True or False? Integration by parts may need to be applied more than once to solve a problem.

True. Integration by parts may need to be applied more than once to solve a problem. This is known as repeated integration by parts.

True or False? In integration by parts , only one overall constant of integration is required.

True. In integration by parts , only one overall constant of integration is required. Even though more than one integral may appear in your working, only one constant of integration is required in the final answer.

True or False? If a quadratic denominator does not factorise, then you should use inverse trigonometric functions to integrate.

True. If a quadratic denominator does not factorise, then you should use inverse trigonometric functions to integrate.

True or False? The result of integrating with partial fractions always involves natural logarithms .

True. The result of integrating with partial fractions always involves natural logarithms .

True or False? The function y = f ( x ) describing the curve must be rearranged into the form x = g ( y ) when finding the area between a curve and the y -axis.

True. The function y = f ( x ) describing the curve must be rearranged into the form x = g ( y ) when finding the area between a curve and the y -axis.

True or False? The limits a and b to be used in the volume of revolution formula are always given directly in the question.

False. The limits a and b to be used in the volume of revolution formula are not always given directly in the question. You may need to work the limits out. They could involve the y -axis ( x =0) or a root of y = f ( x ).

True or False? When calculating the volume of a revolution about the y -axis , the equation of the curve must always be rearranged from y = f ( x ) form to x = g ( y ) form.

True. When calculating the volume of a revolution about the y -axis , the equation of the curve must always be rearranged from y = f ( x ) form to x = g ( y ) form.

True or False? The constant π in the volume of revolution formulas can be pulled out in front of the integral as a multiplier.

True. The constant π in the volume of revolution formulas (like any constant inside an integral) can be pulled out in front of the integral as a multiplier. This can be useful when the integral needs to be evaluated by hand. (Just don't forget to put π back into the final answer at the end!)

What is a key modelling assumption when using volumes of revolution?

A key modelling assumption when using volumes of revolution is that the thickness of the 'walls' of the solid is negligible relative to the size of the object.

True or False? The volume of revolution always includes all parts of an object, such as the handle of a bucket.

False. The volume of revolution does not always include all parts of an object, such as the handle of a bucket. The volume of revolution usually only includes the main shape of the body of the object, not additional parts like the handle of a bucket.

What is the unit conversion between litres and cm 3 that might be needed when calculating the capacity of an object modelled as a volume of revolution?

The unit conversion between litres and cm 3 that might be needed when calculating the capacity of an object modelled as a volume of revolution is 1 litre = 1000 cm 3 .

True or False? Adding or subtracting volumes of revolution is sometimes necessary when modelling complex shapes.

True. Adding or subtracting volumes of revolution is sometimes necessary when modelling complex shapes.

What is a criticism of the volume of revolution model for real-world objects?

A criticism of the volume of revolution model for real-world objects is that it doesn't account for the thickness of the material the object is made from.

True or False? The volume found by a volume of revolution model always accurately represents the capacity of the real object.

False. The volume found by a volume of revolution model may not always accurately represents the capacity of the real object, due to various factors like material thickness or internal components.

IBMathsDPAnalysis & Approaches (AA)HLFlashcards5. CalculusFurther Integration

Further Integration (DP IB Analysis & Approaches (AA))

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What is $\int \sec^{2} x d x$ ?

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What is $\int \sec^{2} x d x$ ?
$\int \sec^{2} x d x = \tan x + C$
This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative ${f (x) = \tan x \Rightarrow f}^{'} (x) = \sec^{2} x$ .
True or False?
$\int {cosec}^{2} x d x = \cot x + C$
False.
$\int {cosec}^{2} x d x = - \cot x + C$
This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative ${f (x) = \cot x \Rightarrow f}^{'} (x) = - {cosec}^{2} x$ .
What is $\int \sec x \tan x d x$ ?
$\int \sec x \tan x d x = \sec x + C$
This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative ${f (x) = \sec x \Rightarrow f}^{'} (x) = \sec x \tan x$ .
What is $\int cosec x \cot x d x$ ?
$\int cosec x \cot x d x = - cosec x + C$
This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative ${f (x) = cosec x \Rightarrow f}^{'} (x) = - cosec x \cot x$ .
What is $\int \frac{1}{\sqrt{1 - x^{2}}} d x$ ?
$\int \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + C$
This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative ${f (x) = \arcsin x \Rightarrow f}^{'} (x) = \frac{1}{\sqrt{1 - x^{2}}}$ .
True of False?
$\int \frac{1}{1 + x^{2}} d x$ can be found using reverse chain rule.
False.
$\int \frac{1}{1 + x^{2}} d x$ can not be found using reverse chain rule.
$\int \frac{1}{1 + x^{2}} d x = arc \tan x + C$
This is not in the exam formula booklet. However the inverse of this is in the formula booklet, i.e. the standard derivative ${f (x) = \arctan x \Rightarrow f}^{'} (x) = \frac{1}{1 + x^{2}}$ .
What is $\int a^{x} d x$ ?
$\int a^{x} d x = \frac{1}{\ln a} a^{x} + C$
This is in the exam formula booklet.
What is $\int \frac{1}{a^{2} + x^{2}} d x$ ?
$\int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \arctan (\frac{x}{a}) + C$
This is in the exam formula booklet.
What is $\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x$ ?
$\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = arc \sin (\frac{x}{a}) + C, |x| < a$
This is in the exam formula booklet.
True or False?
Completing the square may be necessary when integrating with inverse trigonometric functions.
True.
Completing the square may be necessary when integrating with inverse trigonometric functions.
E.g. rewriting $\int \frac{1}{\sqrt{5 - x^{2} + 4 x}} d x$ as $\int \frac{1}{\sqrt{9 - {(x - 2)}^{2}}} d x$ . Then the integral is in a form which can be integrated using the standard integral result $\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = arc \sin (\frac{x}{a}) + C$ from the exam formula booklet.
True or False?
When doing integration by substitution, $\frac{d u}{d x}$ can be treated like a fraction.
True.
When doing integration by substitution, $\frac{d u}{d x}$ can be treated like a fraction.
E.g. $u = x^{2} \Rightarrow \frac{d u}{d x} = 2 x \Rightarrow d u = 2 x d x$
$\frac{d u}{d x}$ is not actually a fraction (it's a derivative, i.e. gradient function), but treating it like a fraction can give correct answers when integrating by substitution.
$\int x {(1 + 3 x)}^{8} d x$ can be integrated by using the substitution $u = 1 + 3 x$ .
It follows from this that $x = \frac{u - 1}{3}$ , and also that $\frac{d u}{d x} = 3 \Rightarrow d x = \frac{1}{3} d u$ .
What is the next step in solving the integral?
$\int x {(1 + 3 x)}^{8} d x$ can be integrated by using the substitution $u = 1 + 3 x$ .
It follows from this that $x = \frac{u - 1}{3}$ , and also that $\frac{d u}{d x} = 3 \Rightarrow d x = \frac{1}{3} d u$ .
The next step in solving the integral is to substitute terms in $u$ for all the terms in $x$ in the integral:
$\int x {(1 + 3 x)}^{8} d x = \int (\frac{u - 1}{3}) u^{8} (\frac{1}{3} d u) = \frac{1}{9} \int (u^{9} - u^{8}) d u$
State the formula for integration by parts.
The formula for integration by parts is $\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$ , which can also be written as $\int u d v = u v - \int v d u$ .
Both versions of the formula are in the exam formula booklet.
True or False?
In integration by parts, the function that becomes simpler when differentiated should be assigned to $\frac{d v}{d x}$ .
False.
In integration by parts, the function that becomes simpler when differentiated should be assigned to u, so that it can then be differentiated.
True or False?
Integration by parts may need to be applied more than once to solve a problem.
True.
Integration by parts may need to be applied more than once to solve a problem. This is known as repeated integration by parts.
True or False?
Integrating expressions like $x^{2} \cos x$ or $e^{x} \sin x$ requires repeated integration by parts.
True.
Integrating expressions like $x^{2} \cos x$ or $e^{x} \sin x$ requires repeated integration by parts.
When using integration by parts to integrate $\ln x$ , what should you choose for $u$ and $\frac{d v}{d x}$ ?
When using integration by parts to integrate $\ln x$ , you should choose $u = \ln x$ and $\frac{d v}{d x} = 1$ .
Then $\int \ln x d x = x \ln x - \int x (\frac{1}{x}) d x = x \ln x - \int d x = x \ln x - x + C$ .
True or False?
In integration by parts, only one overall constant of integration is required.
True.
In integration by parts, only one overall constant of integration is required.
Even though more than one integral may appear in your working, only one constant of integration is required in the final answer.
True or False?
Partial fractions can be used for integration when the denominator of a quotient is of quadratic form.
E.g. to integrate $\int \frac{x - 13}{x^{2} - 5 x - 6} d x$ .
True.
Partial fractions can be used for integration when the denominator of a quotient is of quadratic form.
E.g. to integrate $\int \frac{x - 13}{x^{2} - 5 x - 6} d x$ , first use partial fractions to rewrite the integral as $\int (\frac{2}{x + 1} - \frac{1}{x - 6}) d x$ .
Why should you not use partial fractions to integrate something like $\int \frac{2 x - 5}{x^{2} - 5 x - 6} d x$ ?
If the numerator of what you are trying to integrate is the derivative (or a multiple of the derivative) of the denominator, then partial fractions should not be used because it is much quicker to use the reverse chain rule result $\int \frac{f' (x)}{f (x)} d x = \ln |f (x)| + C$ .
E.g. $\int \frac{2 x - 5}{x^{2} - 5 x - 6} d x = \ln |x^{2} - 5 x - 6| + C$
True or False?
If a quadratic denominator does not factorise, then you should use inverse trigonometric functions to integrate.
True.
If a quadratic denominator does not factorise, then you should use inverse trigonometric functions to integrate.
True or False?
The result of integrating with partial fractions always involves natural logarithms.
True.
The result of integrating with partial fractions always involves natural logarithms.
How can you further simplify an indefinite integral answer like $\ln |\frac{x + 1}{x + 3}| + c$ , where $c$ is the constant of integration?
To further simplify an indefinite integral answer like $\ln |\frac{x + 1}{x + 3}| + c$ , where $c$ is the constant of integration:
- Write $c$ as a logarithm: $c = \ln k$
- Substitute into the answer: $\ln |\frac{x + 1}{x + 3}| + \ln k$
- Use laws of logarithms to combine the two logs: $\ln |k (\frac{x + 1}{x + 3})|$
What is the area between a curve and the y-axis?
The area between a curve and the y-axis is the region bounded by the graph of y=f(x), the y-axis, and two horizontal lines y=a and y=b.
What is the formula for finding the area between a curve and the y-axis?
The formula for finding the area between a curve and the y-axis is $A = \int_{a}^{b} |x| d y$
Where:
- $A$ is the area to be found
- $y = a$ and $y = b$ are the equations of the horizontal lines bounding the area
- $x = g (y)$ is the equation of the curve expressed as a function of $y$
This is in the exam formula booklet.
True or False?
The function y=f(x) describing the curve must be rearranged into the form x=g(y) when finding the area between a curve and the y-axis.
True.
The function y=f(x) describing the curve must be rearranged into the form x=g(y) when finding the area between a curve and the y-axis.
What is a volume of revolution?
A volume of revolution is the volume of a solid formed when an area bounded by a function y=f(x) is rotated 2π radians (or 360°) around an axis.
State the formula for a volume of revolution around the x-axis.
The formula for the volume of revolution around the x-axis is $V = \int_{a}^{b} π y^{2} d x$
Where:
- $V$ is the volume to be found
- $x = a$ and $x = b$ are the equations of the vertical lines bounding the area to be rotated
- $y^{2}$ is the square of the equation of the curve $y = f (x)$
This is in the exam formula booklet.
True or False?
The limits a and b to be used in the volume of revolution formula are always given directly in the question.
False.
The limits a and b to be used in the volume of revolution formula are not always given directly in the question.
You may need to work the limits out. They could involve the y-axis (x=0) or a root of y=f(x).
State the formula for a volume of revolution around the y-axis.
The formula for the volume of revolution around the y-axis is $V = \int_{a}^{b} π x^{2} d y$
Where:
- $V$ is the volume to be found
- $y = a$ and $y = b$ are the equations of the horizontal lines bounding the area to be rotated
- $x^{2}$ is the square of the equation of the curve $x = g (y)$
This is in the exam formula booklet.
(Note that an equation of the curve in $y = f (x)$ form will need to be rearranged into $x = g (y)$ form to use this formula.)
True or False?
When calculating the volume of a revolution about the y-axis, the equation of the curve must always be rearranged from y=f(x) form to x=g(y) form.
True.
When calculating the volume of a revolution about the y-axis, the equation of the curve must always be rearranged from y=f(x) form to x=g(y) form.
True or False?
The constant π in the volume of revolution formulas can be pulled out in front of the integral as a multiplier.
True.
The constant π in the volume of revolution formulas (like any constant inside an integral) can be pulled out in front of the integral as a multiplier.
This can be useful when the integral needs to be evaluated by hand. (Just don't forget to put π back into the final answer at the end!)
What is a key modelling assumption when using volumes of revolution?
A key modelling assumption when using volumes of revolution is that the thickness of the 'walls' of the solid is negligible relative to the size of the object.
True or False?
The volume of revolution always includes all parts of an object, such as the handle of a bucket.
False.
The volume of revolution does not always include all parts of an object, such as the handle of a bucket.
The volume of revolution usually only includes the main shape of the body of the object, not additional parts like the handle of a bucket.
What is the unit conversion between litres and cm³ that might be needed when calculating the capacity of an object modelled as a volume of revolution?
The unit conversion between litres and cm³ that might be needed when calculating the capacity of an object modelled as a volume of revolution is 1 litre = 1000 cm³.
True or False?
Adding or subtracting volumes of revolution is sometimes necessary when modelling complex shapes.
True.
Adding or subtracting volumes of revolution is sometimes necessary when modelling complex shapes.
What is a criticism of the volume of revolution model for real-world objects?
A criticism of the volume of revolution model for real-world objects is that it doesn't account for the thickness of the material the object is made from.
True or False?
The volume found by a volume of revolution model always accurately represents the capacity of the real object.
False.
The volume found by a volume of revolution model may not always accurately represents the capacity of the real object, due to various factors like material thickness or internal components.

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