CP-5 Further calculus — printable coverage pack

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

1. 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.[3 marks]
Answer
- $V=\dfrac{32\pi}{5}$
Method: Using discs, $V=\pi\int_0^2y^2\,dx=\pi\int_0^2x^4\,dx$ . Hence $V=\pi[x^5/5]_0^2=32\pi/5$ .

Tier 2 · Standard

1. 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.[4 marks]
Answer
- $V=\dfrac{206\pi}{15}$
Method: A cross-section perpendicular to the $y$ -axis is a disc of radius $x=y^2+1$ . Therefore $V=\pi\int_0^2(y^2+1)^2\,dy=\pi\int_0^2(y^4+2y^2+1)\,dy$ . Evaluation gives $\pi(32/5+16/3+2)=206\pi/15$ .

Tier 3 · Hard

1. 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.[5 marks]
Answer
- $V=\dfrac{3\pi}{10}$
Method: The curves meet at $(0,0)$ and $(1,1)$ . Rewriting with $x$ in terms of $y$ : the outer boundary is $x=\sqrt{y}$ (from $y=x^2$ ) and the inner is $x=y^2$ (from $y=\sqrt{x}$ ). Then $V=\pi\int_0^1\left((\sqrt{y})^2-(y^2)^2\right)dy=\pi\int_0^1(y-y^4)\,dy=\pi\left(\tfrac12-\tfrac15\right)=3\pi/10$ .

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

1. Evaluate $\displaystyle \int_1^\infty\dfrac{1}{x^2}\,dx$ .[3 marks]
Answer
- The integral converges to $1$ .
Method: Write the integral as $\lim_{b\to\infty}\int_1^b x^{-2}\,dx$ . This is $\lim_{b\to\infty}[-x^{-1}]_1^b=\lim_{b\to\infty}(1-1/b)=1$ .

Tier 2 · Standard

1. Evaluate $\displaystyle \int_0^4\dfrac{1}{\sqrt{x}}\,dx$ .[3 marks]
Answer
- The integral converges to $4$ .
Method: The integrand is undefined at $x=0$ , so use $\lim_{a\to0^+}\int_a^4x^{-1/2}\,dx$ . This equals $\lim_{a\to0^+}[2\sqrt{x}]_a^4=\lim_{a\to0^+}(4-2\sqrt{a})=4$ .

Tier 3 · Hard

1. Evaluate $\displaystyle \int_0^\infty\dfrac{1}{\sqrt{x}(1+x)}\,dx$ , showing explicitly how both improper endpoints are handled.[6 marks]
Answer
- The integral converges to $\pi$ .
Method: For $0<a<b$ , set $x=t^2$ , so $dx=2t\,dt$ and $\sqrt{x}=t$ . Then $\int_a^b dx/(\sqrt{x}(1+x))=2\int_{\sqrt a}^{\sqrt b}dt/(1+t^2)=2[\arctan t]_{\sqrt a}^{\sqrt b}$ . Taking the limits separately gives $2(\pi/2-0)=\pi$ as $a\to0^+$ and $b\to\infty$ .

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

1. Find the mean value of $f(x)=x^2$ on the interval $0\leq x\leq3$ .[3 marks]
Answer
- Mean value $=3$
Method: The interval length is $3$ . Hence the mean is $\frac13\int_0^3x^2\,dx=\frac13[x^3/3]_0^3=\frac13(9)=3$ .

Tier 2 · Standard

1. Determine the exact mean value of $f(x)=\dfrac{1}{x+1}$ on $0\leq x\leq e-1$ .[4 marks]
Answer
- Mean value $=\dfrac{1}{e-1}$
Method: The interval length is $e-1$ . Also $\int_0^{e-1}dx/(x+1)=[\ln(x+1)]_0^{e-1}=\ln e-\ln1=1$ . Dividing by the interval length gives $1/(e-1)$ .

Tier 3 · Hard

1. Find the exact mean value of $f(x)=xe^{-x}$ on $0\leq x\leq2$ .[5 marks]
Answer
- Mean value $=\dfrac{1-3e^{-2}}{2}$
Method: Integration by parts gives $\int xe^{-x}\,dx=-(x+1)e^{-x}+C$ . Therefore $\int_0^2xe^{-x}\,dx=[-(x+1)e^{-x}]_0^2=1-3e^{-2}$ . The interval length is $2$ , so the mean is $(1-3e^{-2})/2$ .

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

1. Find $\displaystyle \int\dfrac{3x+5}{(x+1)(x+2)}\,dx$ .[4 marks]
Answer
- $2\ln|x+1|+\ln|x+2|+C$
Method: Write $(3x+5)/((x+1)(x+2))=A/(x+1)+B/(x+2)$ . Then $3x+5=A(x+2)+B(x+1)$ , giving $A=2$ and $B=1$ . Integrating $2/(x+1)+1/(x+2)$ gives $2\ln|x+1|+\ln|x+2|+C$ .

Tier 2 · Standard

1. Find $\displaystyle \int\dfrac{2x+5}{(x+1)^2}\,dx$ .[4 marks]
Answer
- $2\ln|x+1|-\dfrac{3}{x+1}+C$
Method: Resolve $(2x+5)/(x+1)^2=A/(x+1)+B/(x+1)^2$ . Since $2x+5=A(x+1)+B$ , comparison gives $A=2$ and $B=3$ . Thus the integral is $2\ln|x+1|+3\int(x+1)^{-2}\,dx=2\ln|x+1|-3/(x+1)+C$ .

Tier 3 · Hard

1. Find $\displaystyle \int\dfrac{x^2+1}{x(x^2+4)}\,dx$ .[6 marks]
Answer
- $\dfrac14\ln|x|+\dfrac38\ln(x^2+4)+C$
Method: Use $(x^2+1)/(x(x^2+4))=A/x+(Bx+C)/(x^2+4)$ . Multiplying through gives $x^2+1=A(x^2+4)+x(Bx+C)$ , so $A=1/4$ , $C=0$ and $B=3/4$ . Therefore the integral is $\frac14\int dx/x+\frac34\int x\,dx/(x^2+4)=\frac14\ln|x|+\frac38\ln(x^2+4)+C$ .

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

1. Differentiate $y=\arcsin(2x)$ with respect to $x$ .[2 marks]
Answer
- $\dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-4x^2}}$
Method: Let $u=2x$ , so $u'=2$ . The chain rule gives $dy/dx=u'/\sqrt{1-u^2}=2/\sqrt{1-4x^2}$ .

Tier 2 · Standard

1. Differentiate $y=\arctan\left(\dfrac{x-1}{2}\right)$ .[3 marks]
Answer
- $\dfrac{dy}{dx}=\dfrac{2}{(x-1)^2+4}$
Method: Take $u=(x-1)/2$ , so $u'=1/2$ . Then $dy/dx=(1/2)/(1+u^2)=(1/2)/(1+(x-1)^2/4)=2/((x-1)^2+4)$ .

Tier 3 · Hard

1. Given $y=x\arccos x-\sqrt{1-x^2}$ for $-1<x<1$ , show that $\dfrac{dy}{dx}=\arccos x$ .[5 marks]
Answer
- $\dfrac{dy}{dx}=\arccos x$
Method: The product rule gives $d(x\arccos x)/dx=\arccos x-x/\sqrt{1-x^2}$ . Also $d(-\sqrt{1-x^2})/dx=x/\sqrt{1-x^2}$ . The two rational terms cancel, leaving $dy/dx=\arccos x$ .

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

1. Find $\displaystyle \int\dfrac{1}{\sqrt{25-x^2}}\,dx$ .[2 marks]
Answer
- $\arcsin\left(\dfrac{x}{5}\right)+C$
Method: This matches the standard form with $a=5$ , so the integral is $\arcsin(x/5)+C$ .

Tier 2 · Standard

1. Find $\displaystyle \int\dfrac{1}{9-x^2}\,dx$ .[3 marks]
Answer
- $\dfrac{1}{6}\ln\left|\dfrac{3+x}{3-x}\right|+C$
Method: Since $9-x^2=(3-x)(3+x)$ , partial fractions give $\dfrac{1}{9-x^2}=\dfrac{1}{6}\left(\dfrac{1}{3+x}+\dfrac{1}{3-x}\right)$ . Integrating term by term gives $\dfrac{1}{6}\left(\ln|3+x|-\ln|3-x|\right)+C=\dfrac{1}{6}\ln\left|\dfrac{3+x}{3-x}\right|+C$ .

Tier 3 · Hard

1. Using the substitution $x=4\sin\theta$ , find $\displaystyle \int\dfrac{x^2}{\sqrt{16-x^2}}\,dx$ .[6 marks]
Answer
- $8\arcsin\left(\dfrac{x}{4}\right)-\dfrac{x}{2}\sqrt{16-x^2}+C$
Method: With $x=4\sin\theta$ , $dx=4\cos\theta\,d\theta$ and $\sqrt{16-x^2}=4\cos\theta$ . The integral becomes $16\int\sin^2\theta\,d\theta=8\theta-4\sin2\theta+C$ . Now $\theta=\arcsin(x/4)$ and $\sin2\theta=2(x/4)(\sqrt{16-x^2}/4)=x\sqrt{16-x^2}/8$ . Substitution gives $8\arcsin(x/4)-(x/2)\sqrt{16-x^2}+C$ .