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Methods in Calculus (Edexcel A Level Further Maths: Core Pure): Flashcards

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FrontImproper Integrals
Define an improper integral.

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Define an improper integral.
An improper integral is a definite integral for which one or both limits is $\infty$ or $- \infty$ , or at which the function being integrated is undefined.
For example $\int_{0}^{5} \frac{1}{\sqrt[3]{x}} d x$ is improper because the integrand is undefined at $x = 0$ .
Why can you not just substitute the limits into an improper integral?
Because at least one of the limits is not something you can substitute: it is either $\infty$ , which is not a value, or a point at which the expression is undefined.
A limiting process is used instead, so that the value is approached rather than substituted.
Fill in the rewriting that makes this improper integral possible to evaluate:
$\int_{1}^{\infty} \frac{1}{x^{3}} d x = \lim_{b \to______} \int_{1}^{______} \frac{1}{x^{3}} d x$
The completed rewriting is:
$\int_{1}^{\infty} \frac{1}{x^{3}} d x = \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^{3}} d x$
The infinite limit is replaced by a variable, the integral is worked out in terms of that variable, and only then is the limit taken.
Why does $\int_{0}^{3} \frac{1}{2 \sqrt{x}} d x$ need special treatment, and what is its exact value?
The integrand is undefined at $x = 0$ , so the lower limit is replaced by $a$ and the limit is taken as $a \to 0$ .
Since $\int \frac{1}{2 \sqrt{x}} d x = \sqrt{x}$ , the value is $\lim_{a \to 0} (\sqrt{3} - \sqrt{a}) = \sqrt{3}$ .
True or False?
A definite integral with $0$ as one of its limits is always an improper integral.
False.
It is improper only if the integrand is undefined at $0$ , as $\frac{1}{\sqrt{x}}$ is.
Something like $\int_{0}^{2} x^{2} d x$ is perfectly ordinary, because $x^{2}$ is defined at $0$ .
Find $\int_{3}^{\infty} \frac{1}{2 x^{2}} d x$ .
Integrating between $3$ and $b$ gives ${[- \frac{1}{2 x}]}_{3}^{b} = \frac{1}{6} - \frac{1}{2 b}$ .
As $b \to \infty$ the term $\frac{1}{2 b}$ tends to $0$ , so the integral is equal to $\frac{1}{6}$ .
True or False?
A region that extends infinitely far in one direction can still have a finite area.
True.
The region under a curve can go on forever and still enclose only a finite amount of area, provided the curve approaches the $x$ -axis fast enough.
That is exactly what an improper integral with an infinite limit measures, and it is why such an integral can have an ordinary numerical value at all.
Fill in the formula for the mean value of $f (x)$ over the interval $[a, b]$ :
$\bar{f} =______\int_{a}^{b} f (x) d x$
The completed formula is:
$\bar{f} = \frac{1}{b - a} \int_{a}^{b} f (x) d x$
Dividing by the width of the interval is what turns a total into an average.
Is the mean value of a function over an interval itself a function?
No, it is a single real number, so the graph of $y = \bar{f}$ is a horizontal line.
Its value does depend on the interval chosen, so changing $[a, b]$ can change $\bar{f}$ .
What is the geometrical meaning of the mean value of a function over $[a, b]$ ?
It is the height of the rectangle standing on $[a, b]$ whose area is equal to the area under the curve between those same two values.
Where the curve rises above that height, the extra area exactly makes up for the area lost where the curve dips below it.
True or False?
If the mean value of a function over an interval is $0$ , the areas above and below the $x$ -axis over that interval are equal.
True.
A mean value of $0$ means the integral over the interval is $0$ , and an integral counts area below the $x$ -axis as negative.
The two areas must therefore cancel each other exactly.
Find the exact mean value of $f (x) = \frac{1}{x + 1}$ over the interval $[0, 1]$ .
The width of the interval is $1 - 0 = 1$ , so the mean value is just the integral itself.
That gives $\int_{0}^{1} \frac{1}{x + 1} d x = {[\ln | x + 1 |]}_{0}^{1} = \ln 2 - \ln 1 = \ln 2$ .
A function $f$ has mean value $\bar{f}$ over $[a, b]$ , and $k$ is a constant. What are the mean values of $f (x) + k$ , $k f (x)$ and $- f (x)$ over the same interval?
They are $\bar{f} + k$ , $k \bar{f}$ and $- \bar{f}$ .
Each follows from the integral being linear: adding a constant to every value raises the average by that constant, and multiplying every value by $k$ multiplies the average by $k$ .
A function has mean value $\ln 2$ over $[0, 1]$ . What is the mean value of $6 f (x) + 3$ over the same interval?
It is $6 \ln 2 + 3$ , because the stretch and the translation act on the mean value in exactly the way they act on the function itself.
No new integration is needed, since the mean value of $f$ is already known.
Why is a fraction such as $\frac{8 x^{2} - 9 x}{(x - 3) (4 x^{2} + 9)}$ split into partial fractions before integrating?
Because as it stands the numerator is not a multiple of the derivative of the denominator, and the denominator matches no standard result.
Splitting it turns one unrecognisable integral into a sum of integrals that each match something already known.
Fill in the two numerators of this partial fraction split:
$\frac{x (8 x - 9)}{(x - 3) (4 x^{2} + 9)} = \frac{______}{x - 3} + \frac{______}{4 x^{2} + 9}$
The completed split is:
$\frac{x (8 x - 9)}{(x - 3) (4 x^{2} + 9)} = \frac{A}{x - 3} + \frac{B x + C}{4 x^{2} + 9}$
A denominator that cannot be factorised needs a linear numerator, one degree lower than itself, rather than the constant that a linear denominator takes.
After multiplying up, how do you find the unknown constants when one factor is linear and the other is an irreducible quadratic?
Substitute the root of the linear factor first, which makes the quadratic part vanish and gives one constant immediately.
Then substitute any two further values of $x$ , with $x = 0$ usually the easiest, and solve the resulting equations simultaneously.
Why is a term such as $\frac{4 x + 3}{4 x^{2} + 9}$ split into two separate fractions before integrating?
Because the two halves integrate in completely different ways: the $x$ term gives a logarithm, while the constant term gives an inverse trigonometric function.
Written as $\frac{4 x}{4 x^{2} + 9} + \frac{3}{4 x^{2} + 9}$ , each half can be matched to the result that fits it.
In $\int \frac{4 x}{4 x^{2} + 9} d x$ , how can you tell the answer is a logarithm, and what is the coefficient?
The derivative of the denominator is $8 x$ , and the numerator $4 x$ is exactly half of it, which is the $\frac{f' (x)}{f (x)}$ pattern.
Compensating for that factor of $2$ gives $\frac{1}{2} \ln | 4 x^{2} + 9 | + c$ .
How is $\int \frac{3}{4 x^{2} + 9} d x$ matched to the standard result for $\int \frac{1}{a^{2} + x^{2}} d x$ ?
Take the $4$ out of the denominator, giving $\frac{3}{4} \int \frac{1}{x^{2} + \frac{9}{4}} d x$ , so that the coefficient of $x^{2}$ is $1$ and $a = \frac{3}{2}$ .
The standard result then gives $\frac{3}{4} \times \frac{2}{3} \arctan (\frac{2 x}{3}) = \frac{1}{2} \arctan (\frac{2 x}{3}) + c$ .
True or False?
Integrating three partial fractions separately leaves you with three constants of integration in the final answer.
False.
A sum of arbitrary constants is itself just an arbitrary constant, so the three merge into a single $c$ .
Nothing is lost by doing so, because no one value of $c$ was ever determined by the integration.
Why do the derivatives of the inverse trigonometric functions contain square roots when the functions themselves do not?
Because the derivation replaces a trigonometric function of $y$ by a surd in $x$ , using the Pythagorean identity $\cos^{2} y + \sin^{2} y \equiv 1$ .
The identity turns $\cos y$ into $\sqrt{1 - \sin^{2} y}$ , and since $\sin y = x$ that becomes $\sqrt{1 - x^{2}}$ .
Which identity turns $\frac{1}{\sec^{2} y}$ into the derivative of $\arctan x$ ?
The identity $1 + \tan^{2} y \equiv \sec^{2} y$ , which turns $\frac{1}{\sec^{2} y}$ into $\frac{1}{1 + \tan^{2} y}$ .
Since $\tan y = x$ that is $\frac{1}{1 + x^{2}}$ , and no square root appears here because nothing ever had to be square-rooted.
True or False?
$\frac{1}{1 + x^{2}}$ and $\frac{1}{x^{2} + 1}$ are the derivatives of different functions.
False.
They are the same expression written two ways, since the order of two added terms makes no difference.
Both are the derivative of $\arctan x$ , so a result reached in one form is the same result as one reached in the other.
How do the derivatives of $\arcsin x$ and $\arccos x$ differ, and what does that difference mean for their graphs?
They differ only by a sign: $\frac{1}{\sqrt{1 - x^{2}}}$ for $\arcsin x$ and $- \frac{1}{\sqrt{1 - x^{2}}}$ for $\arccos x$ .
So $y = \arcsin x$ has a positive gradient everywhere on its domain while $y = \arccos x$ has a negative one, which is why one curve rises and the other falls.
Find the derivative of $\arctan (5 x^{3} - 2 x)$ .
Since $5 x^{3} - 2 x$ is not simply $x$ , apply the chain rule to the standard derivative $\frac{1}{1 + x^{2}}$ .
That gives $\frac{1}{1 + {(5 x^{3} - 2 x)}^{2}} \times (15 x^{2} - 2) = \frac{15 x^{2} - 2}{1 + {(5 x^{3} - 2 x)}^{2}}$ .
When integrating $\arcsin x$ by parts, which part do you differentiate and which do you integrate?
Write it as $1 \times \arcsin x$ , then differentiate the $\arcsin x$ and integrate the $1$ .
That is the right way round because an inverse function has a standard derivative but no obvious integral, exactly as with $\ln x$ .
Fill in the missing term in this result, which comes from integrating by parts:
$\int \arcsin x d x = x \arcsin x +______+ c$
The completed result is:
$\int \arcsin x d x = x \arcsin x + \sqrt{1 - x^{2}} + c$
The companion result for $\arccos x$ is the same but with a minus sign in front of the surd.
Why does $\int \arctan x d x$ end in a logarithm while $\int \arcsin x d x$ ends in a surd?
Because the by-parts step leaves an integral with $x$ on top and the inverse function's own derivative underneath, and those two denominators behave quite differently.
With $1 + x^{2}$ underneath, the numerator is half the derivative of the denominator and gives a logarithm, whereas with $\sqrt{1 - x^{2}}$ underneath the integration produces a surd instead.
Which of $\sqrt{a^{2} - x^{2}}$ , $\sqrt{a^{2} + x^{2}}$ and $\sqrt{x^{2} - a^{2}}$ needs a trigonometric substitution, and which need a hyperbolic one?
Only $\sqrt{a^{2} - x^{2}}$ takes a trigonometric substitution, $x = a \sin u$ , because $1 - \sin^{2} u \equiv \cos^{2} u$ is the identity that collapses it.
The other two take hyperbolic substitutions, $x = a \sinh u$ and $x = a \cosh u$ , because $\cosh^{2} u - \sinh^{2} u \equiv 1$ rearranges to fit either of them.
Which substitution turns $\int \frac{1}{a^{2} + x^{2}} d x$ into something you can integrate on sight?
The substitution $x = a \tan u$ , which makes the denominator $a^{2} (1 + \tan^{2} u) = a^{2} \sec^{2} u$ .
Since $d x = a \sec^{2} u d u$ , everything cancels down to $\frac{1}{a} \int 1 d u$ , giving $\frac{1}{a} \arctan (\frac{x}{a}) + c$ .
Why does $\arccos$ never appear in the standard integration results?
Because the derivative of $\arccos (\frac{x}{a})$ is $- \frac{1}{\sqrt{a^{2} - x^{2}}}$ , the negative of the expression the standard result is built on.
Integrating $\frac{1}{\sqrt{a^{2} - x^{2}}}$ would therefore give $- \arccos (\frac{x}{a})$ , which differs from $\arcsin (\frac{x}{a})$ only by a constant that the $+ c$ absorbs.
Which substitution shows that $\int \frac{1}{\sqrt{x^{2} + a^{2}}} d x = arsinh (\frac{x}{a}) + c$ ?
The substitution $x = a \sinh u$ , since ${(a \sinh u)}^{2} + a^{2} = a^{2} (\sinh^{2} u + 1) = a^{2} \cosh^{2} u$ .
With $d x = a \cosh u d u$ the $a \cosh u$ cancels top and bottom, leaving $\int 1 d u = u$ , and $u$ is $arsinh (\frac{x}{a})$ .
The substitution $x = a \cosh u$ is used on $\int \frac{1}{\sqrt{x^{2} - a^{2}}} d x$ . Fill in what $d x$ becomes:
$d x =______d u$
The completed statement is:
$d x = a \sinh u d u$
That $a \sinh u$ is exactly what cancels with the $\sqrt{a^{2} (\cosh^{2} u - 1)} = a \sinh u$ in the denominator, leaving $\int 1 d u$ .
True or False?
$\int \frac{1}{a^{2} - x^{2}} d x$ can be found by partial fractions instead of by using an inverse hyperbolic function.
True.
Since $a^{2} - x^{2}$ is a difference of two squares it factorises as $(a - x) (a + x)$ , so partial fractions turn the integral into two logarithms.
That is why the standard result is printed in an $artanh$ form and a logarithmic form side by side: they are the same answer.
What do adjusting and compensating mean when matching an integral to a standard form?
You rewrite the integrand so that it exactly matches the standard form, then multiply outside the integral by whatever undoes that change.
In $\int \frac{5}{\sqrt{9 x^{2} - 25}} d x$ the denominator becomes $\sqrt{{(3 x)}^{2} - 5^{2}}$ , which wants a numerator of $3$ , so $\frac{1}{3}$ goes outside to compensate.
After completing the square, why is no adjusting or compensating needed for $\int \frac{1}{\sqrt{x^{2} - 6 x + 25}} d x$ ?
Because the bracket is $x - 3$ , whose derivative is $1$ , so the standard form already fits with no extra factor to correct.
The integral is therefore just $arsinh (\frac{x - 3}{4}) + c$ , with $x - 3$ standing in place of the variable.

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