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2.4 Integration

Syllabus
9231–2028–2029
Topic
2.4
Level
A2

Learning objectives

2.4.1Hyperbolic functions• integrate hyperbolic functions and recognise integrals of functions of the form ax 1 22-, xa 1 22+ and xa 1 22-, and integrate associated functions using trigonometric or hyperbolic substitutions as appropriate Including use of completing the square where necessary, e.g. to integrate xx 1 2 +.2.4.2And use reduction formulae for the• derive and use reduction formulae for the evaluation of definite integrals e.g. sin xxdn 0 2 1 r y, x x1edx n 0 1 -- ^hy. In harder cases hints may be given, e.g. sec xxdn 0 4 1 r y by considering tans ecx xxd d n^h.2.4.3How the area under a curve• understand how the area under a curve may be approximated by areas of rectangles, and use rectangles to estimate or set bounds for the area under a curve or to derive inequalities or limits concerning sums Questions may involve either rectangles of unit width or rectangles whose width can tend to zero, e.g. ln r n1 1 1ln n r n 1 22+ + = ^h/, 1 n n r x x1 1 1 d r n 1 1 0 1.+ + - - = c ^m h/ y.2.4.4Arc length and surface area• use integration to find - arc lengths for curves with equations in Cartesian coordinates, including the use of a parameter, or in polar coordinates - surface areas of revolution about one of the axes for curves with equations in Cartesian coordinates, including the use of a parameter. Any questions involving integration may require techniques from Cambridge International A Level Mathematics (9709) applied to more difficult cases, e.g. integration by parts for sin xxedxy, or use of the substitution tantx 2 1=. Surface areas of revolution for curves with equations in polar coordinates will not be required.

Let the sign under the square root choose the substitution

form, a>0a>0 useful substitution primitive on the stated branch
1/a2−x21/\sqrt{a^2-x^2} x=asin⁡θx=a\sin\theta sin⁡−1(x/a)+C\sin^{-1}(x/a)+C, ∣x∣<a|x|<a
1/x2+a21/\sqrt{x^2+a^2} x=asinh⁡ux=a\sinh u sinh⁡−1(x/a)+C\sinh^{-1}(x/a)+C
1/x2−a21/\sqrt{x^2-a^2} x=acosh⁡ux=a\cosh u cosh⁡−1(x/a)+C\cosh^{-1}(x/a)+C, x>ax>a
integrand primitive
sinh⁡x\sinh x cosh⁡x+C\cosh x+C
cosh⁡x\cosh x sinh⁡x+C\sinh x+C
sech⁡2x\operatorname{sech}^2x tanh⁡x+C\tanh x+C
cosech⁡2x\operatorname{cosech}^2x −coth⁡x+C-\coth x+C
sech⁡xtanh⁡x\operatorname{sech}x\tanh x −sech⁡x+C-\operatorname{sech}x+C
cosech⁡xcoth⁡x\operatorname{cosech}x\coth x −cosech⁡x+C-\operatorname{cosech}x+C

Complete the square before choosing. For $x>0$, set $u=x+1$:\int\frac{dx}{\sqrt{x^2+2x}}=\int\frac{du}{\sqrt{u^2-1}}=\cosh^{-1}u+C=\cosh^{-1}(x+1)+C.

Transform the differential and any definite limits as well as the radical. The three signs are not interchangeable: a2−x2a^2-x^2, x2+a2x^2+a^2 and x2−a2x^2-a^2 lead to different inverse families and real-domain conditions.

A reduction formula is a derivation plus a base case

LetLetI_n=\int_0^{\pi/2}\sin^n x,dx,\qquad n\ge2.Integrate by parts with $u=\sin^{n-1}x$ and $dv=\sin x\,dx$.

The boundary term $[-\sin^{n-1}x\cos x]_0^{\pi/2}$ is zero, soI_n=(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x,dx=(n-1)(I_{n-2}-I_n).HenceHenceI_n=\frac{n-1}{n}I_{n-2}.

The base cases are $I_0=\pi/2$ and $I_1=1$. ThereforeI_6=\frac56I_4=\frac56\cdot\frac34I_2=\frac56\cdot\frac34\cdot\frac12I_0=\frac{5\pi}{32}. Even indices end at $I_0$; odd indices end at $I_1$.

Do not quote a recurrence without its valid index, limits and base value. In a different indexed integral, evaluate the integration-by-parts boundary term afresh; it need not vanish.

Rectangle endpoints decide the bound; rectangle width decides the limit

ff on [a,b][a,b] left endpoints right endpoints
increasing underestimate overestimate
decreasing overestimate underestimate

Each rectangle area is f(xr) Δxf(x_r)\,\Delta x. Verify the direction from monotonicity and the rectangles rather than memorising an endpoint label without its interval.

For decreasing $f(x)=1/x$, unit-width rectangles give\int_1^{n+1}\frac{dx}{x}<\sum_{r=1}^{n}\frac1r<1+\int_1^n\frac{dx}{x},hencehence\ln(n+1)<\sum_{r=1}^{n}\frac1r<1+\ln n.Theextraendpointrectangleexplainstheisolated1.The extra endpoint rectangle explains the isolated 1.

With $n$ rectangles on $[0,1]$, $\Delta x=1/n$:\sum_{r=1}^{n}\frac{n}{n^2+r^2}=\frac1n\sum_{r=1}^{n}\frac{1}{1+(r/n)^2}\longrightarrow\int_0^1\frac{dx}{1+x^2}=\frac{\pi}{4}.

Right endpoints do not always overestimate: monotonicity controls the direction. In a Riemann sum, keep the rectangle width outside the function; omitting 1/n1/n changes the scale and usually makes the sum diverge.

Choose the arc element first, then multiply by the rotation radius

curve representation arc element and length surface of revolution
Cartesian y=f(x)y=f(x) ds=1+(dy/dx)2 dxds=\sqrt{1+(dy/dx)^2}\,dx, L=∫dsL=\int ds about x-axis: 2π∫∣y∣ ds2\pi\int |y|\,ds; about y-axis: 2π∫∣x∣ ds2\pi\int |x|\,ds
parametric x(t),y(t)x(t),y(t) ds=(dx/dt)2+(dy/dt)2 dtds=\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt use the same 2π∫(radius) ds2\pi\int(\text{radius})\,ds
polar r(θ)r(\theta) ds=r2+(dr/dθ)2 dθds=\sqrt{r^2+(dr/d\theta)^2}\,d\theta not required by this syllabus

For $x=t^2$, $y=\tfrac23t^3$, $0\le t\le1$,L=\int_0^1\sqrt{(2t)^2+(2t^2)^2},dt=\int_0^1 2t\sqrt{1+t^2},dt=\frac23(2\sqrt2-1).

For $y=x^2$, $0\le x\le1$, rotated about the x-axis, the radius is $y=x^2$ and $ds=\sqrt{1+4x^2}\,dx$. ThusS=2\pi\int_0^1x^2\sqrt{1+4x^2},dx.Thelimitsmusttracetherequiredarconce.The limits must trace the required arc once.

Arc length is ∫ds\int ds, while surface area is 2π∫(radius)ds2\pi\int(\text{radius})ds; neither is a volume formula. Use a non-negative geometric radius. Polar arc length is included, but polar surface area of revolution is explicitly excluded.