- Integration by parts integrates a product of two functions: \( \int u\,\dfrac{dv}{dx}\,dx = uv-\int v\,\dfrac{du}{dx}\,dx \). It is the product rule run backwards.
- Choose \( u \) as the factor that gets simpler when differentiated. The LIATE order (Logs, Inverse trig, Algebraic, Trig, Exponential) is a reliable guide.
- Case: a subscription product earning \( 120t \) thousand dollars a year has a present value over 5 years, at 8% continuous discounting, of about $1.154 million, against $1.5 million of undiscounted revenue.
- Powers like \( x^2 \) need parts twice; the tabular method organises the repeats neatly.
- Exam point: parts is AP Calculus BC only, Cambridge 9709 Paper 3, and IB Analysis and Approaches HL.
The case: what is a growing revenue stream worth today?
A small software company has just launched a subscription app. Sign-ups are growing steadily, and the finance team models the revenue rate as rising in a straight line from zero: after \( t \) years, the app brings in money at a rate of \( R(t) = 120t \) thousand dollars per year. So at the end of year 1 it is earning at $120,000 a year, at the end of year 2 at $240,000 a year, and so on. These figures are illustrative, not taken from a real company.
A larger rival has offered to buy the app outright for $1.1 million today. The founders plan to judge the offer against the first five years of revenue only, because the market may change after that. Over five years the app would earn \( \int_0^5 120t\,dt = 1500 \) thousand dollars, or $1.5 million. That looks like an easy “no” to the offer.
But money received in year 5 is worth less than money received today, because money today could be invested and earn interest in the meantime. The company uses a discount rate of 8% a year, compounded continuously. The real question is: what is the present value of the revenue stream over the next five years, and is it more than $1.1 million?
The concept: discounting a flow, and integrating a product
With continuous discounting at rate \( r \), one dollar received at time \( t \) is worth \( e^{-rt} \) dollars today. A revenue stream arrives bit by bit: in a short interval of length \( dt \) near time \( t \), the app earns about \( R(t)\,dt \), which is worth \( R(t)e^{-rt}\,dt \) today. Adding up all those slices gives the present value as an integral.[5]
For our app, that is \( \int_0^5 120t\,e^{-0.08t}\,dt \). The integrand is a product of two different kinds of function: a polynomial, \( t \), and an exponential, \( e^{-0.08t} \). No basic rule integrates that product directly, and substitution does not help, because the derivative of the exponent, \( -0.08 \), is not a factor of \( t \). This is exactly the job integration by parts was made for.
The formula
Here \( u \) is the factor you choose to differentiate, \( \frac{du}{dx} \) is its derivative, \( \frac{dv}{dx} \) is the factor you choose to integrate, and \( v \) is its integral (no constant needed at this stage). The second line is the version for definite integrals: the \( uv \) part is evaluated between the limits straight away.[1]
Chapter 14 of Mathematics Made Simple covers integration by parts, substitution and partial fractions step by step.Get the textbook · $12
Why this formula, and when not to use it
The formula comes straight from the product rule. If \( u \) and \( v \) are functions of \( x \), then \( \dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx} \). Integrate both sides and rearrange, and you get the formula above. So parts is the product rule in reverse, in the same way that substitution is the chain rule in reverse.[2]
The trade only helps if the new integral, \( \int v\,\frac{du}{dx}\,dx \), is easier than the old one. That happens when differentiating \( u \) makes it simpler and integrating \( \frac{dv}{dx} \) does not make it worse. For \( t\,e^{-0.08t} \), differentiating \( t \) gives 1, and integrating \( e^{-0.08t} \) just gives another exponential. The polynomial disappears, which is the whole point.
The usual guide for choosing \( u \) is LIATE: pick as \( u \) whichever factor comes first in the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Logs come first because they are easy to differentiate (\( \ln x \to \frac{1}{x} \)) but awkward to integrate. Exponentials come last because integrating them is painless.
Parts is not the right tool when the integrand is a product in which one factor is (a multiple of) the derivative of the inside of the other, such as \( 2x\,e^{x^2} \). That is a substitution integral, and it takes one line with \( u = x^2 \). Try substitution first; reach for parts when you see two unrelated types of function multiplied together.
Setting it up
Label each quantity in the case, with its units:
- \( t \): time in years, from \( 0 \) to \( T = 5 \).
- \( R(t) = 120t \): revenue rate in thousands of dollars per year.
- \( r = 0.08 \): continuous discount rate per year, so the discount factor is \( e^{-0.08t} \).
- Target: \( PV = 120\int_0^5 t\,e^{-0.08t}\,dt \), in thousands of dollars.
Take the constant 120 outside the integral. Then, following LIATE (algebraic before exponential), choose \( u = t \) and \( \frac{dv}{dt} = e^{-0.08t} \). It is cleaner to do the general case first, with any rate \( r \) and horizon \( T \), so that the result can be reused for the sensitivity check later.
Solving it
Find \( PV = 120\displaystyle\int_0^5 t\,e^{-0.08t}\,dt \), the present value of the app’s revenue, in thousands of dollars.
The present value of five years of revenue is about $1,154,000. That beats the $1.1 million offer, but only by around $54,000, not by the $400,000 the undiscounted total suggested.
One application of parts was enough there because the polynomial was \( t \), which differentiates to a constant in one go. With \( t^2 \) or \( x^2 \) you need two rounds. Instead of writing the formula out twice, many students use the tabular method: list \( u \) and its derivatives down one column until you reach zero, list \( \frac{dv}{dx} \) and its repeated integrals down another, then multiply along the diagonals with alternating signs \( +, -, +, \dots \)
Find \( \displaystyle\int x^2 e^{2x}\,dx \) using the tabular method.
| Sign | Differentiate \( u \) | Integrate \( \frac{dv}{dx} \) |
|---|---|---|
| \( + \) | \( x^2 \) | \( e^{2x} \) |
| \(-\) | \( 2x \) | \( \tfrac{1}{2}e^{2x} \) |
| \( + \) | \( 2 \) | \( \tfrac{1}{4}e^{2x} \) |
| \( 0 \) | \( \tfrac{1}{8}e^{2x} \) |
The tabular method is just repeated integration by parts written compactly, so it is acceptable in exams as long as your working is clear.
The third example puts the log at the front of the LIATE queue. It is a favourite on Cambridge Paper 3.
Exam style. Show that \( \displaystyle\int_1^e x\ln x\,dx = \frac{e^2 + 1}{4} \).
Choosing \( u = x \) here would force you to integrate \( \ln x \), which is harder than the question you started with.
Practise LIATE, repeated parts and the tabular method with 1,565 questions and fully worked answers.Get 1,565 questions · $9
What the answer tells us
Discounting cuts the value of the five-year stream from $1.5 million to about $1.154 million, so today’s value is about 76.9% of the headline total. The loss is large because this stream is back-loaded: most of the money comes in years 4 and 5, when the discount factor is smallest. At \( t = 5 \) the app earns at $600,000 a year, but each of those dollars is worth only \( e^{-0.4} \approx 0.67 \) dollars today. Figure 1 shows the gap between the two curves growing over time.
A quick reasonableness check: the PV must lie between the discounted value of $1.5 million paid all at once at the end (\( 1500e^{-0.4} \approx 1005.5 \)) and $1.5 million paid today. Our 1154.1 sits comfortably between, and closer to the lower end because the money is weighted towards the end.
Because the general formula from Step 5 holds for any \( r \), the sensitivity to the discount rate takes seconds:
| Discount rate \( r \) | PV ($ thousand) | Share of $1.5m | Accept $1.1m offer? |
|---|---|---|---|
| 5% | 1272.0 | 84.8% | No |
| 8% | 1154.1 | 76.9% | No, narrowly |
| 12% | 1015.8 | 67.7% | Yes |
The decision flips somewhere between 8% and 12%. A riskier outlook for the app would justify a higher discount rate, and then selling looks wise. The calculus does not make the decision, but it shows exactly which assumption the decision depends on.
The Mathematics Made Simple textbook develops integration by parts alongside substitution and partial fractions in Chapter 14, and uses present-value integrals again in Appendix B, Mathematics for economists. For volume practice, the Mathematics Made Simple practice book includes integration questions at every level, each with its working shown.
Revise all of integration with the Mathematics Made Simple bundle: both books for $16, a $5 saving.Both books · $16
Practice questions
EasyQ1. Find \( \displaystyle\int x\,e^{x}\,dx \). Show worked answer
\( \int x\,e^x\,dx = x\,e^x-\int e^x\,dx \)
\( {}= x\,e^x-e^x + c = e^x(x-1) + c \).
EasyQ2. Find \( \displaystyle\int x\cos x\,dx \). Show worked answer
\( \int x\cos x\,dx = x\sin x-\int \sin x\,dx \)
\( {}= x\sin x + \cos x + c \).
MediumQ3. Find \( \displaystyle\int \ln x\,dx \) by writing it as \( \int 1 \cdot \ln x\,dx \). Show worked answer
\( \int \ln x\,dx = x\ln x-\int x \cdot \frac{1}{x}\,dx \)
\( {}= x\ln x-x + c \).
MediumQ4. Evaluate \( \displaystyle\int_0^2 x\,e^{-x}\,dx \), giving an exact answer and a value to 3 decimal places. Show worked answer
\( \Big[-(x + 1)e^{-x}\Big]_0^2 = -3e^{-2}-(-1) = 1-3e^{-2} \).
\( 1-3e^{-2} \approx 0.594 \).
ExamQ5. (a) Use integration by parts to find \( \displaystyle\int t\,e^{-0.1t}\,dt \). [3] (b) A machine produces income at a rate of \( 50t \) thousand dollars per year. Show that, at a continuous discount rate of 10%, the present value of its income over the first 10 years is \( 5000(1-2e^{-1}) \) thousand dollars. [3] (c) Evaluate this to the nearest thousand dollars. [1] Show worked answer
\( \int t\,e^{-0.1t}\,dt = -10t\,e^{-0.1t} + 10\int e^{-0.1t}\,dt \) [1]
\( {}= -10t\,e^{-0.1t}-100e^{-0.1t} + c \) [1].
(b) \( PV = 50\int_0^{10} t\,e^{-0.1t}\,dt \) [1].
\( \Big[-10t\,e^{-0.1t}-100e^{-0.1t}\Big]_0^{10} \)
\( {}= (-100e^{-1}-100e^{-1})-(-100) \)
\( {}= 100-200e^{-1} \) [1].
\( PV = 50(100-200e^{-1}) \)
\( {}= 5000-10000e^{-1} = 5000(1-2e^{-1}) \) [1].
(c) \( 5000 \times 0.264241 \approx 1321.2 \), so about $1,321,000 [1].
Frequently asked questions
How do I choose u in integration by parts?
Use LIATE: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Whichever factor appears first in that list becomes \( u \). The aim is for \( u \) to get simpler when differentiated while \( \frac{dv}{dx} \) stays easy to integrate.
What if integration by parts goes round in a circle?
For integrals like \( \int e^x\cos x\,dx \), applying parts twice brings back the original integral. Call it \( I \), write the equation \( I = e^x\sin x + e^x\cos x-I \), and solve: \( I = \tfrac{1}{2}e^x(\sin x + \cos x) + c \). This “loop” trick appears on Cambridge Paper 3 and in IB HL.
Is the tabular method allowed in exams?
Yes, as long as the working is clear. It is only a compact way of writing repeated integration by parts. It works best when \( u \) is a polynomial that eventually differentiates to zero.
Why does present value use e to the minus rt?
With continuous compounding at rate \( r \), one dollar grows to \( e^{rt} \) dollars after \( t \) years. Turning that round, a dollar due at time \( t \) is worth \( e^{-rt} \) today. Multiplying a revenue rate by \( e^{-rt} \) and integrating adds up the present value of every small slice of income.
References
- Strang, G. and Herman, E. (2016). Calculus Volume 2. Houston, TX: OpenStax. Section 3.1, Integration by Parts. https://openstax.org/details/books/calculus-volume-2
- Stewart, J. (2016). Calculus: Early Transcendentals, 8th edition. Boston, MA: Cengage Learning.
- College Board (2020). AP Calculus AB and AP Calculus BC Course and Exam Description. New York: College Board.
- Cambridge International. Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. Cambridge: Cambridge University Press & Assessment.
- Chiang, A. C. and Wainwright, K. (2005). Fundamental Methods of Mathematical Economics, 4th edition. New York: McGraw-Hill.
