is s = b a + [f (x)] 2 dx. Example 0.2. Find the arc length s of the graph of f(x) = 2 x3 + x over [, 3]. Fact 0.3. Let f(x) be a differentiable function, and assume f(x) 0 and f (x) is continuous on [a, b]. The surface area S of the surface obtained by rotation the graph of f(x) about the x-axis for a x b is S = 2π b a f(x) + [f (x)] 2 dx. Example 0.4. Calculate the surface area of a sphere of radius R. Example 0.5. Find the surface area S of the surface obtained by rotating the graph of f(x) = x /2 3 x3/2 about the x-axis for x 3. Sequences Definition.. A sequence is an infinite list of numbers, enumerated by the natural numbers starting from. We use the following notations to refer to the sequence whose n-th term is the number a n : (a, a 2, a 3,..., a n,...) = (a n ) n= = (a n ). We should start with a few examples of sequences. Example.2. For each sequence defined below, write out the first 5 terms of the sequence. () Let a =, and for each integer n, let a n+ = 3 + a n. (2) Let f(x) = x 3. For each integer n, let a n = f(n). (3) For each integer n, let a n = 2 n. (4) For each integer n, let a n be the n-th digit after the decimal point in the decimal expansion of the number π 3. Each of the examples above clearly defines a sequence, but they are all defined by different means. For the first example, we say that the sequence (a n ) is defined by a recurrence relation, or we say that it is defined implicitly. For the second and third examples, the sequence (a n ) is defined explicitly, i.e. each term a n may be computed exactly as a well-defined function f of n. We have not defined the fourth example via a recurrence relation nor by an explicit formula, but it is still a valid sequence nonetheless. Example.3. Let (a n ) = ( 2, 5, 2, 9,...) and let (b n ) = (3, 6, 2, 24, 48,...). () Find recurrence relations that could generate the terms of the sequences (a n ) and (b n ).

16 6 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES (2) Using the relations you found, compute what the next two terms could be in each sequence. (3) For each sequence, find an explicit formula for the n-th term generated by your recurrence relation. Definition.4. If the terms of a sequence (a n ) n= become arbitrarily close to some fixed number L as n grows sufficiently large, then we say that L is the limit of (a n ), or that (a n ) converges to L. We also write lim a n = L, or (a n ) L, n so long as the latter notation does not introduce confusion about what limit we are taking. If there is no such number L then we say the sequence diverges. Example.5. Write the first four terms of each sequence. diverges? If you believe it converges, then to what value? ( 2n 2 ) () 2n 2 n= ( ) ( ) n (2) n 2 + n= (3) (cos(πn)) n= Do you believe the limit converges or (4) (cos(2πn)) n= (5) (n) n= (6) (n!) n= (7) (a n ) n=, where a = and a n+ = 2a n for all integers n. (8) (a n ) n=, where a n is the n-th digit after the decimal point in the decimal expansion of π 3 The next theorem makes it easier to compute limits of explicitly-defined sequences. Theorem.6. Suppose f is a function for which f(n) = a n for all positive integers n. If lim x f(x) = L, then lim n a n = L. Example.7. Does the converse of the above theorem hold? That is, is it true that if f(n) = a n for some function f, and lim a n = L, then lim f(x) = L? n x In addition to the above theorem, the following nice properties all hold for any convergent sequences (a n ), (b n ) with lim a n = A and lim b n = B.. n n () lim n (a n ± b n ) = A ± B; (2) lim n ca n = ca, where c is any constant; (3) lim a nb n = AB; and n a n (4) lim = A, provided B 0. n b n B Example.8. Compute the limits of the following sequences, if they exist. () (a n ) where a n = 3n3 n 3 +.

17 (2) (b n ) where b n = n + ln n n 3/2. CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 7 (3) (c n ) where c n = e n n 0. ( Theorem.9. e = lim + n ( ) n n + = lim. n n) n n Proof. We wish to compute the limit as n of the sequence (( + n) n ). Note that it is conceptually non-nontrivial to figure out what this limit should be, if it even exists, since + n shrinks to as n, but we are raising this term to the nth power, which causes the sequence to grow this can confuse our intuition. To see these conflicting processes, write out the first few terms of the sequence: 2, 9 4, 64 27, , e? ( Set L = lim + n, if this limit exists. The trick we will employ is to take the natural logarithm n n) of both sides of the equation, then use L Hospital s rule. First, since the natural log is a continous function, we get ( ln L = lim n ln + ) ln( + /n) ln( + n ) = lim = lim n n n /n n n. Note that both numerator and denominator in the last two expressions above go to ±, so it is valid for us to apply L Hospital s rule. We do so, and get: ln L = lim n n 2 +n n 2 = lim =. n + n (Let us briefly remark here that by applying L Hospital s rule above, we have implicitly used Theorem.6 in order to analyze the sequence in terms of the differentiable function which generates it.) Exponentiating both sides of the above, we get L = e ln L = e = e, which proves the theorem. Theorem.0 (Squeeze Theorem for Sequences). Let (a n ), (b n ), and (c n ) be sequences such that for some integer M, b n a n c n for all n > M. If lim n b n = lim n c n = L, then so too does lim n a n = L. R n Example.. Prove that lim = 0 for all real numbers R > 0. n n! Definition.2. If (a n ) is a sequence for which a n+ a n for every n, then we say (a n ) is nondecreasing. If (a n ) satisfies a n+ a n for every n, then we say (a n ) is nonincreasing. If (a n ) is either nonincreasing or nondecreasing, we say (a n ) is monotone. If there exists any real number M for which a n M for every integer n, then we say the sequence (a n ) is bounded. Example.3. Are the following sequences monotone? Bounded? () ( n ) n= (2) (5 3n) n= Theorem.4. Every bounded monotone sequence converges to some limit.

18 8 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 2 Relative Growth Rates Definition 2.. Let (a n ) and (b n ) be two sequences for which (a n ) and (b n ). b n M = lim, if this limit exists. n a n () If 0 < M <, we say that (a n ) and (b n ) have comparable growth rates. Let (2) If M = 0, we say that (a n ) grows faster than (b n ). (3) If M =, we say that (b n ) grows faster than (a n ). If (b n ) grows faster than (a n ), we will write (a n ) << (b n ). If (a n ) and (b n ) have comparable growth rates, and moreoever the limit M in the above definition is actually exactly, then we say (a n ) and (b n ) are asymptotic and we write (a n ) (b n ). Fact 2.2. The following sequences are ranked in order, from the slowest growth rate to the fastest growth rate. () (ln n) n=. (Logarithmic Growth) (2) ((ln n) r ) n= where r >. (Logarithmic Growth) (3) (n q ) n= where 0 < q <. (Polynomial Growth) (4) (n) n=. (Special Case of Polynomial Growth: Linear Growth) (5) (n p ) n= where p >. (Polynomial Growth) (6) (b n ) n= where b >. (Exponential Growth) (7) (n!) n=. (Factorial Growth) (8) (n n ) n=. (Superexponential Growth) Example 2.3. For any pair of sequences (a n ) and (b n ) in the list above, verify the order of growth b n rates by computing lim. (L Hospital s rule is a helpful tool here.) n a n Example 2.4 (Stirling s Formula). Just how fast does (n!) really grow? According to the above, it lies somewhere in between exponential growth and superexponential growth. But since the factorial function is defined only on the natural numbers, it is difficult to comprehend its growth analytically (for instance, it does not appear that one can use L Hospital s rule to compare its growth rate with that of other sequences). The following asymptotic estimate for the growth of (n!) sequence is called Stirling s formula. It is named after James Stirling, but was actually first discovered by Abraham de Moivre in 730. It says that factorial growth is approximately the same as superexponential growth divided by exponential growth, times square-root growth (up to the constant 2π). 3 Zeno s Paradox ( 2πn n n ) (n!) Here is a variation of one of Zeno of Elea s paradoxes (5th century BCE): Achilles is chasing a tortoise. The tortoise is 000 meters away, but Achilles runs 0 times as fast as the tortoise, so he expects to do well. Unfortunately, by the time Achilles runs the first 000 meters, the tortoise has advanced another 00 meters. By the time Achilles runs this additional 00 meters, the tortoise has advanced 0 more meters. By the time Achilles goes the next 0, the tortoise has gone another meter. The tortoise continually moves away from Achilles: next. meter, then.0 meter, then.00 meter, etc., ad infinitum. So Achilles never catches the tortoise, because each time he attains the tortoise s previous position, the tortoise has escaped some positive distance. e n

19 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 9 Can this paradox be resolved? Our intuition says that surely Achilles catches the tortoise, but where and when? It seems that the distance Achilles must travel is given by the sum But is it possible to compute such an infinite sum? 4 Infinite Series Now we arrive at one of the major developments of the course- the extension of the basic arithmetical operation of addition, which up until now in our education has formally applied only to finite collections of numbers, to encompass infinite collections of numbers. We have already seen during our study of improper integrals (see, for example, the comments following Example 6.) that it is possible and natural for an infinite process to yield only a finite result. We involve limits to formalize this idea. Let s look at some explicit examples. Example 4.. Find a geometric way to visualize the infinite sum n What should the value of this sum be? Consider the area of a unit square in the xy-plane. Example 4.2. Does it make sense to compute the infinite sum ? Now we will formalize the concept of infinite addition by applying the technology of sequences which we developed in the previous section. Definition 4.3. Let (a n ) n= be an infinite sequence. We refer to the infinite sum as an infinite series. We denote this series by a + a 2 + a n= which is a natural extension of our sigma notation for finite sums. Note here that sequences and series are intimately connected, but that they represent different concepts. A sequence is always just an infinite list of numbers without any additional structure; a series always refers to an infinite addition of numbers. Definition 4.4. Let a n be an infinite series. Define a new sequence (S n ) n= as follows: S = a S 2 = a + a 2 a n S 3 = a + a 2 + a 3 S 4 = a + a 2 + a 3 + a 4 In general, define S n by S n = a + a a n for every integer n. We call this sequence (S n ) the sequence of partial sums of L, then we say that the series... a n. If (S n ) converges to some limit a n converges to L, and we write

20 20 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES a n = L. If the sequence (S n ) diverges, then we say that the series Example 4.5. (2) Compute a n diverges. n= () Compute, if the series converges. 2n n=, if the series converges. n= Example 4.6. Consider the infinite series n= n(n + ). () Find the first four terms in the sequence of partial sums. (2) Do you think the series converges? If so, to what sum? Example 4.7. Compute ( ) n, if the sum converges. (Be careful!) k= 5 Geometric Series and Telescoping Series Definition 5.. A geometric series is a series of the form ar n = a + ar + ar 2 + ar where a and r are some fixed real numbers. Example 5.2. Consider the series Is the series geometric? Does it converge, and if so to what value? It is not hard to see that if ar n is a geometric series where r, then the series diverges. On the other hand, if we take r <, then we can prove the following theorem. Theorem 5.3 (Convergence of Geometric Series). Let ar n be a geometric series with r <. Then the series always converges, and moreover we can compute the sum by the following formula: Proof. For each n set ar n = a r. S n = a + ar ar n + ar n So (S n ) is the sequence of partial sums. We must check if (S n ) converges.

21 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 2 We can do this by making the following clever observation. Multiply both sides of the equation above by r. Then we get rs n = r(a + ar ar n ) = ar + ar ar n + ar n+. Subtracting the two equalities we ve obtained, and then solving for S n, we get the following. S n rs n = (a + ar ar n ) (ar + ar ar n + ar n+ ) S n rs n = a ar n+ ( r)s n = a( r n+ ) S n = a rn+. r So we have an explicit formula for the value of each partial sum S n = a rn+ r. All that remains to do is take the limit. Since r <, we have lim n rn+ = 0, and hence So the series converges to a r a n = lim n S n = lim a rn+ n r = a r = a r. as we hoped. Example 5.4. Compute the following geometric sums, if they converge. (). n (2) (3) (4) (5) e n ( ( ( n=2 ) n ) n ) n Example 5.5. Compute n 5 n, if the series converges.

22 22 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 2 Solution. Note that 5 n = ( ) n 2 and 5 convergent. So it makes sense to write: Example 5.6. Compute n 5 n = = 2 3 n 5 n = ( ) n + 5 ( ) n 3 are both geometric series and hence 5 ( ) n (/5) + (3/5) = = 5. e 3 2n, if the series converges. n=2 Solution. Note that e 3 2n = e 3 ( e 2 ) n. Note also that the series starts from index 2 instead of 0. So we have: Example 5.7. Compute n= 3 [ ] e 3 2n = e 3 2n [e 3 + e] n=2 [ ( ) ] n = e 3 e 2 [e 3 + e] = e 3 (/e 2 ) [e3 + e] = e5 e 2 e3 e. 2 n, if the series converges. 3n Theorem 5.8 (Divergence Test). If n=a n converges, then lim a n = 0. Equivalently, if (a n ) does not n converge to 0, then a n diverges. n= Example 5.9. Write out the first few terms of the series n= n 2 2n 2. Does the series converge? + 5n Solution. The first few terms make it look like it might have a chance of converging. n 2 lim n 2n 2 + 5n = 0, this series diverges by the divergence test. 2 Example 5.0. Determine the convergence or divergence of ( ) n n n +. Example 5. (Telescoping Series). Evaluate the following series. ( () 3 k ) 3 k+ k= (2) (k + )(k + 2) k= n= But since

23 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 23 Solution. (a) This series could be evaluated by breaking it up into two geometric series. But instead let s just observe the behavior of the sum: k= ( 3 k ) ( 3 k+ = ) ( ) ( ) ( ) If one computes the sequence of partial sums, one notices that many natural cancellations occur and in general the k-th partial sum is S k =. (These repeated cancellations are why we use the term 3k+ telescoping series.) So taking k, we get k= ( 3 k ) ( 3 k+ = lim S k = lim ) k k 3 k+ =. (b) This series is another telescoping series in disguise. Use the method of partial (k + )(k + 2) k= fractions to reveal it. Write: (k + )(k + 2) = A k + + B k + 2. Solving for A and B yields A =, B =. Therefore, k= Example 5.2. Compute Example 5.3. Compute (k + )(k + 2) = ( k + n=4 = k= = 2. ( 2 3 ) + ) k + 2 ( 3 4, if the series converges. n(n 4) 4n2 +, if the series converges. 3n n= 6 The Harmonic Series Consider the infinite series: n= n = ) ( + 4 ) The series above is called the harmonic series. Does the harmonic series converge or diverge? Since the sequence ( n ) n= converges to 0, it seems plausible that the harmonic series might converge. Computing the first few terms in the sequence of partial sums, we may obtain:

24 24 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES S = S 2 = 3 2 S 3 = 6 S 4 = The trend of convergence or divergence may not be clear from the first few terms. What about if we consider very large n-th partial sums? It can be shown that the sum of the first, 000, 000 terms of the harmonic series is less than 5. The number of atoms comprising the planet Earth is (roughly) 0 50 ; the sum of the first 0 50 terms of the harmonic series is approximately 5.6. The number of atoms in the observable universe is (very roughly) 0 80 ; the sum of the first 0 80 terms of the harmonic series is approximately It may be unclear from this information whether the series in fact converges or diverges. In fact, we can show through an elementary argument that the harmonic series diverges. Simply make the following observation: n= n = }{{} /2 + 2 }{{} / }{{ 4 } }{{ 8 } /2 /2 Since computing the infinite sum n must involve adding terms whose sum is 2 n= times, the sum must diverge to infinity. It just diverges very, very slowly! infinitely many Example 6.. Prove that lim ln x =. x Theorem 6.2 (Integral Test). Suppose f is a continuous, positive, decreasing function for x and suppose a n = f(n) for every positive integer n. Then the infinite sum a n and the improper integral f(x)dx either both converge, or else both diverge. However, if they both converge, their values need not in general be the same. Example 6.3. Determine whether the following series converge or diverge. () n n + (2) k k 2 + k= (3) 2k 5 (4) k=3 n=, where p > is some fixed real number. np Theorem 6.4 (Convergence of p-series). n= n p n= always converges if p >, and diverges otherwise.

25 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 25 Proof. This theorem is the direct consequence of Theorem 6.0 and Theorem 6.2. In the next few sections we will develop many methods for determining whether a given series converges or diverges. In general it will not always be completely clear which test, if any, will be the appropriate one to apply when faced with any particular series. It is up to the student to develop the intuition and problem-solving skills for attacking such problems. 7 The Direct Comparison Test Theorem 7. (Direct Comparison Test). Let a n and b n be series with positive terms, and such that 0 < a n b n for every integer n. () If b n converges, then a n converges. (2) If n= a n diverges, then n= n= n= b n diverges. n= n= Example 7.2. Determine whether the following series converge. () (2) (3) n= k= k= n3 n k 3 2k 4 ln k k 3 8 The Limit Comparison Test Theorem 8. (Limit Comparison Test). Let () If 0 < L <, then (2) If L = 0 and (3) If L = and a n and n= a n and n= lim n a n b n = L. b n be series with positive terms, and let n= b n either both converge or both diverge. n= b n converges, then n= b n diverges, then n= a n converges. n= a n diverges. n= Example 8.2. Determine whether the following series converge. () (2) k= k= k 4 2k k 6 k + 5 ln k k 2

26 26 CALCULUS II MATH 66 SPRING 206 (COHEN) LECTURE NOTES 9 Alternating Series Example 9.. Determine if the series alternating harmonic series.) ( ) k+ converges or diverges. k k= k ( ) k+ Solution. For each k let S k = = k n= of partial sums of the given series. (We call this series the ( )k , so S k is the sequence k First consider the sequence which consists only of the even terms (S 2, S 4, S 6,..., S 2k,...). Notice that since 3 > 4, we have 3 4 > 0, and hence S 4 = S > S 2. In fact for any even number 2k, we have 2k+ 2k+2 > 0 and hence S 2k+2 = S 2k + 2k+ 2k+2 > S 2k. It follows that the sequence of even terms (S 2, S 4,...) is strictly increasing. In addition, the sequence (S 2, S 4,...) is bounded above by = S. So the sequence (S 2, S 4,...) is a bounded monotone sequence, and hence it converges to some limit L by our theorem from earlier. If we consider the sequence (S, S 3, S 5,..., S 2k,...) of odd partial sums, then arguments which are very similar to the above will show that the sequence is strictly decreasing and bounded below by S 2 ; so the sequence of odd partial sums also converges to some limit, say M, by our same theorem from earlier. If we can check that L = M, then we will have shown the series converges; otherwise we will have shown it diverges. To see the relationship between the two limits, simply observe that S 2k = S 2k 2k for every positive integer k. Hence we have L = lim S 2k k [ = lim S 2k ] k 2k = lim S 2k lim k k 2k = M 0 = M. So the series converges! Question 9.2. To what value does the alternating harmonic series converge? We don t have the tools to compute this right now, but we will see the answer in the future. Notice that the argument above depends only on three facts: that the sequence of terms in the series alternates signs; that the terms of the (non-alternating) harmonic series are a decreasing sequence; and that the terms converge to 0. So the solution above may be easily modified to prove the following theorem (which we leave to the reader): Theorem 9.3 (Alternating Series Test). Suppose (a n ) n= is a sequence of positive terms for which () a n+ a n for every positive integer n, and (2) lim n a n = 0. Then the alternating series ( ) n+ a n converges. n=