BC Unit 6 built on its AB foundations: integration by parts as the product rule reversed, partial fractions as common-denominator algebra reversed, improper integrals as the Fundamental Theorem plus a limit, and a thirty-second triage routine for choosing between them.
Subject: Calculus BC · 65 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus BC · Session 2
Parts, partial fractions and improper integrals — every one an AB rule run backwards
Objectives
Every technique here is a rule you already met in AB, reversed. That framing is the point: there is nothing new to believe, only something new to run in the other direction.
College Board, AP Calculus AB and BC Course and Exam Description, Unit 6 Unit 6 — the BC integration unit in full
Section
Section 1
Warm-up
Before anything new, recall the forward direction.
Discussion prompt
State the product rule from memory, then say what you would get if you integrated both sides of it.
Hint: The left side integrates to something with no integral sign left on it.
Answer:
The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.
\[ (uv)' = u'v + uv' \]
Integrating both sides gives the product back on the left and two integrals on the right. Solving for one of those two integrals is the whole of integration by parts.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) §3.1
Concept
Rearranging the integrated product rule isolates one integral in terms of the other.
\[ \int u\,dv = uv - \int v\,du \]
You are trading one integral for another. The trade is only worth making when the new integral is easier than the old one, which is what choosing the pieces well means.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) §3.1
Picture it
One identity, rearranged. There is nothing else in it.
Figure (svg): The product rule in a box above, an arrow labelled integrate both sides and solve, and the integration by parts formula in a box below
If you ever forget the formula on an exam, you can rebuild it from the product rule in about fifteen seconds.
Intuition
The whole art is choosing which factor to differentiate and which to integrate. Differentiate the factor that simplifies under differentiation; integrate the one that does not get worse.
A polynomial factor loses a degree every time you differentiate it, and eventually hits zero. A logarithm turns into an algebraic fraction. An exponential is unchanged either way, which makes it the natural candidate for the other slot.
The usual mnemonic orders the candidates for u as logarithmic, then inverse trigonometric, then algebraic, then trigonometric, then exponential. It is a heuristic and not a theorem, but it is right far more often than it is wrong.
Paul's Online Math Notes, Calculus II — Integration Techniques Calculus II, Integration by Parts
Prediction
The integral is x times the exponential of x.
Predict first
Which choice of u leads somewhere?
Correct: u = x.
Why: Differentiating x gives 1, which removes the polynomial from the new integral entirely and leaves only the exponential, which you can integrate on sight. Choosing the exponential as u leaves the polynomial to be integrated, raising its degree and making the new integral harder than the original.
Worked example
Find the antiderivative of x times the exponential of x.
Split the integrand into u and dv
Why: The polynomial gets differentiated; the exponential gets integrated.
\[ u = x, \; du = dx, \qquad dv = e^x dx, \; v = e^x \]
Apply the formula
Why: The product uv, minus the integral of v du.
\[ \int x e^x dx = x e^x - \int e^x dx \]
Finish the remaining integral
Why: The exponential is its own antiderivative, which is why this choice of pieces was the right one.
\[ x e^x - e^x + C \]
Verify: by differentiating the answer
Why: The product rule gives the exponential plus x times the exponential, and the last term contributes minus the exponential. Two of the three cancel, leaving x times the exponential, which is the integrand.
Picture it
Between zero and one the same antiderivative gives an area of exactly one.
Figure (svg): The curve y equals x times the exponential of x with the region between zero and one shaded, labelled area equals one
Trap
The same integral, with the pieces swapped.
\[ u = e^x, \qquad dv = x\,dx, \; v = \frac{x^2}{2} \]
Apply the formula anyway
Why: Nothing here is illegal. Parts is valid for any split of the integrand.
\[ \frac{x^2 e^x}{2} - \int \frac{x^2 e^x}{2}\,dx \]
Notice the new integral is worse
Why: The polynomial went up a degree instead of down. Applying parts again makes it a cube, and so on forever.
Choose the factor that simplifies when differentiated.
\[ u = x, \qquad dv = e^x dx, \; v = e^x \]
The polynomial degree drops to zero
Why: One application removes x from the problem completely.
\[ x e^x - e^x + C \]
Sanity-check the direction
Why: If the new integral looks harder than the one you started with, swap the pieces and start again rather than pushing on.
Explain it to yourself
A question worth two minutes, because the answer explains something about antiderivatives generally.
\[ dv = e^x dx \;\Rightarrow\; v = e^x \]
Discussion prompt
Every antiderivative comes with a constant. Why can you drop it when you build v inside integration by parts?
Hint: Write v plus k in both terms of the formula and see what survives.
Answer:
Because a constant added to v contributes the constant times u in the first term, and minus the integral of the constant times du in the second — and those two contributions cancel exactly.
So any choice of the constant gives the same final answer. Choosing zero is simply the least work.
It is worth checking this once by hand. Doing so also makes it obvious why you keep a single constant at the very end.
MIT OpenCourseWare 18.01, Single Variable Calculus, Lectures 27-29 Lecture 27
Worked example
Find the antiderivative of x squared times the exponential of x.
First pass: differentiate the polynomial
Why: The degree drops from two to one, which is progress even though an integral remains.
\[ x^2 e^x - \int 2x e^x dx \]
Second pass on the remaining integral
Why: This is the integral from the previous example, times two.
\[ \int 2x e^x dx = 2(x e^x - e^x) \]
Substitute back and collect
Why: Watch the sign: the whole second result is being subtracted.
\[ x^2 e^x - 2x e^x + 2e^x + C \]
Verify: by differentiating
Why: The three product rules give two x times the exponential plus x squared times the exponential, minus two times the exponential minus two x times the exponential, plus two times the exponential. Everything cancels except x squared times the exponential.
Picture it
When one factor differentiates down to zero, you can run all the passes at once.
Figure (svg): A two column table with x squared differentiating down to 2x then 2 then 0 on the left and the exponential repeated on the right, with alternating plus and minus signs and diagonal arrows
This is not a different method. It is the same two passes laid out so you cannot lose a sign, which is where almost all the marks go.
Fill the middle
There is only one factor here, so the second one has to be invented.
Fill in the blanks
\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C
Why: Take u to be the logarithm and dv to be dx, so v is x. The remaining integral is x times one over x, which is just 1, and the integral of 1 is x. This is the standard trick for integrating a lone function whose derivative is nicer than itself.
Pattern
Six steps. The fourth is the one that decides whether the method was worth using.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) §3.1
Check
Solve it on paper before you click.
Check your understanding
For the integral of x times the cosine of x, which split of the integrand leads to a simpler second integral?
Answer: A
Why: Differentiating x gives 1, so the second integral becomes the integral of the sine, which is elementary. The answer is x times sine x, plus cosine x, plus a constant.
Section
Section 2
Concept
Adding two simple fractions produces one complicated one. Partial fractions runs that backwards.
\[ \frac{5x - 3}{x^2 - 2x - 3} \]
On its own this is not an integral you know. Split into pieces with linear denominators and every piece becomes a logarithm.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) §3.4
Picture it
The technique does nothing except reverse the algebra that made the fraction complicated in the first place.
Figure (svg): A single rational expression on the left splitting via two arrows into two simpler fractions with linear denominators on the right
Step zero
The rational function above.
Discussion prompt
Before finding any coefficients: what has to be true of the degrees, and what has to be done to the denominator first?
Hint: Two checks, both about the denominator.
Answer:
The numerator degree must be strictly lower than the denominator degree. Here it is one against two, so no long division is needed.
The denominator must be factored. This one factors as x minus three times x plus one.
Only then do you write the decomposition with unknown constants over each factor. Skipping straight to the constants is how students end up solving the wrong system.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) §3.4
Worked example
Integrate the rational function whose numerator is five x minus three and whose denominator factors into x minus three and x plus one.
Write the decomposition with unknown numerators
Why: One unknown constant per linear factor.
\[ \frac{5x-3}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1} \]
Clear the denominators
Why: Multiply through by the full denominator; the fractions vanish and a polynomial identity is left.
\[ 5x - 3 = A(x+1) + B(x-3) \]
Substitute the roots one at a time
Why: Each root kills one unknown, which is far faster than expanding and matching coefficients.
\[ x = 3: \; 12 = 4A \Rightarrow A = 3 \]
\[ x = -1: \; -8 = -4B \Rightarrow B = 2 \]
Integrate each piece
Why: Each is a constant over a linear factor, so each is a logarithm with no substitution needed beyond the obvious one.
\[ 3\ln|x-3| + 2\ln|x+1| + C \]
Verify: by recombining the two fractions
Why: Three over x minus three plus two over x plus one has numerator three x plus three plus two x minus six, which is five x minus three. That is the original numerator, so the decomposition was right.
Picture it
The logarithm is only defined for positive inputs, but the fraction is defined on both sides of each root.
Figure (svg): Graph of the natural logarithm of the absolute value of x, defined on both sides of zero and symmetric about the vertical axis
On a definite integral the bars usually cancel, which is why students get away with dropping them until they do not.
Trap
The same rational function, attacked directly.
\[ \frac{5x-3}{x^2-2x-3} = \frac{A}{x^2} + \frac{B}{-2x-3} \]
Split the denominator term by term
Why: Denominators do not distribute over addition. This is the fraction equivalent of claiming that one over a plus b equals one over a plus one over b.
Get a system with no solution
Why: The identity cannot hold for all inputs, so the coefficients come out inconsistent — usually after a page of algebra.
Factor the denominator first, always.
\[ x^2 - 2x - 3 = (x-3)(x+1) \]
One unknown per factor of the factored denominator
Why: The decomposition is over factors, not over terms.
\[ \frac{A}{x-3} + \frac{B}{x+1} \]
Check the factoring by expanding it back
Why: Two seconds, and it removes the possibility of building the whole decomposition on a wrong factorisation.
Elimination
The denominator is x minus one, times x plus two squared.
Eliminate the wrong options
Which template would you write down?
Survives elimination: A
Why: A linear factor repeated twice contributes two terms: one over the factor and one over its square, each with a constant numerator. The rule is one term per power up to the multiplicity, and a numerator one degree below its denominator.
Comparison
Every partial-fractions problem you will meet is one of these. Fill in the gaps.
Comparison matrix
| denominator factor | term it contributes |
|---|---|
| distinct linear, x minus a | A over (x minus a) |
| repeated linear, (x minus a) squared | A over (x-a) plus B over (x-a) squared |
| irreducible quadratic | (Ax + B) over the quadratic |
| numerator degree at least denominator degree | long division first |
The last row is the one that gets skipped. If the top is not lower degree than the bottom, no decomposition exists until you divide.
Section
Section 3
Concept
An integral is improper when a limit of integration is infinite, or when the integrand blows up somewhere on the interval. The Fundamental Theorem does not apply directly to either.
\[ \int_1^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_1^{b} f(x)\,dx \]
The fix is to make the bad endpoint a variable, integrate normally, and then take a limit. That limit either exists, in which case the integral converges, or it does not, in which case it diverges.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) §3.7
Prediction
Both curves stretch to infinity along the horizontal axis, and both shrink towards zero.
Predict first
Can a region of infinite length enclose a finite area?
Correct: Yes, if the curve shrinks fast enough.
Why: The area added between one input and the next depends on the height there. If the heights shrink quickly enough, the added areas form a convergent total. One over x squared shrinks fast enough; one over x does not, even though both approach zero.
Worked example
Evaluate the integral of one over x squared, from one to infinity.
Replace the infinite limit with a variable
Why: Never write infinity as a limit of integration and substitute it. Infinity is not a number you can substitute.
\[ \lim_{b \to \infty} \int_1^{b} x^{-2}\,dx \]
Integrate normally
Why: Raise the exponent by one and divide, giving negative one over x.
\[ \left[-\frac{1}{x}\right]_1^{b} = -\frac{1}{b} + 1 \]
Take the limit
Why: The reciprocal of b goes to zero, leaving the constant.
\[ \lim_{b \to \infty}\left(1 - \frac{1}{b}\right) = 1 \]
Verify: that the value is plausible
Why: The region sits under a curve that starts at height one and falls away. A total area of exactly one for an infinitely long region is surprising but consistent with a curve that drops that quickly.
Picture it
The shaded region never ends, and its total area is one.
Figure (svg): The curve one over x squared with the region from one rightwards shaded green, labelled total area equals one
Worked example
The same setup, but with one over x.
Replace the infinite limit and integrate
Why: The antiderivative of the reciprocal is the natural logarithm.
\[ \lim_{b \to \infty}\left[\ln|x|\right]_1^{b} = \lim_{b \to \infty}\left(\ln b - 0\right) \]
Take the limit
Why: The logarithm grows without bound, slowly but without ever stopping.
\[ \lim_{b \to \infty} \ln b = \infty \]
The integral diverges. There is no number to report; the correct answer is the word.
Verify: against the previous example
Why: The only change was the exponent, from two to one, and the outcome flipped completely. That sensitivity is exactly what the p-test summarises.
Picture it
One over x also falls towards zero. It does not fall fast enough.
Figure (svg): The curve one over x with the region from one rightwards shaded red, labelled area grows without bound
Both curves go to zero. Going to zero is necessary for convergence and nowhere near sufficient — the same distinction that decides whether a BC series converges.
Concept
For the reciprocal of a power, starting at one and running to infinity, the outcome depends only on the exponent.
\[ \int_1^{\infty} \frac{1}{x^p}\,dx \text{ converges exactly when } p > 1 \]
At the boundary itself, where the exponent equals one, the integral diverges. Memorise the strict inequality — the boundary case is the one exam questions choose.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) §3.7
Picture it
One number decides it, and the boundary belongs to the divergent side.
Figure (svg): A number line from zero to three with the region below one shaded red for divergence and above one shaded green for convergence, with the boundary at one marked as divergent
Edge cases
Not at 0.9, not at 1.1.
Discussion prompt
What is special about the exponent one that makes it the dividing line, and why does the boundary case itself diverge?
Hint: What is the antiderivative of the reciprocal, and how is it different from every other case?
Answer:
Because one is the only exponent whose antiderivative is not a power. Every other exponent gives a power of x, and at infinity that power either dies or explodes depending on its sign.
At the exponent one the antiderivative is the logarithm, which does neither: it grows without bound, but slower than any positive power. Slow is not the same as bounded, so it diverges.
That is why the boundary sits on the divergent side rather than being a special convergent case.
MIT OpenCourseWare 18.01, Single Variable Calculus, Lectures 27-29 Lecture 29
Worked example
Evaluate the integral of one over the square root of x, from zero to one. The limits are both finite, so what makes it improper?
Identify the bad endpoint
Why: At the input zero the integrand is undefined and grows without bound. The interval is finite but the function is not.
Replace the bad endpoint with a variable approaching it from inside
Why: The variable approaches zero from the right, because that is the side the interval is on.
\[ \lim_{a \to 0^{+}} \int_a^{1} x^{-1/2}\,dx \]
Integrate and evaluate
Why: Raising the exponent by one gives one half, and dividing by one half doubles it.
\[ \left[2\sqrt{x}\right]_a^{1} = 2 - 2\sqrt{a} \]
\[ \lim_{a \to 0^{+}}\left(2 - 2\sqrt{a}\right) = 2 \]
Verify: by differentiating the antiderivative
Why: The derivative of twice the square root is one over the square root, which is the integrand. And the answer is finite despite the vertical asymptote, which the picture makes believable.
Picture it
The curve shoots up at the left edge and the shaded region still totals two.
Figure (svg): The curve one over the square root of x rising steeply towards a vertical asymptote at zero with the region from zero to one shaded, labelled area equals two
An integral can be improper at either end, at both, or in the middle. If it is improper in the middle, split it there and test each half separately.
Trap
The integral of one over x squared from one to infinity.
\[ \left[-\frac{1}{x}\right]_1^{\infty} = -\frac{1}{\infty} + 1 = 1 \]
Write infinity as a limit of integration and substitute it
Why: The answer happens to be right, which is exactly what makes this habit dangerous.
Get no marks, and get the next one wrong
Why: The same move on a divergent integral produces a confident wrong number, because infinity does not obey the arithmetic being used on it.
The same integral, done as a limit.
\[ \lim_{b \to \infty}\left[-\frac{1}{x}\right]_1^{b} \]
Make the bad endpoint a variable first
Why: Now every step is ordinary arithmetic on a real number, and the only limit is taken at the very end.
\[ \lim_{b \to \infty}\left(1 - \frac{1}{b}\right) = 1 \]
Say the word
Why: Write converges or diverges explicitly. On the AP exam that word is worth a point on its own.
Two truths and a lie
Three statements, one survivor.
Eliminate the wrong options
Which is true?
Survives elimination: A
Why: The only safe direction is the contrapositive: heights that do not shrink to zero keep adding area forever, so convergence is impossible. The converse fails, and that failure is the whole reason the p-test has to exist.
Section
Section 4
Concept
Some integrands never simplify under parts. An exponential times a sine returns an exponential times a cosine, and a second pass returns the original integrand.
\[ \int e^x \sin x\,dx \]
That looks like failure. It is not: the original integral has reappeared on the right-hand side, which means you can solve for it as if it were an unknown.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) 3.1
Picture it
Two applications of parts take you around the circle and back to the integral you started with.
Figure (svg): A circular diagram showing an integral leading to parts once, then parts twice, and returning to the original integral
Worked example
Call the integral I so it can be treated as an unknown.
First pass: differentiate the sine, integrate the exponential
Why: Neither factor simplifies, but be consistent about which one you differentiate on both passes.
\[ I = e^x \sin x - \int e^x \cos x\,dx \]
Second pass on the new integral, same choices
Why: Switching which factor you differentiate here undoes the first pass and returns a trivial identity.
\[ \int e^x \cos x\,dx = e^x \cos x + \int e^x \sin x\,dx \]
Substitute back and recognise the original integral
Why: The last term is I again, with a sign.
\[ I = e^x \sin x - e^x \cos x - I \]
Solve for I algebraically
Why: Add I to both sides and halve. No further integration happens.
\[ I = \frac{e^x(\sin x - \cos x)}{2} + C \]
Verify: by differentiating
Why: The product rule gives the exponential times sine minus cosine, plus the exponential times cosine plus sine, all over two. The cosines cancel and the sines double, leaving the exponential times the sine.
Picture it
The purple curve is the integrand; the green curve is its antiderivative. Where the purple crosses zero, the green is flat.
Figure (svg): The function e to the x times sine x plotted with its antiderivative, showing that the antiderivative is flat wherever the integrand crosses zero
Trap
Second application of parts, with the roles reversed.
\[ \int e^x \cos x\,dx: \; u = e^x, \; dv = \cos x\,dx \]
Differentiate the exponential this time
Why: Legal, but it exactly undoes the first pass.
\[ I = I \]
Arrive at a true statement carrying no information
Why: Nothing is wrong and nothing was learned. Students often take this as evidence that the method fails.
Keep differentiating the same kind of factor on both passes.
\[ \int e^x \cos x\,dx: \; u = \cos x, \; dv = e^x dx \]
Differentiate the trigonometric factor both times
Why: Now the second pass advances the cycle rather than reversing it.
\[ I = e^x \sin x - e^x \cos x - I \]
Solve for the unknown
Why: An equation in I with I appearing once on each side always has a solution.
Section
Section 5
Concept
The logistic differential equation says a population grows in proportion both to its size and to the room it has left.
\[ \frac{dP}{dt} = kP(1 - P) \]
Separating the variables puts a rational function on one side, and that rational function is exactly a partial-fractions problem. Without the technique from Section 2 there is no closed-form solution.
College Board, AP Calculus AB and BC Course and Exam Description, Unit 6 Unit 7
Worked example
Separate, then decompose the left-hand side.
\[ \int \frac{dP}{P(1-P)} = \int k\,dt \]
Decompose the rational function
Why: Two linear factors, so two constants. Substituting the roots gives both immediately.
\[ \frac{1}{P(1-P)} = \frac{1}{P} + \frac{1}{1-P} \]
Integrate each piece
Why: The second picks up a sign from the chain rule, and the two logarithms combine into one.
\[ \ln|P| - \ln|1-P| = kt + C \]
Exponentiate and solve for the population
Why: The ratio of population to remaining room grows exponentially, which rearranges into the familiar S-curve.
\[ P = \frac{Ae^{kt}}{1 + Ae^{kt}} \]
Verify: by recombining the decomposition
Why: One over P plus one over one minus P has numerator one minus P plus P, which is 1, over the product. That is the original integrand, so the decomposition was correct.
Picture it
The solution never exceeds the carrying capacity, and it grows fastest exactly halfway there.
Figure (svg): A logistic S-shaped curve rising from near zero towards a dashed horizontal carrying capacity line, steepest at half the capacity
Free-response questions often ask for the maximum growth rate without solving the equation at all: it happens at half the carrying capacity, which you can read off the product form.
Concept
For a positive, decreasing function, the sum of its values at the whole numbers and the integral of it converge or diverge together. That is the integral test.
\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges exactly when } p > 1 \]
This is the same boundary, at the same place, for the same reason. The p-test for integrals and the p-series test are one theorem wearing two hats.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) 5.3
Picture it
Each term of the series is the area of a rectangle, and every rectangle fits under the curve.
Figure (svg): Rectangles of decreasing height at the whole numbers sitting beneath the curve one over x squared, illustrating the integral test
The comparison runs both ways with a shift of one, which is what makes the test an if-and-only-if rather than a one-way bound.
Analogy
Match each improper-integral fact to the series fact it is really the same statement as.
Match the pairs
Why: Every row is one theorem stated twice. Recognising that halves what you have to memorise for the series unit, and it means a solid grasp of improper integrals is worth marks in a unit you have not started yet.
Explain it
Say it out loud, in terms of the picture.
Discussion prompt
Why does the area under a positive decreasing curve tell you anything about the sum of its values at the whole numbers?
Hint: What is the width of each rectangle, and why does decreasing matter?
Answer:
Because each term of the series is the area of a rectangle of width one, and a decreasing curve means each rectangle either fits under the curve or contains the piece of curve above it.
So the total of the rectangles is trapped between the integral and the integral plus the first term. Trapped between two finite things means finite; trapped above something infinite means infinite.
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) 5.3
Check
Solve it on paper before you click.
Check your understanding
A free-response question asks you to solve a logistic differential equation for the population as a function of time. Which technique from this deck is unavoidable?
Answer: A
Why: Separating the logistic equation leaves the integral of one over P times one minus P. That rational function has to be decomposed before either piece can be integrated.
Section
Section 6
Picture it
Read the integrand top to bottom against this list. The first match is usually right.
Figure (svg): A vertical decision flow listing substitution, integration by parts, partial fractions, improper integrals and algebraic rewriting as successive checks
Discrimination
This is the skill the exam actually tests under time pressure.
Sort into buckets
Which technique does each integrand call for?
Ranking
The order matters: checking for substitution first saves time on problems that would otherwise get parts applied to them.
Put in order
Why: Standard forms first because they cost nothing to check. Substitution next because it is the cheapest real technique and it often turns a hard integrand into a standard form. The improper check comes last only in the sense of technique — but you should notice an infinite limit the moment you read the problem, because it changes how you write every line.
Pattern
Three techniques, one triage routine, and one habit that catches every mistake.
| signal in the integrand | technique | the AB rule it reverses |
|---|---|---|
| inside derivative present as a factor | substitution | the chain rule |
| product of unlike factors | integration by parts | the product rule |
| polynomial times exponential, repeatedly | tabular parts | the product rule, several times |
| rational, denominator factors | partial fractions | adding fractions over a common denominator |
| infinite limit, or integrand blows up | improper: rewrite as a limit | the Fundamental Theorem, plus a limit |
College Board, AP Calculus AB and BC Course and Exam Description, Unit 6 Unit 6
Check
Solve it on paper before you click.
Check your understanding
Which of these improper integrals from one to infinity converges?
Answer: A
Why: The p-test says convergence requires the exponent in the denominator to be strictly greater than one. Only 1.5 clears that bar; its value is 2.
Check
Solve it on paper before you click.
Check your understanding
Using the tabular layout for the integral of x squared times the exponential of x, what is the antiderivative?
Answer: A
Why: The signs alternate down the table: plus, minus, plus. Differentiating the answer gives six terms in which everything cancels except x squared times the exponential.
Exit ticket
One honest answer.
Predict first
Which of the three techniques would you least want to see first on a timed free-response question?
Correct: Whichever you named opens the next session.
Why: If the answer is the fourth option, that is genuinely the most common one and the most fixable: triage is a drill, not a concept, and twenty integrands sorted by technique without solving any of them is usually enough to fix it.
Connect it up
This is the page you want beside you while doing homework, and it has to be in your handwriting to be any use.
Draw it
Write the five signals down the left of a page and the technique beside each. Then add one integrand of your own for each row, and do not solve any of them.
Bring it to the next session and we will work through your own examples rather than mine.
Recap
Three techniques, all of them AB rules run in reverse, plus the triage that picks between them.
| integral | value |
|---|---|
| x times e to the x, from 0 to 1 | 1 |
| x squared times e to the x | x squared minus 2x plus 2, all times e to the x |
| (5x-3) over (x-3)(x+1) | 3 ln|x-3| + 2 ln|x+1| + C |
| one over x squared, from 1 to infinity | 1, converges |
| one over x, from 1 to infinity | diverges |
| one over root x, from 0 to 1 | 2, converges |
OpenStax, Calculus Volume 2, Ch. 3 (Techniques of Integration) Ch. 3 — every technique above with a full problem set
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