CP-5 Further calculus — 9FM0 spec coverage

CP-5.1

Derive formulae for and calculate volumes of revolution.

A thin strip perpendicular to the axis of rotation forms a disc or washer: summing its volume and taking a limit gives $V=\pi\int y^2\,dx$ about the $x$ -axis and $V=\pi\int x^2\,dy$ about the $y$ -axis.
For a region between two curves, subtract the inner cross-sectional area from the outer one before integrating: $V=\pi\int(R^2-r^2)\,dx$ or the corresponding integral with respect to $y$ .
For rotation about the $y$ -axis use $V=\pi\int x^2\,dy$ ; between two curves subtract volumes, $V=\pi\int(x_{\text{outer}}^2-x_{\text{inner}}^2)\,dy$ . When the curve is given parametrically, substitute to integrate with respect to the parameter, e.g. $V=\pi\int y^2\,\frac{dx}{dt}\,dt$ .
Sketch the region and mark the axis before choosing $dx$ or $dy$ . A common error is to integrate the radius rather than its square, or to omit the factor $\pi$ .

Tier 1 · Easy

3 marks

ORIGINAL

The region bounded by $y=x^2$ , the $x$ -axis and the line $x=2$ is rotated through $2\pi$ radians about the $x$ -axis. Determine the exact volume generated.

Tier 2 · Standard

4 marks

ORIGINAL

The region bounded by $x=y^2+1$ , the $y$ -axis, $y=0$ and $y=2$ is rotated through $2\pi$ radians about the $y$ -axis. Find the exact volume.

Tier 3 · Hard

5 marks

ORIGINAL

The finite region between $y=\sqrt{x}$ and $y=x^2$ is rotated through $2\pi$ radians about the $y$ -axis. Determine the exact volume of the solid formed.

CP-5.2

Evaluate improper integrals where either the integrand is undefined at a value in the range of integration or the range of integration extends to infinity.

An improper integral is defined by a limit: replace an infinite endpoint by a variable, or approach a point where the integrand is undefined from the appropriate side.
If the integrand is undefined at an interior point $c$ , split the integral at $c$ and test the two one-sided integrals separately; both must converge.
For an infinite range, write $\int_a^\infty f(x)\,dx=\lim_{b\to\infty}\int_a^b f(x)\,dx$ and evaluate the limit only after finding an antiderivative.
Never substitute the singular endpoint directly into an antiderivative. A finite-looking cancellation between two divergent pieces does not make the original integral convergent.

Tier 1 · Easy

3 marks

ORIGINAL

Evaluate $\displaystyle \int_1^\infty\dfrac{1}{x^2}\,dx$ .

Tier 2 · Standard

3 marks

ORIGINAL

Evaluate $\displaystyle \int_0^4\dfrac{1}{\sqrt{x}}\,dx$ .

Tier 3 · Hard

6 marks

ORIGINAL

Evaluate $\displaystyle \int_0^\infty\dfrac{1}{\sqrt{x}(1+x)}\,dx$ , showing explicitly how both improper endpoints are handled.

CP-5.3

Understand and evaluate the mean value of a function.

The mean value of an integrable function $f$ on $[a,b]$ is $\dfrac{1}{b-a}\int_a^bf(x)\,dx$ .
Geometrically, the mean value is the height of a rectangle of width $b-a$ having the same signed area as the region represented by the integral.
Find the definite integral first, then divide by the interval length; for instance, if the integral is $I$ , the mean is $I/(b-a)$ .
A common error is to divide by $b$ instead of $b-a$ , or to average only the endpoint values when the function is not linear.

Tier 1 · Easy

3 marks

ORIGINAL

Find the mean value of $f(x)=x^2$ on the interval $0\leq x\leq3$ .

Tier 2 · Standard

4 marks

ORIGINAL

Determine the exact mean value of $f(x)=\dfrac{1}{x+1}$ on $0\leq x\leq e-1$ .

Tier 3 · Hard

5 marks

ORIGINAL

Find the exact mean value of $f(x)=xe^{-x}$ on $0\leq x\leq2$ .

CP-5.4

Integrate using partial fractions.

Before resolving into partial fractions, make the rational function proper by polynomial division whenever the numerator degree is at least the denominator degree.
A distinct linear factor $ax+b$ contributes $A/(ax+b)$ ; a repeated factor $(ax+b)^2$ requires both $A/(ax+b)$ and $B/(ax+b)^2$ .
After finding the constants, integrate each term separately, remembering that $\int dx/(ax+b)=a^{-1}\ln|ax+b|+C$ .
Check the decomposition by recombining it. Common errors are omitting a repeated-factor term and forgetting the chain-rule factor inside a logarithm.

Tier 1 · Easy

4 marks

ORIGINAL

Find $\displaystyle \int\dfrac{3x+5}{(x+1)(x+2)}\,dx$ .

Tier 2 · Standard

4 marks

ORIGINAL

Find $\displaystyle \int\dfrac{2x+5}{(x+1)^2}\,dx$ .

Tier 3 · Hard

6 marks

ORIGINAL

Find $\displaystyle \int\dfrac{x^2+1}{x(x^2+4)}\,dx$ .

CP-5.5

Differentiate inverse trigonometric functions.

The standard derivatives are $\dfrac{d}{dx}(\arcsin x)=(1-x^2)^{-1/2}$ , $\dfrac{d}{dx}(\arccos x)=-(1-x^2)^{-1/2}$ and $\dfrac{d}{dx}(\arctan x)=(1+x^2)^{-1}$ .
For a composite argument $u(x)$ , apply the chain rule: for example $\dfrac{d}{dx}(\arcsin u)=u'/\sqrt{1-u^2}$ .
Products involving inverse trigonometric functions still need the product rule; simplify only after differentiating every factor.
Do not write $1/\cos x$ for the derivative of $\arccos x$ : inverse-function notation and reciprocal notation represent different functions.

Tier 1 · Easy

2 marks

ORIGINAL

Differentiate $y=\arcsin(2x)$ with respect to $x$ .

Tier 2 · Standard

3 marks

ORIGINAL

Differentiate $y=\arctan\left(\dfrac{x-1}{2}\right)$ .

Tier 3 · Hard

5 marks

ORIGINAL

Given $y=x\arccos x-\sqrt{1-x^2}$ for $-1<x<1$ , show that $\dfrac{dy}{dx}=\arccos x$ .

CP-5.6

Integrate functions of the form (a^2 - x^2)^(-1/2) and (a^2 - x^2)^(-1) and be able to choose trigonometric substitutions to integrate associated functions.

The standard results are $\int dx/\sqrt{a^2-x^2}=\arcsin(x/a)+C$ and $\int dx/(a^2-x^2)=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C$ for $a>0$ .
For expressions containing $\sqrt{a^2-x^2}$ , the substitution $x=a\sin\theta$ changes the root to $a\cos\theta$ on a suitable interval.
For $\dfrac{1}{a^2-x^2}$ , either split into partial fractions $\frac{1}{2a}\left(\frac{1}{a+x}+\frac{1}{a-x}\right)$ or substitute $x=a\sin\theta$ ; both routes lead to the logarithmic form.
Choose the substitution from the quadratic form and convert the final answer back to $x$ . A common error is to lose the factor $dx/d\theta$ .

Tier 1 · Easy

2 marks

ORIGINAL

Find $\displaystyle \int\dfrac{1}{\sqrt{25-x^2}}\,dx$ .

Tier 2 · Standard

3 marks

ORIGINAL

Find $\displaystyle \int\dfrac{1}{9-x^2}\,dx$ .

Tier 3 · Hard

6 marks

ORIGINAL

Using the substitution $x=4\sin\theta$ , find $\displaystyle \int\dfrac{x^2}{\sqrt{16-x^2}}\,dx$ .