Question 1.
The Taylor series formula means that a given function can be written as a power series, i.e an infinite sum of powers (x-a) ⁿ with certain coefficients. The coefficients are numbers, given by evaluating the derivatives of the function at the given point a divided by the factorials. In particular, the Taylor series at 0 (aka Maclaurin series) is the sum of powers of x, with coefficients given by derivatives at 0, divided by the factorials. The answer for such Taylor series would be an expression of the form 5- x + 1/2 x ² + 7/15 x ³ -115 x ⁴ +... , with coefficients being NUMBERS. If you got an f(x), square roots, sines, or whatever else on the right hand side of your Taylor series expansion, your answer is WRONG. The parameter a should also not be present (i.e. it must be replaced by the appropriate number) if you are writing an expansion at a specific point. In other words, you would be getting formulas like 5- (x+3) + 1/2 (x+3) ² + 7/15 (x+3) ³ -115 (x+3) ⁴ +... , for example, if x=-3. General formulas (like those for sine and cosine) include expressions like n!, but you must understand that any given coefficient can be computed explicitly.

(a) The question asks for the first three terms of the Taylor series at 0; the answer will be the sum of some number (0-th term), x with some coefficient, and x ² with some coefficient. Indeed, (x-a) ⁿ = x ⁿ when a=0, and we only need three terms, with n=0, n=2, n=3. To find the coefficients, we compute the first and second derivatives of f(x)= √(x+4) and plug in 0; the answer is
f(0) + f'(0) x + 1/2 f"(0) x ² = 2 + 1/4 x - 1/64 x ²

(b) Many people just didn't know where to start on this one; some expanded a part of the expression (maybe using the formula for sine) and stopped there. Using L'Hospital three times might give the right answer if you manage to get there without mistakes, but this question is about series. So how do you use series for limits?

Limits are very easy to compute when you're looking at a rational function, say you are computing limit of (x ²-4)/(x-2) when x goes to 2, or limit of (x⁵+7x³ -x²)/(x³ + 2 x²) when x goes to 0. (In the latter example, you would just cancel x² at top and bottom, and plug in x=0).

Series allow us to use the same strategy for arbitrary functions. If you expand ALL of the functions (sine, cosine, and radical) as power series. Some first terms will cancel - note that there will be no constant terms and no x in it. Thus you will get 0/0 if you just try to plug in x=0 right away. Instead, cancel a power of x at top and bottom (all the terms will be powers of x) and plug in 0 to get the answer. If you didn't get this question right, DO THIS EXERCISE NOW.

Question 2.
When computing definite integrals, be sure to plug in the limits of integration CAREFULLY. In particular, when doing subtitution, don't forget to change the limits of integration as well. Don't forget to evaluate f(x)g(x) at the endpoints when doing integration by parts. When computing improper integrals, you need to consider the limit of proper integrals (where you have to be careful with the limits of integration as well.)

When using partial fractions, compute integrals of 1/(something) - you might need a substitution if the thing is a linear expression like 2-3x.

When using comparison test for improper integrals, be sure to check ALL the conditions - in particular, your functions have to be positive for the test to work.

Question 3.
There was a lot of confusion on what the Integral Test means. Some people thought that this is about deciding whether some improper integral is convergent. Others claimed that the sum of the series equals to the integral (INCORRECT). In reality, the Integral Test is about COMPARING the integral and the series: they are usually not equal, but in the appropriate setup you can say that one is greater than the other. If the integral is CONVERGENT, and you set up the comparison so that the series is SMALLER that the integral, you can conclude that the series is also convergent. If the integral is DIVERGENT, you set up comparison so that the series is LARGER, to conclude that the series is divergent. (By setting aside some initial terms, it is possible to make the comparison work in the desired direction -- see textbook and the comment below.)

The first step is to study the improper integral. You need to use the limit definition. If computing the proper integral by substitution, make sure to do substitution in the limits of integration as well.

In this question the integral turns out to be convergent. Thus you want to show that the series is smaller. The terms of the series can be represented by rectangles UNDER the graph. (A few students used rectangles ABOVE the graph; while it is possible to represent the series this way, in this case this is wrong strategy, because you would show that the series is BIGGER than the integral, leading to no conclusion.) A staircase-shaped area of all these rectangles represents the sum of the series. (A number of people just shaded the area under graph instead. To get the correct answer, it is crucial to understand exactly which term of the series each rectangle represents: the area equals to the height (base of rectangle is always 1), and the height is given by the value at the point where the rectangle touches the graph. REVIEW 8.3 TO MAKE SURE YOU UNDERSTAND THIS. Once you know that the sum of the series is represented by the staircase-shaped area that fits INSIDE the area of the graph and thus is SMALLER that the area representing the CONVERGENT improper integral, you can conclude that the series is convergent.

(To make comparison work, sometimes you have to set aside a term in the series or consider an improper integral with a different limit of integration. In the exam question, this extra issue didn't arise.)

Question 4. (a), (b) Plugging into the Trapezoidal Rule formula should be straightforward, but many students got confused about how many points x_i to use and where those points are. We have [a, b] = [-1/2, 1/2] , which has length 1; taking 3 small intervals means that each has length Δ= 1/3. The enpoints of these intervals are -1/2, -1/6, 1/6, 1/2. (Some people started at 0, working with [0,1] or [0, 1/2] instead of the given interval; some included 0 as an endpoint; some took 2 intervals instead of three. All of this is incorrect.) Plug these into the formula CAREFULLY, otherwise it is easy to miss something. For the final computation, a mistake in the trig cost 1 point. To illustrate the approximation in (b), mark the four points we found above on the x-axis, mark corresponding points on the graph, and connect them by straight lines. The sum of areas of the resulting trapezoids is the approximation. SHADE THIS AREA. (In our case the "trapezoids" are actually two triangles and a rectangle, but that's okay.) Because the area you shaded lies INSIDE the area under graph (expressed by the precise value of the integral), you got an UNDERAPPROXIMATION. (Note that if you don't make a careful picture, you can't compare anything).

(c) The main mistake here was to use the error estimate incorrectly: people assumed E=0.01, plugged into the formula E ≤ K (b-a)³/12 n² , and tried to solve for n . This is wrong: in fact you don't know what the actual error will be; you are hoping to get E ≤.01. How can you guarantee this? The expression K (b-a)³/12 n² is always GREATER than the error, so if you make sure that K (b-a)³/12 n² is LESS than .01, the error (which is even smaller) will be LESS than .01 as well. This strategy works every time we do estimates: you take the expression which gives the error bound and make it LESS than the required precision; then the error is guaranteed to be even smaller.

Thus we need to solve for n such that K (b-a)³/12 n² ≤ .01. We know that b-a = 1. The parameter K is any number that is greater than the absolute value of the second derivative of the function. The second derivative of cos π x is π² cos π x. (Many people lost π².) Thus we can take K= 10. Solving for n , we get that n should be 10 or GREATER. 10 intervals suffice to get the required precision.

Notice two important points here: when you solve for n , the answer you get should be of the form n ≥ something. This is because the more intervals you take, the better the approximation; a very large n will typically work, a very small one won't. So if you're getting that n is LESS than something, it means you made a mistake somewhere. Another point is that n is always an integer. be sure to ROUND UP your answer if you get a non-integer n.

Question 5. There were many mistakes at the very start of solution - if you get the rotation axis or the original graphs wrong, your whole solution breaks down. So please follow the strategy for computing volumes step-by-step - do careful pictures, think about the direction of rotation and how the slicing goes and what the slices look like, etc.