Integrating the natural exponential and its substitutions, general exponentials with other bases, the reciprocal with its absolute value, the logarithmic form for a quotient whose numerator is the denominator's derivative, and applications to growth and decay.
Subject: Calculus I · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Calculus I · Chapter 5 — Integration
Integrals Involving Exponential and Logarithmic Functions
Objectives
Five outcomes. Every formula here is Section 3.9 reversed, and the technique is Section 5.5's.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 516-525 — the section these objectives are drawn from
Warm-up
Section 3.9 differentiated the exponential and logarithmic family, and Section 5.5 supplied substitution.
Discussion prompt
What is the integral of the exponential of 3x? The exponential is its own antiderivative, but the inner function is not a bare x.
Hint: Substitute, or recognise the pattern for a linear inside.
Answer:
Setting u equal to 3x gives du equal to 3 dx, so dx is du over 3 and the answer carries a factor of one third.
\[ \int e^{3x}dx = \frac{e^{3x}}{3}+C \]
Differentiating confirms it: the chain rule brings a 3 down, cancelling the third. The pattern generalises — any linear inner function contributes the reciprocal of its slope — and it is worth recognising directly rather than substituting every time, because linear inner functions are extremely common in this family.
Concept
Every integration formula in this section is a Section 3.9 derivative read backwards. What is new is recognising which integrands have that shape, especially quotients whose numerator is the denominator's derivative.
the logarithmic form — An integrand that is a quotient whose numerator is the derivative of its denominator integrates to the natural logarithm of the denominator's absolute value.
\[ \int e^{x}dx = e^{x}+C, \qquad \int\frac{u'}{u}\,dx = \ln|u|+C \]
The second formula is the section's workhorse and is nothing more than substitution with the denominator as the new variable. Recognising the shape is the whole skill.
Figure (svg): The exponential and logarithmic integration formulas, each reversed from Section 3.9
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 556-565
Section
Section 1
Concept
The natural exponential integrates to itself. With anything but a bare x in the exponent, substitution supplies the necessary factor — provided the exponent's derivative is present.
exponential integrals — The natural exponential is its own antiderivative. A composition requires the exponent's derivative as a factor, which substitution then absorbs.
\[ \int e^{u}u'\,dx = e^{u}+C \]
For a linear exponent the derivative is a constant and can always be adjusted. For a non-linear one it must actually be present, and when it is not the integral is generally non-elementary.
Figure (svg): The exponential and logarithmic integration formulas, each reversed from Section 3.9
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 556-563 — integrals of exponential functions
Picture it
Six entries, five of them direct reversals.
Figure (svg): The exponential and logarithmic integration formulas, each reversed from Section 3.9
The last entry — the logarithm's own integral — is not a reversal of anything in the table and needs integration by parts, a technique beyond this course. The others follow immediately from Section 3.9.
Worked example
Example 5.49. The derivative is present.
\[ \text{Evaluate } \int x e^{4x^{2}}dx. \]
Choose the exponent as the substitution
Why: The inner function.
\[ u = 4 x ^{2} \]
Compute the differential
Why: Differentiate.
\[ \,du = 8 x \,dx \]
Compare with the integrand
Why: Only x dx is present.
\[ x \,dx = \,du / 8 \]
Rewrite
Why: The constant comes outside.
\[ (\frac{1}{8}) \int e ^{u} \,du \]
Integrate and substitute back
Why: The exponential is its own antiderivative.
\[ e ^{4 x ^{2}} / 8 + C \]
Figure (svg): The exponential and logarithmic integration formulas, each reversed from Section 3.9
\[ \frac{e^{4x^{2}}}{8}+C \]
Verify: differentiate the answer back
Why: The chain rule gives one eighth times the exponential times 8x, which is x times the exponential — the original integrand. Note that without the factor of x the integral would be the exponential of 4x squared alone, which has no elementary antiderivative: it is the error function's relative. A single factor of x is again the difference between routine and impossible.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 558-559
Fill the middle
An exponential with a linear exponent.
Fill in the blanks
\int e^3dx = \frac___}___}+C
Why: The chain rule would bring a 3 down, so dividing by 3 cancels it. Any linear inner function contributes the reciprocal of its slope, which is worth recognising directly.
Worked example
Checkpoint 5.49. Limits converted.
\[ \text{Evaluate } \int_{0}^{1}e^{-2x}dx. \]
Recognise the linear exponent
Why: Its derivative is a constant.
\[ \text{factor of } -\frac{1}{2} \]
Antidifferentiate
Why: Divide by the slope.
\[ -e ^{-2 x} / 2 \]
Evaluate at 1
Why: Substitute.
\[ -e ^{-2} / 2 \]
Evaluate at 0 and subtract
Why: Substitute.
\[ -\frac{1}{2} \]
Combine
Why: Upper minus lower.
\[ \frac{1 - e ^{-2}}{2} \]
Figure (svg): The solution to Worked example a definite exponential integral shown as a ladder of expressions, one row per legal move
\[ \frac{1-e^{-2}}{2} \approx 0.432 \]
Verify: sanity-check against the integrand's range
Why: The integrand falls from 1 at x equal to 0 to about 0.135 at x equal to 1, so over a width of 1 the answer must lie between those values — and 0.432 sits inside, nearer the lower end because the decay is fastest early. Note that this integral over the whole positive axis would converge to one half, since the tail contributes an ever smaller amount, which is the kind of question Chapter 6 takes up.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 559-561
Trap
\[ \int e^{3x}dx = e^{3x}+C \]
Treat the exponential as its own antiderivative regardless
Why: The student ignores the inner function.
\[ \frac{d}{dx}e^{3x} = 3e^{3x} \ne e^{3x} \]
Differentiating the proposed answer gives three times the integrand, so the answer is three times too large.
\[ \int e^{3x}dx = \frac{e^{3x}}{3}+C \]
Divide by the inner function's derivative when it is a constant
Why: The chain rule would otherwise multiply by it.
The check is immediate: differentiate and see whether the integrand comes back. Because that check is always available, this is another mistake that need never survive to a final answer.
Sorting
Is the exponent's derivative present?
Sort into buckets
Sort each integrand.
The last is the subtle one: an x squared is present but the derivative needed is 2x, and the extra factor cannot be absorbed. Having too much of the variable is as fatal as having none.
Two truths and a lie
All three are about exponential integrals.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. That property holds only for a bare x in the exponent. Anything else needs the chain rule accounted for, and for a non-linear exponent the required factor must actually be present in the integrand or nothing can be done.
Prediction
Commit before reasoning.
Predict first
Why can an exponential with any linear exponent always be integrated, while a quadratic one usually cannot?
Correct: Because a linear function's derivative is a constant.
\[ u=kx: \; du=k\,dx \;\checkmark\; \qquad u=x^{2}: \; du=2x\,dx \;\text{must be present} \]
Why: Substitution requires the inner derivative to be present, and a numerical factor may be supplied by multiplying and dividing. A linear inner function has a constant derivative, so the requirement is always satisfiable. A quadratic's derivative contains x, which cannot be created — so unless the integrand already contains it, the integral is non-elementary. The boundary is exactly the one Section 5.5 established.
Section
Section 2
Concept
An exponential with any base is the natural exponential with a rescaled exponent. Rewriting it that way integrates it in two lines and derives the general formula.
the base-change identity — Any positive base raised to a power equals the natural exponential of that power times the base's natural logarithm, which converts every exponential into a natural one.
\[ a^{x}=e^{x\ln a} \;\Longrightarrow\; \int a^{x}dx = \frac{a^{x}}{\ln a}+C \]
The formula is worth deriving rather than memorising: two lines reproduce it, and the derivation makes clear where the logarithm in the denominator comes from.
Figure (svg): A general exponential integrated by rewriting it in base e
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 560-567 — exponentials with other bases
Picture it
Base change, integrate, convert back.
Figure (svg): A general exponential integrated by rewriting it in base e
The logarithm appears in the denominator because the base change put it in the exponent, where it acts as a linear slope — and a linear slope always contributes its reciprocal.
Worked example
Example 5.51. Rewrite, then integrate.
\[ \text{Evaluate } \int 2^{x}dx. \]
Rewrite in base e
Why: The base-change identity.
\[ e ^{x \ln 2} \]
Identify the exponent's slope
Why: A constant.
\[ \ln 2 \]
Integrate
Why: Divide by the slope.
\[ e ^{x \ln 2} / \ln 2 \]
Convert back
Why: The identity in reverse.
\[ 2 ^{x} / \ln 2 \]
Add the constant
Why: As always.
\[ +C \]
Figure (svg): A general exponential integrated by rewriting it in base e
\[ \frac{2^{x}}{\ln 2}+C \]
Verify: differentiate the answer back
Why: Section 3.9 gives the derivative of 2 to the x as 2 to the x times the logarithm of 2, so dividing by that logarithm returns exactly 2 to the x. Note that base e is the only base with no such factor, since its logarithm is 1 — which is the whole reason it is called natural, and why every other base carries this correction.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 562-563
Fill the middle
An exponential rewritten as a natural one.
Fill in the blanks
a^\ln a = e^___}}
Why: The identity converts every exponential into a natural one with a rescaled exponent. That rescaling is a constant, so integrating contributes its reciprocal — which is where the logarithm in the denominator comes from.
Worked example
Checkpoint 5.51. Both complications at once.
\[ \text{Evaluate } \int x\,3^{x^{2}}dx. \]
Rewrite in base e
Why: The identity.
\[ e ^{x ^{2} \ln 3} \]
Choose the substitution
Why: The exponent.
\[ u = x ^{2} \ln 3 \]
Compute the differential
Why: Differentiate.
\[ \,du = 2 x \ln 3 \,dx \]
Adjust and rewrite
Why: The integrand has x dx.
\[ (\frac{1}{2 \ln 3}) \int e ^{u} \,du \]
Integrate and convert back
Why: Both substitutions undone.
\[ 3 ^{x ^{2}} / (2 \ln 3) + C \]
Figure (svg): The solution to Worked example a base with a composition shown as a ladder of expressions, one row per legal move
\[ \frac{3^{x^{2}}}{2\ln 3}+C \]
Verify: differentiate back and account for both factors
Why: Differentiating gives 3 to the x squared, times the logarithm of 3, times 2x, all divided by twice that logarithm — leaving x times 3 to the x squared, the original integrand. The two corrections are independent: one for the base and one for the quadratic exponent, and both are constants so both are adjustable. Recognising that they multiply rather than interact is what keeps such problems manageable.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 563-565
Error analysis
A student integrates an exponential with base two.
Annotate
On: \( \int 2^{x}dx = 2^{x}\ln 2 + C \)
The confusion is with the derivative formula, where the logarithm does multiply. Integration reverses that, so it divides — and one differentiation of the answer resolves the direction every time.
Two truths and a lie
All three are about other bases.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Differentiating multiplies by the base's logarithm, so integrating divides by it — reversing an operation reverses what it does to constants. Differentiating the proposed answer settles the direction in one line.
Sorting
Base and exponent both contribute.
Sort into buckets
Sort each integrand by what its antiderivative is divided by.
The two corrections are independent and simply multiply, which is what keeps the fourth case from being any harder than the others. Recognising that saves treating it as a new type of problem.
Prediction
Commit before reasoning.
Predict first
Why does base e alone need no correction factor?
Correct: Because the logarithm of e is 1.
\[ \ln e = 1 \;\Longrightarrow\; \int e^{x}dx = \frac{e^{x}}{1}+C \]
Why: Every base contributes a factor of its natural logarithm on differentiating, and e's is exactly 1 — so the factor is invisible. That is precisely what makes e the natural base: it is chosen so this correction disappears, which simplifies every derivative and integral in the family. Nothing about its irrationality or its size is relevant.
Section
Section 3
Concept
The reciprocal integrates to the natural logarithm of the absolute value, which supplies an antiderivative on either side of the origin. It does not make the integral across the origin meaningful.
the absolute value in the logarithm — The logarithm is undefined for negative inputs, but the reciprocal is not, so the antiderivative uses the absolute value to cover both sides of zero with one formula.
\[ \int\frac{dx}{x} = \ln|x|+C, \quad x \ne 0 \]
The restriction is easy to forget. An interval containing zero makes the integrand unbounded, so the integral is not defined there at all and no antiderivative formula rescues it.
Figure (svg): Why the logarithm carries an absolute value
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 563-570 — integrals of logarithmic functions
Picture it
The reciprocal and the logarithm of the absolute value.
Figure (svg): Why the logarithm carries an absolute value
The formula works on either branch and the two branches are genuinely separate. Nothing connects them, and an integral crossing the origin is undefined rather than merely difficult.
Worked example
Example 5.53. The absolute value earning its place.
\[ \text{Evaluate } \int_{-3}^{-1}\frac{dx}{x}. \]
Check the interval
Why: It avoids zero.
Antidifferentiate
Why: With the absolute value.
\[ \ln | x | \]
Evaluate at -1
Why: The absolute value is 1.
\[ \ln 1 = 0 \]
Evaluate at -3 and subtract
Why: The absolute value is 3.
\[ 0 - \ln 3 \]
State
Why: The value.
\[ -\ln 3,\text{ about } -1.099 \]
Figure (svg): Why the logarithm carries an absolute value
\[ \int_{-3}^{-1}\frac{dx}{x} = -\ln 3 \approx -1.099 \]
Verify: confirm the sign against the integrand
Why: The reciprocal is negative throughout this interval, so a negative integral is correct. Without the absolute value the antiderivative would be undefined at both limits and the computation impossible — yet the area is perfectly well defined, which is exactly why the absolute value belongs in the formula. Note the answer's magnitude matches the integral from 1 to 3, as the symmetry of the reciprocal about the origin would suggest.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 566-567
Sorting
Check for singularities inside the interval.
Sort into buckets
Sort each integral of the reciprocal.
The fourth is the dangerous one, because the singularity is not at a symmetric position and the formula produces a plausible non-zero number. Checking the interval for the integrand's singularities should come before any antiderivative is written.
Worked example
Checkpoint 5.53. What the formula cannot do.
\[ \text{What is wrong with } \int_{-1}^{1}\frac{dx}{x}? \]
Check the integrand at zero
Why: It is undefined.
Recall the integrability condition
Why: Section 5.2.
Note what a formula would give
Why: ln 1 minus ln 1.
Explain why that is wrong
Why: The antiderivative is not defined on the whole interval.
\[ \text{Part } 2\text{ does not apply} \]
State
Why: The integral does not exist.
Figure (svg): The solution to Worked example an interval containing zero shown as a ladder of expressions, one row per legal move
\[ \text{undefined, though the formula appears to give } 0 \]
Verify: see why the apparent answer is so misleading
Why: Blindly applying the antiderivative gives zero, which looks like a legitimate answer and even seems to fit the reciprocal's symmetry. But Part 2 requires the antiderivative to be defined and continuous on the whole interval, and the logarithm of the absolute value is not defined at zero. The two halves are separately infinite, and infinity minus infinity is precisely Section 4.8's indeterminate form — no cancellation is legitimate.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 567-569
Trap
\[ \int_{-1}^{1}\frac{dx}{x} = \ln|x|\Big|_{-1}^{1} = 0 \]
Apply the antiderivative regardless
Why: The student trusts the formula.
Part 2 requires the antiderivative to be continuous on the whole interval, and it is undefined at zero. The apparent cancellation is between two infinities.
\[ \text{the integral does not exist on } [-1,1] \]
Check the integrand is defined and bounded on the interval first
Why: An unbounded integrand is not integrable, by Section 5.2.
Checking the integrand for singularities inside the interval takes a moment and prevents an answer that looks entirely reasonable. The formula's absolute value covers each side of zero; it does not bridge them.
Fill the middle
The reciprocal's antiderivative.
Fill in the blanks
\int\frac|___ = \ln___x| + C
Why: The absolute value gives an antiderivative on both sides of zero with one formula. It does not, however, make an integral across zero meaningful — the integrand is unbounded there.
Two truths and a lie
All three are about the reciprocal.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The absolute value covers each side of the origin separately and bridges nothing — the two branches are genuinely disconnected. Applying the formula across zero gives an apparent answer that is a cancellation between two infinities, which is Section 4.8's indeterminate form and not a legitimate value.
Prediction
Commit before reasoning.
Predict first
Why does the reciprocal's antiderivative carry an absolute value?
Correct: Because the reciprocal is defined for negative inputs.
\[ \frac{d}{dx}\ln(-x) = \frac{-1}{-x} = \frac1x \quad \text{for } x<0 \]
Why: On the negative side the reciprocal is a perfectly ordinary continuous function, so it must have an antiderivative there — and the plain logarithm does not exist for negative inputs. Differentiating the logarithm of negative x by the chain rule gives one over negative x times negative one, which is the reciprocal, so the absolute-value form works on both branches. It is a necessity rather than a tidying convention.
Section
Section 4
Concept
A quotient whose numerator is the derivative of its denominator integrates to the logarithm of the denominator's absolute value. It is substitution with the denominator as the new variable.
the logarithmic form — An integrand of the shape derivative over function. Substituting the denominator turns it into the reciprocal, whose antiderivative is the logarithm.
\[ \int\frac{u'(x)}{u(x)}\,dx = \ln|u(x)|+C \]
Checking whether the numerator is the denominator's derivative costs one differentiation and immediately settles whether this form applies. A missing constant factor is adjustable; a missing variable factor is not.
Figure (svg): The logarithmic form: numerator is the denominator's derivative
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 565-572 — the logarithmic integration formula
Picture it
Whether the numerator matches.
Figure (svg): The logarithmic form: numerator is the denominator's derivative
The third row is the trap: it differs from the first by a single factor of x in the numerator, and the answer changes from a logarithm to an inverse tangent — an entirely different function.
Worked example
Example 5.55. A classic logarithmic form.
\[ \text{Evaluate } \int\tan x\,dx. \]
Rewrite as a quotient
Why: The definition of the tangent.
\[ \sin x / \cos x \]
Check the numerator against the denominator's derivative
Why: Differentiate the cosine.
Choose the substitution
Why: The denominator.
\[ u = \cos x \]
Adjust the sign
Why: The differential is negative.
\[ -\int \,du / u \]
Integrate and substitute back
Why: The logarithm.
\[ -\ln | \cos x | + C \]
Figure (svg): The logarithmic form: numerator is the denominator's derivative
\[ -\ln|\cos x|+C = \ln|\sec x|+C \]
Verify: differentiate back and reconcile the two forms
Why: Differentiating negative the logarithm of the absolute cosine gives negative one over cosine times negative sine, which is the tangent — correct. The alternative form follows from the logarithm law: the negative of a logarithm is the logarithm of the reciprocal, and the reciprocal of cosine is the secant. Both appear in tables and both are right, which is worth knowing before assuming a book disagrees.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 568-569
Sorting
Differentiate the denominator and compare.
Sort into buckets
Sort each integrand.
The fourth needs only a factor of one half, which is adjustable, so it gives half a logarithm. The second needs a factor of x, which is not adjustable, and gives an inverse tangent instead — a completely different function.
Worked example
Checkpoint 5.55. One factor of x apart.
\[ \text{Compare } \int\frac{2x\,dx}{x^{2}+1} \text{ with } \int\frac{dx}{x^{2}+1}. \]
Check the first numerator
Why: Against the denominator's derivative.
\[ \text{exactly } 2 x:\text{ it matches} \]
Integrate the first
Why: The logarithmic form.
\[ \ln(x ^{2} + 1) + C \]
Check the second numerator
Why: Against 2x.
Recognise the second
Why: From Section 4.10's table.
Integrate the second
Why: Directly.
\[ \arctan x + C \]
Figure (svg): The solution to Worked example two similar quotients shown as a ladder of expressions, one row per legal move
\[ \ln(x^{2}+1)+C, \qquad \arctan x+C \]
Verify: note how far apart the answers are
Why: Two integrands differing by a single factor of x give a logarithm and an inverse tangent — functions with entirely different shapes, one unbounded and one approaching a horizontal asymptote. Nothing about the integrands' appearance suggests such a gap, and the only reliable way to tell them apart is to check the numerator against the denominator's derivative. Note that no absolute value is needed on the first, since the denominator is always positive.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 569-571
Error analysis
A student integrates a quotient.
Annotate
On: \( \int\frac{dx}{x^{2}+1} = \ln|x^{2}+1|+C \)
One differentiation of the denominator settles whether this form applies, and it takes a few seconds. Assuming it because the integrand is a quotient produces an answer that is wrong in kind, not merely in a constant.
Fill the middle
A quotient whose denominator has a known derivative.
Fill in the blanks
\frac2x___(x^___+1) = ___, \text___
Why: Differentiating the denominator and comparing with the numerator is the whole test. It takes one line and distinguishes a logarithm from an inverse tangent, which are not close.
Matching
Similar quotients, different answers.
Match the pairs
Why: The first two share a denominator and give unrelated functions; the last two are reciprocals of each other and differ only by a sign. Checking the numerator against the denominator's derivative is what separates all four.
Prediction
Commit before reasoning.
Predict first
Two quotients have the same denominator and give a logarithm and an inverse tangent. What decides which?
Correct: Whether the numerator is the denominator's derivative.
\[ \frac{2x}{x^{2}+1} \to \ln(x^{2}+1); \qquad \frac{1}{x^{2}+1} \to \arctan x \]
Why: With 2x on top, substituting the denominator turns the integral into the reciprocal and gives a logarithm. With 1 on top there is nothing to absorb the differential, so a different formula applies entirely — the inverse tangent's, from Section 4.10's table. The denominator is identical in both cases and decides nothing on its own; the numerator decides everything.
Section
Section 5
Concept
Exponential rate laws describe any quantity whose rate of change is proportional to its current amount. Integrating them gives accumulated totals, and the constant ratio is what makes half-lives meaningful.
exponential decay — A quantity whose rate of decrease is proportional to its amount, so it loses the same FRACTION in equal intervals. The time to halve is constant and independent of the starting amount.
\[ Q(t)=Q_{0}e^{-kt}, \qquad \int_{0}^{T}Q = \frac{Q_{0}}{k}\left(1-e^{-kT}\right) \]
Section 6.8 develops this into a full treatment. Here the point is that the integrals of this section are exactly what such models require, which is why the family is worth its own section.
Figure (svg): Exponential decay: the same fraction lost in equal intervals
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 568-575 — applications
Picture it
Exponential decay at equal intervals.
Figure (svg): Exponential decay: the same fraction lost in equal intervals
Each unit of time removes half of whatever remains, so the amount lost shrinks while the fraction lost does not. That is what distinguishes exponential from linear decay.
Worked example
Example 5.57. Integrating a decay.
\[ \text{A source emits at } r(t)=50e^{-0.2t} \text{ units per hour. Find the total over } 10 \text{ hours.} \]
Set up the integral
Why: The Net Change Theorem.
\[ \int\text{ from } 0\text{ to } 10\text{ of } 50 e ^{-0.2 t} \]
Antidifferentiate
Why: Divide by the exponent's slope.
\[ -250 e ^{-0.2 t} \]
Evaluate at 10
Why: The exponent is -2.
\[ -250 e ^{-2} \]
Evaluate at 0 and subtract
Why: The exponent is zero.
\[ -250 \]
Combine
Why: Upper minus lower.
\[ 250(1 - e ^{-2}) \]
Figure (svg): Exponential decay: the same fraction lost in equal intervals
\[ 250\left(1-e^{-2}\right) \approx 216 \]
Verify: compare with the total over an unbounded time
Why: Letting the upper limit grow without bound makes the exponential vanish and the total approaches 250 units — so ten hours has already captured 86 percent of everything the source will ever emit. That a decaying source has a finite lifetime total despite never quite stopping is characteristic of exponential decay, and it is why such sources are described by a total yield rather than a duration.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 571-573
Fill the middle
A decay constant from a known half-life.
Fill in the blanks
\tfrac12 = e^5 \;\Longrightarrow\; k = \frac______}
Why: Taking logarithms brings the constant down and the initial amount cancels, so the half-life determines the constant regardless of the starting quantity.
Worked example
Checkpoint 5.57. The logarithm doing the work.
\[ \text{A substance has a half-life of } 5 \text{ hours. Find its decay constant.} \]
Write the model
Why: Exponential decay.
\[ Q = Q _{0} e ^{-k t} \]
Impose the half-life
Why: Half remains at t = 5.
\[ Q _{0} / 2 = Q _{0} e ^{-5 k} \]
Cancel the initial amount
Why: It divides out.
\[ \frac{1}{2} = e ^{-5 k} \]
Take logarithms
Why: To bring k down.
\[ -\ln 2 = -5 k \]
Solve
Why: Divide.
\[ k = \ln(2) / 5,\text{ about } 0.139 \]
Figure (svg): The solution to Worked example finding a rate constant from a half-life shown as a ladder of expressions, one row per legal move
\[ k = \frac{\ln 2}{5} \approx 0.139 \]
Verify: note that the initial amount cancelled
Why: The starting quantity divided out at step three, so the half-life does not depend on how much there was — which is the defining feature of exponential decay and the reason half-lives are quoted as properties of a substance rather than of a sample. For linear decay the time to lose half WOULD depend on the starting amount, so no such constant would exist.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 573-575
Trap
\[ \text{half-life } 5\text{ h} \;\Longrightarrow\; \text{it is gone after } 10\text{ h} \]
Assume two half-lives empty it
Why: The student reasons as though a fixed amount is lost each period.
After ten hours a quarter remains, not nothing. Each half-life removes half of what is LEFT, not half of the original.
\[ Q(10) = Q_{0}\cdot\tfrac12\cdot\tfrac12 = \tfrac{Q_{0}}{4} \]
Halve the remaining amount each period
Why: Exponential decay has a constant ratio, not a constant difference.
The quantity never reaches zero, approaching it asymptotically — which is why contaminated sites are described by how long until a level is safe rather than until it is gone.
Sorting
Constant fraction, or constant amount?
Sort into buckets
Sort each situation.
The test is whether the wording says a fraction or an amount per unit time. Exponential quantities approach zero without reaching it; linear ones hit zero at a definite moment, which is a sharply different prediction.
Two truths and a lie
All three are about decay.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Two half-lives leave a quarter, three leave an eighth, and the quantity never reaches zero. Each period removes half of what remains rather than half of the original, which is the difference between a constant ratio and a constant difference.
Prediction
Commit before reasoning.
Predict first
Why can a single half-life describe a substance regardless of how much there is?
Correct: Because the initial amount cancels.
\[ \frac{Q_{0}}{2}=Q_{0}e^{-kt} \;\Longrightarrow\; \frac12 = e^{-kt}, \; Q_{0} \text{ gone} \]
Why: Setting half the initial quantity equal to the initial quantity times the exponential lets the starting amount divide out, leaving an equation in the decay constant alone. So the time to halve is a property of the substance rather than of the sample — which is why half-lives are tabulated in reference books. For linear decay the initial amount would not cancel and no such constant would exist.
Comparison
Fill the blanks. Small differences give unrelated answers.
Comparison matrix
| Integrand | Numerator matches? | Antiderivative |
|---|---|---|
| 2x/(x^2+1) | yes, exactly | ln(x^2+1) |
| x/(x^2+1) | up to a factor of 2 | (1/2) ln(x^2+1) |
| 1/(x^2+1) | no: an x is missing | arctan x |
| 1/x | trivially, the derivative of x is 1 | ln|x| |
The middle two differ by one factor of x and give a logarithm and an inverse tangent — functions with entirely different shapes. Nothing but the numerator check distinguishes them.
Pattern
Given an exponential or logarithmic integrand.
Step three is the one that prevents the section's characteristic error. Assuming a quotient gives a logarithm because it is a quotient produces an answer wrong in kind, and one differentiation would have settled it.
Stewart, Calculus: Early Transcendentals 8e, §5.5 The Substitution Rule §5.5, pp. 412-420
Check
Exponentials.
Check your understanding
What is the integral of e to the 3x?
Answer: A
Why: A linear exponent contributes the reciprocal of its slope.
Check
Quotients.
Check your understanding
What is the integral of 1/(x^2+1)?
Answer: A
Why: The numerator is not the denominator's derivative, so the logarithmic form does not apply.
Check
Decay.
Check your understanding
A substance has a half-life of 5 hours. How much remains after 10 hours?
Answer: A
Why: Each half-life removes half of what remains, so two leave a quarter.
Real world
A radioactive tracer with a six-hour half-life is injected for a diagnostic scan. The radiologist needs the total radiation dose the patient receives, and the scanner needs a window during which the signal is strong enough to image.
Discussion prompt
Explain which integral gives the dose, why the total over unbounded time is finite, and how the imaging window is chosen.
Hint: Dose is activity accumulated over time.
Answer:
Activity decays exponentially with a constant found from the half-life: the logarithm of 2 divided by 6, about 0.1155 per hour. Total dose is that activity integrated over time — the Net Change Theorem with an exponential rate.
\[ D = \int_{0}^{T}A_{0}e^{-kt}\,dt = \frac{A_{0}}{k}\left(1-e^{-kT}\right) \]
The total over unbounded time is finite, equal to the initial activity divided by the decay constant, because the tail contributes an ever-smaller amount. That is the lifetime dose, and it is the figure that determines whether the procedure is within safe limits — a genuinely useful number precisely because it does not depend on how long the patient is observed.
The imaging window is bounded below by the time for the tracer to distribute through the tissue and above by when activity falls below the scanner's detection threshold. Setting the model equal to that threshold and taking logarithms gives the upper bound directly — which is the same computation as the half-life conversion, with a different target value.
Note why the half-life is quoted as a property of the isotope rather than of the dose: the initial activity cancels when the model is solved, so the same six hours applies whatever quantity was injected. That independence is what makes a single tabulated number useful across every procedure.
Commit first
Answer, then rate your confidence honestly.
Predict first
How do you tell whether a quotient integrates to a logarithm?
Correct: Differentiate the denominator and compare.
\[ \frac{2x}{x^{2}+1} \to \ln(x^{2}+1); \qquad \frac{1}{x^{2}+1} \to \arctan x \]
Why: The logarithmic form requires the numerator to be the denominator's derivative, since that is exactly what substitution needs to absorb the differential. A numerical mismatch is adjustable; a missing factor of the variable is not. The test costs one line and distinguishes the logarithm of x squared plus one from the inverse tangent — two functions with completely different shapes that arise from integrands differing by a single factor of x.
Explain it
They integrated one over x squared plus one and wrote the logarithm of x squared plus one.
Discussion prompt
In four sentences or fewer, show them the check.
Hint: Have them differentiate their answer.
Answer:
Ask them to differentiate what they wrote: the chain rule gives 2x over x squared plus one, which is not the integrand — it has an extra 2x on top. The logarithmic form needs the numerator to be the denominator's derivative, and here the numerator is just 1.
This one is in the basic table instead: it is the derivative of the inverse tangent. The habit worth building is differentiating the denominator first and comparing, before assuming a quotient gives a logarithm.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For exponents, check whether the derivative is present before committing. For other bases, rewrite in base e rather than recalling a formula. For the logarithmic form, differentiate the denominator and compare with the numerator. For decay, remember each half-life removes half of what remains. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, list the six formulas of this section, marking which are direct reversals of Section 3.9 and which is not. Below, derive the general-base formula in three lines from the base-change identity, circling where the logarithm in the denominator comes from. In the middle of the page, sketch the reciprocal and the logarithm of the absolute value on the same axes, mark the origin, and write one line on what the absolute value does and one on what it does not do. Beside that, write the four similar quotients with their antiderivatives, drawing an arrow between the two that differ by a single factor of x and naming both answers. In the lower half, sketch an exponential decay with the amount marked at four equal intervals, write the constant ratio beside it, and convert a half-life of 5 hours into a decay constant in four lines. At the bottom, write why the initial amount cancels in that conversion.
If your reciprocal sketch shows the two branches connected, redraw them — they are genuinely separate, and that separation is exactly why an integral across the origin does not exist.
Recap
Five things, and the third is a one-line check that prevents the commonest error.
| If you see | Then |
|---|---|
| An exponential with a linear exponent | Divide by the slope |
| An exponential with a quadratic exponent | Check the derivative is present |
| A base other than e | Rewrite as e to the exponent times ln of the base |
| The reciprocal | Logarithm of the absolute value |
| An interval containing zero | The reciprocal's integral does not exist |
| A quotient | Differentiate the denominator and compare |
| A half-life | Take logarithms; the initial amount cancels |
Section 5.7 completes the chapter's formula list with the integrals producing inverse trigonometric functions — including the one this section's third row could not handle.
OpenStax Calculus Volume 1, §5.6 Integrals Involving Exponential and Logarithmic Functions §5.6, pp. 516-525 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Calculus I — $55/session, free consultation.