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.

True or False? Evaluating a definite integral always requires you to find an antiderivative first.

False. Where the region between the curve and the axis is a triangle, a rectangle or part of a circle, its area can be found from geometry alone. The properties of definite integrals can also give a value directly, as when the two limits are equal or when adjacent intervals are being combined.

When would you approximate a definite integral instead of evaluating it exactly?

When the integrand has no antiderivative that can be written down, or when the function is given only as a table of values or a graph. A Riemann sum or a trapezoidal sum then produces a numerical value with no antidifferentiation at all.

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

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

1/34

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 (34)

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.
What relationship between differentiation and integration do all antidifferentiation techniques rely on?
They rely on differentiation and integration being inverse operations.
That is what lets you check any antiderivative by differentiating it, and it is why knowing $\frac{d}{d x} (\sin x) = \cos x$ gives $\int \cos x d x = \sin x + C$ straight away.
Which technique do $\int {(x^{2} + 2)}^{2} d x$ and $\int \frac{5 x^{3} - 3}{x^{2}} d x$ need?
Neither needs a special technique: both can be rewritten as sums of powers of $x$ and integrated term by term.
Expanding the first gives $x^{4} + 4 x^{2} + 4$ and dividing through the second gives $5 x - 3 x^{- 2}$ , and each of those integrates directly.
Fill in the two missing words in this test for choosing between long division and completing the square:
Divide first when the $______$ has degree at least as large as the denominator, and complete the square instead when the denominator has no real $______$ to take out.
The completed test is: divide first when the numerator has degree at least as large as the denominator, and complete the square instead when the denominator has no real factors to take out.
Comparing the two degrees is the quickest way to tell the two situations apart.
Which method suits each of $\int {(3 x + 1)}^{8} d x$ , $\int \frac{x^{2} - 4}{x + 1} d x$ and $\int \frac{1}{x^{2} + 4 x + 9} d x$ ?
The first is a composite whose inside function differentiates to a constant, so reverse the chain rule and compensate for the factor of 3.
The second calls for long division since the numerator has the higher degree, and the third calls for completing the square since $x^{2} + 4 x + 9$ has no real factors.
True or False?
Evaluating a definite integral always requires you to find an antiderivative first.
False.
Where the region between the curve and the axis is a triangle, a rectangle or part of a circle, its area can be found from geometry alone.
The properties of definite integrals can also give a value directly, as when the two limits are equal or when adjacent intervals are being combined.
When would you approximate a definite integral instead of evaluating it exactly?
When the integrand has no antiderivative that can be written down, or when the function is given only as a table of values or a graph.
A Riemann sum or a trapezoidal sum then produces a numerical value with no antidifferentiation at all.

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