1. Find a power series representation of

This function differentiates to a fraction we can interpret as the sum of a geometric series.

To complete Step 4, note that for x=0, = 0 and the series also equals 0, so C must be 0. The radius of convergence for the series obtained in Step 2 is 1, with an interval of (-1, 1). When you integrate the interval of convergence can gain or lose endpoints, but that’s a small thing to check after this fairly quick conversion.

2. Find a power series representation of

Instead of differentiating, this time we integrate to get the sum of a geometric series.

In Step 3 note that from the perspective of the derivative operator is constant. There’s no constant of integration to find so once you’ve rounded the bases you’re done – well, except for thinking about the radius and interval of convergence.

One more after the jump.

3. Find a power series representation for

This time we differentiate twice.

Once again you do not have to find C at the end, but in this case it is because C stays undefined: the series is intended to equal an indefinite integral, not a specific function. C does not affect the radius or interval of convergence of the series.

We can use the series above to find a power series for the definite integral Since plugging 0 into the series obtained in Step 6 gives 0, the answer is the result of plugging in 1:

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For a line, there is only one way to be parallel, but infinitely many ways to be orthogonal (think: every vector parallel to the xy-plane is orthogonal to the z-axis). Therefore the vector we want, the direction vector, is parallel to the line.

For a plane, conversely, there are infinitely many ways to be parallel, but only one way to be orthogonal (any vector orthogonal to the xy-plane is parallel to the z-axis). Therefore the vector we want, the normal vector, is orthogonal to the plane.

You may obtain these two pieces in many ways. In addition to being given the point and direction vector immediately, the following are enough to determine a line:

• two points on the line
• a point and a parallel line
• a point and two nonparallel vectors orthogonal to the line
• a point and two nonparallel lines orthogonal to the desired line
• a point and an orthogonal plane
• two intersecting (nonidentical) planes (the line of intersection)
• two intersecting (nonidentical) lines (the line through their point of intersection and orthogonal to both)

The following are enough to define a plane, in addition to being given a point and normal vector directly:

• three points in the plane
• a point and an orthogonal line
• a point and a line in the plane not containing that point
• two lines in the plane
• a point and a parallel plane
• a point and two planes orthogonal to the desired plane

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1. For example, the sum from n=1 to infinity of (n!)/(n^n). Once n is at least 2, we can compare to a nice series we know about.

The sum from n=1 to infinity of 2/(n^2) is easily shown to converge, and, from a finite point on, it sits on top of our original series, which has only positive terms and hence must also converge.

2. We could use the ratio test for the sum , but we can also use the alternating series test. A similar trick to the previous example shows the magnitudes of the terms have limit 0. Once n is at least 3, we have the following comparison on the absolute value of each term.

That last fraction has a limit of 0 as n goes to infinity.

However, for the alternating series test we need more than a limit of zero; the magnitudes must decrease to 0, for which we use the fact that (n+1)! can be written in terms of n!:

As long as n is at least 3, the n+1st term is the nth term times a value less than 1. The series doesn’t decrease right from the start, but from a finite point on it always decreases, and that is enough.

3. How about the sum from n=1 to infinity of ? You might be tempted to use the ratio test for this example, but it would waste your time. The terms simplify to [(n+1)(n+2)]/n^2, and have a limit of 1 as n goes to infinity, and therefore the series diverges by the test for divergence.

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Test	Value to find	Convergent	Divergent	Inconclusive
test for divergence		N/A	≠ 0	= 0
p-series	p in	> 1	≤ 1	N/A
geometric series	|r| in	< 1	≥ 1	N/A
ratio test		< 1	> 1	= 1
root test		< 1	> 1	= 1

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1. Directly/by rules.
2. After algebraic manipulation.
3. Using substitution.
4. By parts.

Using limits for improper bounds or internal discontinuities should be mentioned here, but it’s not really in the same category since you don’t use the limit to find an antiderivative.

1. is the method you learn first.
2. includes completing the square, partial fractions, trigonometric integrals where you apply trig identities to whittle down powers or make substitution possible. In a (strong) sense it also includes integration by Taylor series.
3. comes from the chain rule for differentiation, and includes u-substitution and inverse substitution, such as trig substitution.
4. comes from the product rule for differentiation, and is one of the most powerful techniques you learn in calc 2.

When should you apply each technique? There are rules of thumb, though it is also possible to solve many integrals in more than one way.

Completing the square comes in when you have a quadratic polynomial in an awkward spot such as the denominator of a fraction or under a square root. In the latter case it is often followed by trig substitution. In the former case it may lead to a u-substitution where the expression that becomes u has a constant derivative.

Partial fractions are for rational functions, to try to get down to simpler fractions which are amenable to u-substitution or trig substitution.

Trig integrals are easy to identify; they are powers and products of trig functions: sine with cosine, tangent with secant, cotangent with cosecant. They appear frequently after making a trig substitution.

Inverse substitution is generally used when there is an ugly function of x such as a square root, but the rest of the integrand is not simply a constant times the derivative of that function. You set u as a function of x, but solve for x in order to complete the substitution. Square roots of linear functions are a common place for this: if the integrand were , you could let , and then let x = u² – 1 and dx = 2u du. The integrand becomes 2u⁴ – 2u². For trig substitution the traditional variable is theta, and here we exploit similarities to trig identities to collapse binomials into monomials and eliminate problematic square roots or fractions: 1 – sin² x collapses to cos² x, 1 + tan² x to sec² x, and sec² x – 1 to tan² x. [Why introduce sine into the binomial instead of cosine? Because the derivative of your chosen function comes into play in the substitution, and using sine means no negatives are introduced. That is, no reason except it makes for simpler bookkeeping.]

Integration by parts is essential when your integrand is the product of two functions of different “types”: polynomial, trigonometric, exponential, logarithmic.

These techniques can be chained together, of course. Completing the square leading to trig substitution leading to a trig integral is common. Making a substitution to make partial fractions possible is easy to come by (e.g., ).

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Imagine when you have f(g(x)) that x is a little sealed box. It is inside a second sealed box marked g, and that box is in a third sealed box marked f.

To differentiate you open the box marked f, and remove and open the box marked g, and remove and open the box marked x. The open boxes are the derivatives of the closed boxes, and containment still represents composition.

All put together, the drawing above represents the derivative of f(g(x)):

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A beginner’s guide to quadric surfaces (pdf), by me, for my calculus students and intended to supplement the textbook rather than be used as a replacement.

I found the first paragraph of this article on seat belt use to be a great example of awkward statistics writing.

Chapter 23 of The Art of Public Speaking by Carnegie and Esenwein is a thorough exploration of logic in debate.

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1. There are many ways for a limit to fail to exist. It could be infinite, oscillate within a finite interval, completely devolve into radio static, or simply give you different values when you approach from the right or the left.

2. It is worth restating that a limit of infinity is just shorthand for “the limit is undefined, but in a special way.”

3. Proving the existence of a limit using the delta-epsilon definition involves working backwards. In the actual proof, you say “given epsilon, take delta < [function of epsilon]" and then show that value of delta works. Behind the scenes, though, you have to determine the appropriate function of epsilon by reverse-engineering it from the desired inequality on the function.

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2. Advice I’ve given students: If the series does not look like anything but you’re being asked to evaluate it, try partial fractions and see if you get something telescoping.

3. The limit comparison test asks whether the terms of two series are “proportional in the limit.” The ratio and root tests ask, in two ways, whether the terms of one series are “geometric in the limit.” The geometric series with terms gives ratio r between successive terms (), leading to the ratio test, and the nth root of its nth term is r times the nth root of c (assuming c is positive, and otherwise taking the negation of the series, which has the same convergence behavior), which limits to r, leading to the root test. This is why the cutoff point for convergence and divergence is 1 – that is what it is for geometric series. The distinction, that at limit 1 we don’t know the behavior, is because this is something only “geometric in the limit.”

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