What shape must an integrand have for the reverse chain rule to work?

A composite function multiplied by the derivative of its inside function . If the coefficient is not quite right, a constant can be supplied and compensated for, since constants come outside an integral freely.

True or False? Once the square is completed you can write down the answer with no further adjusting.

False. Differentiate the candidate answer by the chain rule first, because the inside function brings out a constant factor. Only then do you know whether the result needs scaling.

Why use polynomial long division before integrating a rational function?

It rewrites the function as a polynomial plus a simple remainder fraction, and both of those parts integrate easily. A quotient of polynomials cannot be integrated term by term as it stands.

How do you start a polynomial long division?

Divide the highest power in the dividend by the highest power in the divisor, and write the result above the line. Then subtract that result times the whole divisor, and repeat on whatever is left.

What kind of integrand does integration by parts handle?

A product of two functions, though not every product yields to it. The technique swaps the original integral for a different one, which has to be easier than the original for it to help at all.

What is a DI table used for?

It sets out repeated integration by parts in two columns, one differentiating and one integrating, with the products taken diagonally and the signs alternating. Stop at the row where the D column reaches zero, and add a constant.

When can a rational function be written as partial fractions?

When the numerator's degree is less than the denominator's, and the denominator factors. If the numerator's degree is not lower, polynomial long division comes first and leaves a proper fraction that can then be split.

How do you find the unknown constants in a partial fraction split?

Multiply both sides by the original denominator to clear the fractions, then substitute the roots of the factors one at a time. Each root kills one unknown, leaving the other on its own.

True or False? On this course, every partial fraction has a constant numerator.

True. The denominators met here are products of distinct linear factors, so each partial fraction has a linear denominator. A numerator's degree is always lower than its denominator's, which for a linear denominator leaves only a constant.

What is the alternative to substituting the roots?

Comparing coefficients: collect like terms on the right-hand side and match them against the left. That gives simultaneous equations in the unknowns, which are then solved together.

Define an improper integral .

An improper integral is one where at least one limit of integration is infinite, or where the integrand is unbounded somewhere on the interval. It either converges to a finite value or diverges.

What is the first step with any improper integral?

Rewrite it as the limit of an ordinary definite integral, replacing the offending limit of integration with a variable. Only then do you evaluate the integral and take the limit.

How do you handle an integral with both limits infinite?

Split it at any convenient value into two integrals, and write each as its own limit with its own variable. If either of those limits fails to exist, the whole integral diverges.

When is an improper integral divergent?

When the limit it has been rewritten as does not exist, or is infinite. Where the integral has been split into two, one bad limit is enough to make the whole thing divergent.

What should you try before reaching for any integration technique?

Recognising the integrand as the derivative of a standard function, so that an antiderivative can be written down at once. Only when that fails is a technique needed at all.

True or False? If an integrand is a product, integration by parts is the only option.

False. A product may simplify by expanding, or may turn out to be a composite function times the derivative of its inside, which the reverse chain rule handles far more quickly. Integration by parts is for the products where neither of those applies.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 6: Integration & Accumulation of ChangeMethods of Integration

Methods of Integration (College Board AP® Calculus BC): Flashcards

1/58

0Still learning

Know0

FrontIntegrals of Composite Functions
Fill in the missing composite function in the reverse chain rule:
$\int f^{'} (g (x)) \cdot g^{'} (x) d x =______+ C$ for differentiable $f$ and $g$

Was this flashcard helpful?

Cards in this collection (58)

Fill in the missing composite function in the reverse chain rule:
$\int f^{'} (g (x)) \cdot g^{'} (x) d x =______+ C$ for differentiable $f$ and $g$
The completed rule is $\int f^{'} (g (x)) \cdot g^{'} (x) d x = f (g (x)) + C$ .
It is the chain rule read backwards, which is why the technique is often called integration by inspection.
What shape must an integrand have for the reverse chain rule to work?
A composite function multiplied by the derivative of its inside function.
If the coefficient is not quite right, a constant can be supplied and compensated for, since constants come outside an integral freely.
Find $\int \cos x \sin^{5} x d x$ .
It is $\frac{1}{6} \sin^{6} x + C$ .
The integrand is ${[\sin x]}^{5}$ times the derivative of $\sin x$ , so the power rule applies with $\sin x$ standing where $x$ usually stands.
True or False?
Every integral that can be done by inspection can also be done by substitution.
True.
Inspection and $u$ -substitution handle the same composite integrands, and substitution simply writes out in full what inspection does in one step.
Substitution is the safer choice whenever the pattern is awkward to spot.
How do you handle a coefficient that is not quite the derivative of the inside function?
Adjust and compensate: multiply inside the integral by whatever constant is needed, and divide outside by the same constant.
For instance $5 x^{2} = \frac{5}{12} (12 x^{2})$ , so a missing factor of 12 can be supplied and paid for at once.
Find $\int x {(5 x^{2} - 2)}^{6} d x$ .
It is $\frac{1}{70} {(5 x^{2} - 2)}^{7} + C$ .
Differentiating ${(5 x^{2} - 2)}^{7}$ gives $70 x {(5 x^{2} - 2)}^{6}$ , so the integrand needs a factor of 70 supplied and compensated for.
Fill in the missing function in the rule for a fraction whose numerator is the derivative of its denominator:
$\int \frac{f^{'} (x)}{f (x)} d x = \ln |______| + C$ for any differentiable $f$
The completed rule is $\int \frac{f^{'} (x)}{f (x)} d x = \ln | f (x) | + C$ .
The absolute value bars matter, and leaving them off is the commonest slip on this form.
Find $\int \frac{x^{2} + 1}{x^{3} + 3 x} d x$ .
It is $\frac{1}{3} \ln | x^{3} + 3 x | + C$ .
The denominator differentiates to $3 x^{2} + 3$ , which is three times the numerator, so a factor of $\frac{1}{3}$ compensates.
Find $\int \frac{2}{5 x \ln x} d x$ .
It is $\frac{2}{5} \ln | \ln x | + C$ .
Writing it as $\frac{2}{5} \cdot \frac{\frac{1}{x}}{\ln x}$ shows that the numerator is the derivative of the denominator, since $\ln x$ differentiates to $\frac{1}{x}$ .
What does $u$ -substitution do to an integral?
It replaces the original variable with a new one, chosen so that the integral in $u$ is easier than the integral in $x$ .
The substitution is reversed at the end to give the answer back in terms of $x$ .
Fill in the two missing parts of the second step of a substitution, given $u = 2 x^{3} - 6 x + 7$ :
$\frac{d u}{d x} =______$ so $(x^{2} - 1) d x =______d u$ after rearranging
The completed working is $\frac{d u}{d x} = 6 x^{2} - 6$ so $(x^{2} - 1) d x = \frac{1}{6} d u$ .
Treat $\frac{d u}{d x}$ like a fraction and multiply through by $d x$ , then divide by whatever constant is needed to isolate the part that appears in the integral.
Which part of a composite integrand do you choose as $u$ ?
The inside function, so if the integrand involves $f (g (x))$ then take $u = g (x)$ .
For $\int 3 x \cos (x^{2} - 5) d x$ that means $u = x^{2} - 5$ .
True or False?
When a definite integral is done by substitution, the original limits can be kept.
False.
The limits are values of $x$ , so they must be converted into the corresponding values of $u$ .
Forgetting to change them is one of the commonest errors on this technique.
What must be replaced when you carry out a substitution?
Every part of the integral, including the $d x$ .
No $x$ may be left anywhere in the integrand once the substitution has been made.
Find $\int x \sqrt{x - 4} d x$ .
It is $\frac{2}{5} {(x - 4)}^{\frac{5}{2}} + \frac{8}{3} {(x - 4)}^{\frac{3}{2}} + C$ .
With $u = x - 4$ you also need $x = u + 4$ , giving $\int (u + 4) u^{\frac{1}{2}} d u$ , which integrates term by term.
After changing the limits of a definite integral, why need you not substitute $x$ back in?
Because the integral is now written entirely in $u$ , with $u$ limits, so it evaluates to the same number either way.
Substituting back would add work and a chance of error for no gain.
Evaluate $\int_{1}^{2} \frac{2 x + 3}{3 x^{2} + 9 x - 5} d x$ .
It is $\frac{1}{3} \ln (\frac{25}{7})$ .
With $u = 3 x^{2} + 9 x - 5$ the limits become $7$ and $25$ , and $(2 x + 3) d x = \frac{1}{3} d u$ , leaving $\frac{1}{3} \int_{7}^{25} \frac{1}{u} d u$ .
Fill in the two missing parts of the completed square form:
$x^{2} + b x + c = {(x +______)}^{2} + c -______$ after completing the square
The completed identity is $x^{2} + b x + c = {(x + \frac{b}{2})}^{2} + c - {(\frac{b}{2})}^{2}$ .
For example $x^{2} - 6 x + 2 = {(x - 3)}^{2} - 7$ .
How does completing the square change when there is a coefficient in front of $x^{2}$ ?
It becomes $a x^{2} + b x + c = a {(x + \frac{b}{2 a})}^{2} + c - a {(\frac{b}{2 a})}^{2}$ .
For example $3 x^{2} + 12 x - 7 = 3 {(x + 2)}^{2} - 19$ .
Why complete the square before integrating a quadratic denominator?
It turns the denominator into a constant plus a square, which is the form the inverse trigonometric standard integrals need.
That identifies $\arctan$ or $\arcsin$ as the shape of the answer before any working is done.
True or False?
Once the square is completed you can write down the answer with no further adjusting.
False.
Differentiate the candidate answer by the chain rule first, because the inside function brings out a constant factor.
Only then do you know whether the result needs scaling.
Find $\int \frac{1}{x^{2} - 6 x + 13} d x$ .
It is $\frac{1}{2} \arctan (\frac{x - 3}{2}) + C$ .
Completing the square gives $4 + {(x - 3)}^{2}$ , so the integrand is $\frac{1}{4} \cdot \frac{1}{1 + {(\frac{x - 3}{2})}^{2}}$ , and differentiating the answer returns exactly that.
Find $\int \frac{1}{\sqrt{21 - x^{2} + 4 x}} d x$ .
It is $\arcsin (\frac{x - 2}{5}) + C$ .
Completing the square gives $25 - {(x - 2)}^{2}$ , so the integrand is $\frac{1}{5} \cdot \frac{1}{\sqrt{1 - {(\frac{x - 2}{5})}^{2}}}$ , which is exactly the derivative of the answer.
Why use polynomial long division before integrating a rational function?
It rewrites the function as a polynomial plus a simple remainder fraction, and both of those parts integrate easily.
A quotient of polynomials cannot be integrated term by term as it stands.
Fill in the two missing terms in this polynomial division:
$\frac{x^{3} + 6 x^{2} - 9 x - 11}{x - 2} = x^{2} +______+ 7 + \frac{______}{x - 2}$ after dividing
The completed division is $\frac{x^{3} + 6 x^{2} - 9 x - 11}{x - 2} = x^{2} + 8 x + 7 + \frac{3}{x - 2}$ .
The 3 left at the foot of the division is the remainder, and it sits over the original divisor.
How do you start a polynomial long division?
Divide the highest power in the dividend by the highest power in the divisor, and write the result above the line.
Then subtract that result times the whole divisor, and repeat on whatever is left.
True or False?
A missing power in the dividend can simply be skipped over.
False.
Put the missing term in with a coefficient of zero, such as $+ 0 x^{2}$ , so that the columns line up.
Skipping it is how terms end up being subtracted from the wrong power.
Find $\int \frac{x^{3} + 2 x + 13}{x + 3} d x$ .
It is $\frac{1}{3} x^{3} - \frac{3}{2} x^{2} + 11 x - 20 \ln | x + 3 | + C$ .
Long division gives $x^{2} - 3 x + 11 - \frac{20}{x + 3}$ , and each term of that integrates in the usual way.
Fill in the two missing parts of the integration by parts formula:
$\int u \cdot \frac{d v}{d x} d x =______- \int______\cdot v d x$ for suitable $u$ and $v$
The completed formula is $\int u \cdot \frac{d v}{d x} d x = u \cdot v - \int \frac{d u}{d x} \cdot v d x$ .
It is the product rule for differentiation integrated through and then rearranged.
What kind of integrand does integration by parts handle?
A product of two functions, though not every product yields to it.
The technique swaps the original integral for a different one, which has to be easier than the original for it to help at all.
How do you decide which factor to call $u$ ?
Take $u$ in the order logarithms, inverse trigonometric functions, polynomials, exponentials, then trigonometric functions, which the letters LIPET record.
The other factor has to be integrated, so anything without a straightforward antiderivative should be $u$ .
True or False?
You need a constant of integration when you find $v$ from $\frac{d v}{d x}$ .
False.
Any antiderivative will do for $v$ , so the simplest one is used and a single constant is added at the very end.
Carrying a constant through the formula only cancels itself out.
Find $\int 5 x e^{3 x} d x$ .
It is $\frac{5}{3} x e^{3 x} - \frac{5}{9} e^{3 x} + C$ .
With $u = 5 x$ and $\frac{d v}{d x} = e^{3 x}$ you get $v = \frac{1}{3} e^{3 x}$ , and the remaining integral $\int \frac{5}{3} e^{3 x} d x$ is straightforward.
How do you find $\int \ln x d x$ ?
Use integration by parts with $u = \ln x$ and $\frac{d v}{d x} = 1$ , which gives $x \ln x - x + C$ .
The same trick handles the inverse trigonometric functions, whose antiderivatives are found no other way.
When do you need integration by parts twice?
When one application leaves an integral that is still a product, as $\int x^{2} \sin x d x$ does.
A second application then finishes it off.
What is a DI table used for?
It sets out repeated integration by parts in two columns, one differentiating and one integrating, with the products taken diagonally and the signs alternating.
Stop at the row where the D column reaches zero, and add a constant.
What do you do when integration by parts returns the original integral?
Treat the original integral as an unknown and rearrange to make it the subject.
For $\int e^{x} \cos x d x$ two applications give $I = e^{x} \sin x + e^{x} \cos x - I$ , so $I = \frac{1}{2} (e^{x} \sin x + e^{x} \cos x) + C$ .
Define partial fractions.
Partial fractions are the simpler fractions a rational function is split into, each carrying one factor of the original denominator beneath it.
For instance $\frac{8 x + 10}{(x + 3) (2 x - 1)} = \frac{2}{x + 3} + \frac{4}{2 x - 1}$ .
When can a rational function be written as partial fractions?
When the numerator's degree is less than the denominator's, and the denominator factors.
If the numerator's degree is not lower, polynomial long division comes first and leaves a proper fraction that can then be split.
How do you find the unknown constants in a partial fraction split?
Multiply both sides by the original denominator to clear the fractions, then substitute the roots of the factors one at a time.
Each root kills one unknown, leaving the other on its own.
True or False?
On this course, every partial fraction has a constant numerator.
True.
The denominators met here are products of distinct linear factors, so each partial fraction has a linear denominator.
A numerator's degree is always lower than its denominator's, which for a linear denominator leaves only a constant.
Fill in the two missing parts of the integral of a single partial fraction:
$\int \frac{1}{a x + b} d x = \frac{1}{______} \ln |______| + C$ for a non-zero constant $a$
The completed result is $\int \frac{1}{a x + b} d x = \frac{1}{a} \ln | a x + b | + C$ .
The $\frac{1}{a}$ compensates for the factor $a$ that differentiating $a x + b$ would produce.
What is the alternative to substituting the roots?
Comparing coefficients: collect like terms on the right-hand side and match them against the left.
That gives simultaneous equations in the unknowns, which are then solved together.
Find $\int \frac{2 (x - 4)}{2 x^{2} + 5 x - 3} d x$ as a single logarithm.
It is $\ln | \frac{{(x + 3)}^{2}}{2 x - 1} | + C$ .
The split is $\frac{2}{x + 3} - \frac{2}{2 x - 1}$ , which integrates to $2 \ln | x + 3 | - \ln | 2 x - 1 | + C$ , and the laws of logarithms combine those into one.
Define an improper integral.
An improper integral is one where at least one limit of integration is infinite, or where the integrand is unbounded somewhere on the interval.
It either converges to a finite value or diverges.
What is the first step with any improper integral?
Rewrite it as the limit of an ordinary definite integral, replacing the offending limit of integration with a variable.
Only then do you evaluate the integral and take the limit.
Fill in the two missing parts of the rewriting of an integral with an infinite upper limit:
$\int_{a}^{\infty} f (x) d x = \lim_{k \to______} \int_{a}^{______} f (x) d x$ for a new variable $k$
The completed rewriting is $\int_{a}^{\infty} f (x) d x = \lim_{k \to \infty} \int_{a}^{k} f (x) d x$ .
The infinite limit is replaced by a variable, and that variable is then sent to infinity.
Evaluate $\int_{1}^{\infty} \frac{1}{x^{2}} d x$ .
It is $1$ .
Writing it as $\lim_{k \to \infty} {[- \frac{1}{x}]}_{1}^{k}$ gives $\lim_{k \to \infty} (1 - \frac{1}{k})$ , which is 1.
How do you handle an integral with both limits infinite?
Split it at any convenient value into two integrals, and write each as its own limit with its own variable.
If either of those limits fails to exist, the whole integral diverges.
True or False?
$\int_{- 1}^{1} \frac{1}{x^{2}} d x$ can be evaluated as ${[- \frac{1}{x}]}_{- 1}^{1} = - 2$ .
False.
The integrand is unbounded at $x = 0$ , which lies inside the interval, so the integral has to be split there and written as two limits.
Both of those limits are infinite, so the integral in fact diverges.
Which one-sided limit do you use when the integrand is unbounded at the lower limit $a$ ?
The limit as the new variable approaches $a$ from above, written $\lim_{h \to a^{+}}$ .
At the upper limit it is the other way round, approaching from below.
When is an improper integral divergent?
When the limit it has been rewritten as does not exist, or is infinite.
Where the integral has been split into two, one bad limit is enough to make the whole thing divergent.
What should you try before reaching for any integration technique?
Recognising the integrand as the derivative of a standard function, so that an antiderivative can be written down at once.
Only when that fails is a technique needed at all.
Fill in the two missing methods for these shapes of integrand:
a fraction whose numerator is the derivative of its denominator integrates to a $______$ , while a fraction whose denominator factors calls for $______$ instead
The completed rule is: a fraction whose numerator is the derivative of its denominator integrates to a logarithm, while a fraction whose denominator factors calls for partial fractions instead.
Both are worth checking for before reaching for anything heavier.
Which method does an irreducible quadratic denominator suggest?
Completing the square, which brings the integrand into the form of one of the inverse trigonometric standard results.
The answer will then involve $\arctan$ or $\arcsin$ .
Which method does a product of two unrelated functions suggest?
Integration by parts, which trades the integral for a different one that should be easier.
The factor chosen as $u$ is the one whose derivative simplifies the remaining integral.
True or False?
If an integrand is a product, integration by parts is the only option.
False.
A product may simplify by expanding, or may turn out to be a composite function times the derivative of its inside, which the reverse chain rule handles far more quickly.
Integration by parts is for the products where neither of those applies.
What three routes are there to the value of a definite integral?
Find an antiderivative and use $F (b) - F (a)$ , use a geometric area formula, or use the properties of definite integrals.
Which is quickest depends on whether you are given a formula, a graph, or the values of other integrals.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Riemann Sums & Definite IntegralsNext:First-Order Differential Equations