Anti-derivatives

Differentiation takes a function and produces its rate of change. Anti-differentiation runs the machine backwards: you are given the rate of change and must recover the function. That single idea — reversing the derivative — turns out to compute areas, volumes, distances travelled from a velocity, and the total from a rate. At 16 hours this is the largest chapter of Grade 11 Mathematics, and it carries the most marks in the calculus area of the paper.

19.1 Basic (standard) integrals

The anti-derivative

Why a constant? Because \frac{d}{dx}(C) = 0, so if F'(x) = f(x) then (F(x)+C)' = f(x) too. Two anti-derivatives of the same function can differ only by a constant, so F(x)+C is the complete answer, not just one answer.

Geometrically, the anti-derivatives of f form a family of curves, each a vertical translate of the others. At any fixed x_0 every curve in the family has the same gradient f(x_0).

Two rules that do all the splitting

There is no product rule and no quotient rule for integration. A product must be turned into something integrable — by multiplying out, by a trigonometric identity, by substitution, or by parts.

The standard integrals

Two generalisations follow at once from the chain rule. For any constants a \neq 0 and b,

so, for example, \int \cos(3x+1)\,dx = \frac{1}{3}\sin(3x+1)+C and \int (2x-5)^{7}dx = \frac{(2x-5)^{8}}{16}+C.

Why integration is harder than differentiation

Differentiation is mechanical. Given any combination of the standard functions, the product, quotient and chain rules will grind out the derivative in a finite number of steps, and there is never any doubt about what to do next. Integration has no such algorithm. There is no rule for the integral of a product, and there are perfectly ordinary-looking functions — e^{-x^{2}}, \frac{\sin x}{x}, \sqrt{1+x^{3}} — whose anti-derivatives cannot be written down in terms of the functions you know at all. So integration is a matter of pattern recognition: you learn a table of standard forms, and you learn three or four manoeuvres (substitution, parts, partial fractions, trigonometric identities) for pushing an unfamiliar integral into that table. When you meet a new integral, the first question is never "what is the rule?" but "which standard form is this trying to be?".

Finding one particular anti-derivative

An indefinite integral gives the whole family F(x)+C. In an applied problem one extra piece of information — the value of the function at one point, called an initial condition or boundary condition — pins down C and selects one member of the family. This is the pattern for every "given the rate, find the quantity" question in the paper.

19.2 Integration by substitution

Substitution is the chain rule reversed. If x = g(t) is a differentiable function with dx = g'(t)\,dt, then

In practice you go the other way: you spot a function u = g(x) inside the integrand whose derivative g'(x) is also present (up to a constant), put du = g'(x)dx, and the integral collapses to a standard form in u.

The two workhorse forms

A substitution that needs the inverse

Sometimes the substitution does not clear out the whole integrand in one stroke and you must also express the leftover x in terms of u. That is allowed, and it is often the neatest route when a linear expression sits under a root.

Deriving the standard forms 18, 16 and 15

These three appear in the table, but NEB regularly asks you to derive them.

The substitutions used in (ii) and (iii) are the trigonometric substitutions. Choose them from the shape of the surd, and read the other sides off a right-angled triangle when you have to convert back to x.

Expression in the integrand	Substitution	It becomes
\sqrt{a^{2}-x^{2}} or a^2-x^2	x = a\sin\theta	a\cos\theta
\sqrt{a^{2}+x^{2}} or a^2+x^2	x = a\tan\theta	a\sec\theta
\sqrt{x^{2}-a^{2}} or x^2-a^2	x = a\sec\theta	a\tan\theta
\sqrt{\frac{a-x}{a+x}}	x = a\cos 2\theta	\tan\theta form

Quadratics in the denominator: complete the square

Any ax^{2}+bx+c can be written as a\left[(x+\frac{b}{2a})^{2} + \frac{4ac-b^2}{4a^2}\right]. After the shift u = x + \frac{b}{2a}, the integral is one of forms 15, 16, 18, 19, 20 or 21.

The form \frac{px+q}{ax^{2}+bx+c}

Write the numerator as (a multiple of the derivative of the denominator) + (a constant):

The \lambda part integrates by form 25 to a logarithm; the \mu part is handled by completing the square as above. The same split works with the denominator under a square root.

Rational functions: partial fractions

If the integrand is \frac{P(x)}{Q(x)} with P,Q polynomials, first make it a proper fraction (degree of P less than degree of Q) by dividing out. Then split \frac{P}{Q} into partial fractions using the four denominator cases.

Factor in Q(x)	Contributes
distinct linear (x-a)	\frac{A}{x-a}
repeated linear (x-a)^{k}	\frac{A_1}{x-a}+\frac{A_2}{(x-a)^{2}}+\cdots+\frac{A_k}{(x-a)^{k}}
distinct quadratic (x^{2}+px+q)	\frac{Ax+B}{x^{2}+px+q}
repeated quadratic (x^{2}+px+q)^{k}	\frac{A_1x+B_1}{x^{2}+px+q}+\cdots+\frac{A_kx+B_k}{(x^{2}+px+q)^{k}}

Partial fractions always work on a rational function, because every polynomial with real coefficients factorises into linear factors and irreducible quadratic factors. Those are precisely the two kinds of piece we know how to integrate: a linear denominator gives a logarithm, and an irreducible quadratic gives a logarithm plus an inverse tangent after completing the square. So a rational function can always, in principle, be integrated — which is not true of most other families of functions.

19.3 Integration by parts

Integrate the product rule \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} over x:

Rearranging gives the rule.

The whole skill is choosing which factor is u (to be differentiated) and which is dv (to be integrated). Choose u to be whichever comes first in this list:

Repeated parts

If the algebraic factor is x^{n}, apply the rule n times.

The e^{x}[f(x)+f'(x)] form

Integration by parts produces one standard form so often that it is quicker to recognise it than to derive it each time.

The cyclic case

When both factors are of type T or E, two applications bring back the original integral. Treat it as an unknown and solve for it algebraically.

A result worth memorising, which follows from the same trick:

19.4 Definite integral

Definition as the limit of a sum

Divide [a,b] into n equal sub-intervals of width h = \frac{b-a}{n}. On each sub-interval build a rectangle whose height is the value of f at the right-hand end. The total rectangle area is \sum_{r=1}^{n} f(a+rh)\,h. As n grows the rectangles hug the curve better and better.

The two kinds of integral are different objects and are marked differently, so keep them apart.

	Indefinite integral	Definite integral
Written	\int f(x)\,dx	\int_{a}^{b}f(x)\,dx
Result is	a family of functions	a single number
Constant C	compulsory	cancels, so omitted
Means	all anti-derivatives of f	signed area from a to b
On substituting	change back to x at the end	change the limits instead

The fundamental theorem of calculus

Computing every definite integral from the definition would be unbearable. The fundamental theorem converts it into an anti-derivative evaluation.

Geometrically \int_a^b f(x)\,dx is the signed area between the curve and the x-axis from x=a to x=b: the limit of the thin strips of height f(x) and width dx.

Properties of the definite integral

#	Property	Note
P1	\int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx	swapping limits changes the sign
P2	\int_{a}^{a}f(x)dx = 0	zero width
P3	\int_{a}^{b}f(x)dx = \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx	split at any c
P4	\int_{a}^{b}f(x)dx = \int_{a}^{b}f(t)dt	the variable is a dummy
P5	\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx	reflect in the midpoint
P6	\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx	general reflection
P7	\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx if f is even	f(-x)=f(x)
P8	\int_{-a}^{a}f(x)dx = 0 if f is odd	f(-x)=-f(x)
P9	\int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)dx + \int_{0}^{a}f(2a-x)dx	split at a, then P5

19.5 Area under a curve; area between two curves

Signed area and true area

The definite integral counts area below the x-axis as negative. That is exactly what you want for a displacement, and exactly what you do not want when a question asks for "the area of the region".

Area with respect to the x-axis

The area bounded by y = f(x), the x-axis and the ordinates x=a, x=b is

Area with respect to the y-axis

When the region is bounded by the y-axis and horizontal lines, use horizontal strips of length x and width dy:

Area between two curves

If y = f(x) lies above y = g(x) on [a,b], a vertical strip has height f(x)-g(x), so

The limits a and b are the x-coordinates of the points of intersection, found by solving f(x) = g(x).

This formula does not care where the x-axis is. Even if part of the region lies below the axis, the height of a strip is still (top curve) minus (bottom curve), and the negative parts cancel correctly on their own. That is why you do not split an "area between two curves" question at the x-axis — you split it only where the two curves cross each other, because that is where "top" and "bottom" swap over.

Where areas are used

The area idea is the reason integration appears everywhere outside mathematics. If a graph plots a rate against time, the area under it is the total: the area under a velocity-time graph is the distance travelled, the area under a force-displacement graph is the work done, the area under a current-time graph is the charge that has flowed, and the area under a discharge-time graph for a river such as the Bagmati is the total volume of water carried. In each case the units of the area are (units of the y-axis) \times (units of the x-axis), which is a quick way to check that you have set up the right integral.

Chapter summary

• \int f(x)dx = F(x)+C means F'(x)=f(x). The constant C is compulsory in an indefinite integral and cancels in a definite one.
• Integration is linear: constants come out, sums split. There is no product or quotient rule.
• Standard forms to know cold: \int x^n dx = \frac{x^{n+1}}{n+1} (n\neq-1), \int\frac{dx}{x}=\ln|x|, \int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}\frac{x}{a}, \int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}, \int\frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\frac{|x-a|}{|x+a|}.
• Substitution: u=g(x), du=g'(x)dx. Two workhorses: \int\frac{f'}{f}dx=\ln|f|+C and \int f^n f'dx = \frac{f^{n+1}}{n+1}+C. Surds \sqrt{a^2-x^2}, \sqrt{a^2+x^2}, \sqrt{x^2-a^2} call for x=a\sin\theta, a\tan\theta, a\sec\theta.
• Quadratic denominators: complete the square; for \frac{px+q}{ax^2+bx+c} write px+q=\lambda(2ax+b)+\mu. Rational functions: divide if improper, then split into partial fractions by the four denominator cases.
• By parts: \int u\,dv = uv - \int v\,du, choosing u by ILATE. Apply twice for x^2 factors; for \int e^{ax}\sin bx\,dx apply twice and solve for I.
• \int_a^b f(x)dx = \lim_{n\to\infty}h\sum_{r=1}^{n}f(a+rh) with h=\frac{b-a}{n}, and equals F(b)-F(a) by the fundamental theorem. Change the limits whenever you change the variable.
• Useful properties: \int_0^a f(x)dx=\int_0^a f(a-x)dx; \int_{-a}^{a}f=2\int_0^a f for even f and 0 for odd f.
• Area under a curve is \int_a^b y\,dx (or \int_c^d x\,dy for horizontal strips); area between curves is \int_a^b (y_{top}-y_{bot})dx. Split at every crossing of the axis and take moduli before adding.

Practice questions

Group A — Multiple choice (1 mark each)

1. \int \frac{dx}{x} equals [1]a-x^{-2}+Cb\ln|x|+Cc\frac{x^{0}}{0}+Cdx\ln x+C
2. \int_{0}^{\pi/2}\cos x\,dx equals [1]a0b\frac{1}{2}c1d\frac{\pi}{2}
3. \int \frac{dx}{4+x^{2}} equals [1]a\tan^{-1}\frac{x}{2}+Cb\frac{1}{2}\tan^{-1}\frac{x}{2}+Cc\frac{1}{4}\tan^{-1}\frac{x}{4}+Cd\ln(4+x^{2})+C
4. The area bounded by y = x^{2}, the x-axis and x = 3 is [1]a3b6c9d27
5. \int e^{x}(\sin x + \cos x)\,dx equals [1]ae^{x}\cos x+Cbe^{x}\sin x+Cce^{x}(\sin x-\cos x)+Cd2e^{x}\sin x+C
6. \int_{-1}^{1} x^{3}\,dx equals [1]a\frac{1}{2}b\frac{1}{4}c2d0
7. In \int x\ln x\,dx, ILATE says u should be [1]axb\ln xceitherdx\ln x
8. \int \frac{2x}{x^{2}+5}\,dx equals [1]a\ln(x^{2}+5)+Cb\frac{1}{2}\ln(x^{2}+5)+Cc2\tan^{-1}\frac{x}{5}+Cd\frac{x^{2}}{x^{2}+5}+C
9. The value of \int_{2}^{2}(x^{5}+7x)\,dx is [1]a1b0c36dundefined
10. \int \frac{dx}{\sqrt{9-x^{2}}} equals [1]a\sin^{-1}\frac{x}{9}+Cb\frac{1}{3}\sin^{-1}\frac{x}{3}+Cc\sin^{-1}\frac{x}{3}+Cd\ln|x+\sqrt{9-x^{2}}|+C
11. The area between y = f(x) and y = g(x) from x=a to x=b, where f \ge g throughout, is [1]a\int_a^b [f(x)+g(x)]dxb\int_a^b [f(x)-g(x)]dxc\int_a^b f(x)g(x)dxd\int_a^b \frac{f(x)}{g(x)}dx

Answers to Group A

1. (b) — rule 1 fails for n=-1; the derivative of \ln|x| is 1/x.

2. (c) — [\sin x]_0^{\pi/2} = 1-0 = 1.

3. (b) — form 16 with a=2 gives the factor \frac1a = \frac12.

4. (c) — \int_0^3 x^2dx = [x^3/3]_0^3 = 9.

5. (b) — form 27 with f(x)=\sin x, since f'(x)=\cos x.

6. (d) — x^3 is odd and the limits are symmetric, so by P8 the value is 0.

7. (b) — L comes before A in ILATE.

8. (a) — the numerator is exactly the derivative of the denominator, so form 25 applies with no extra factor.

9. (b) — the limits are equal, so by P2 the integral is zero whatever the integrand.

10. (c) — form 15 with a=3; there is no \frac1a factor in the inverse sine form.

11. (b) — a vertical strip has height (top curve) minus (bottom curve).

Group B — Short answer (5 marks each)

1. Evaluate \int \frac{x+1}{(x+2)(x+3)}\,dx. [5]
2. Evaluate \int x\cos x\,dx and state which factor you took as u and why. [5]
3. Evaluate \int \frac{dx}{x^{2}+6x+13}. [5]
4. Show that \int_{0}^{\pi/2}\frac{\sin x}{\sin x + \cos x}\,dx = \frac{\pi}{4}. [5]
5. Find the area bounded by the curve y = x^{2}, the x-axis and the lines x=1 and x=3. [5]
6. Evaluate \int e^{2x}\cos x\,dx. [5]
7. Evaluate \int \sqrt{4-x^{2}}\,dx using a trigonometric substitution. [5]
8. Evaluate \int \sin^{-1}x\,dx. [5]

Answers to Group B

1. Put \frac{x+1}{(x+2)(x+3)} = \frac{A}{x+2}+\frac{B}{x+3}, so x+1 = A(x+3)+B(x+2). At x=-2: -1 = A. At x=-3: -2 = -B, so B = 2. Hence the integral is

\int\left(\frac{-1}{x+2}+\frac{2}{x+3}\right)dx = -\ln|x+2| + 2\ln|x+3| + C

2. A (algebraic) comes before T (trigonometric) in ILATE, so u = x and dv = \cos x\,dx, giving du = dx and v = \sin x:

\int x\cos x\,dx = x\sin x - \int\sin x\,dx = x\sin x + \cos x + C

Choosing u=\cos x instead would leave \int\frac{x^2}{2}\sin x\,dx, which is worse.

3. Complete the square: x^{2}+6x+13 = (x+3)^{2}+4. With u=x+3 and a=2 in form 16,

\int\frac{du}{u^{2}+2^{2}} = \frac{1}{2}\tan^{-1}\frac{u}{2}+C = \frac{1}{2}\tan^{-1}\frac{x+3}{2}+C

4. Let I = \int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx. By P5 with a=\frac{\pi}{2}, replacing x by \frac{\pi}{2}-x swaps \sin and \cos: I = \int_0^{\pi/2}\frac{\cos x}{\cos x+\sin x}dx. Adding,

2I = \int_0^{\pi/2}\frac{\sin x+\cos x}{\sin x+\cos x}dx = \int_0^{\pi/2}dx = \frac{\pi}{2}

so I = \frac{\pi}{4}.

5. The curve is above the axis on [1,3], so

A = \int_1^3 x^2dx = \left[\frac{x^3}{3}\right]_1^3 = 9 - \frac13 = \frac{26}{3} \ \text{square units}

6. Let I = \int e^{2x}\cos x\,dx. Take u = \cos x, dv = e^{2x}dx, so v = \frac{e^{2x}}{2}:

I = \frac{e^{2x}\cos x}{2} + \frac{1}{2}\int e^{2x}\sin x\,dx

Apply parts again to the new integral with u = \sin x: \int e^{2x}\sin x\,dx = \frac{e^{2x}\sin x}{2} - \frac{1}{2}I. Substituting,

I = \frac{e^{2x}\cos x}{2} + \frac{e^{2x}\sin x}{4} - \frac{I}{4} \;\Longrightarrow\; \frac{5I}{4} = \frac{e^{2x}(2\cos x + \sin x)}{4}

so I = \frac{e^{2x}(2\cos x+\sin x)}{5}+C.

7. Put x = 2\sin\theta, dx = 2\cos\theta\,d\theta, \sqrt{4-x^2}=2\cos\theta:

\int 4\cos^{2}\theta\,d\theta = 2\int(1+\cos2\theta)d\theta = 2\theta + \sin2\theta + C = 2\theta + 2\sin\theta\cos\theta + C

Since \sin\theta = \frac{x}{2} and \cos\theta = \frac{\sqrt{4-x^{2}}}{2},

\int\sqrt{4-x^{2}}\,dx = 2\sin^{-1}\frac{x}{2} + \frac{x}{2}\sqrt{4-x^{2}} + C

8. There is only one factor, so supply a second factor of 1. By ILATE the inverse trigonometric function is u: take u = \sin^{-1}x and dv = dx, so du = \frac{dx}{\sqrt{1-x^{2}}} and v = x:

\int\sin^{-1}x\,dx = x\sin^{-1}x - \int\frac{x\,dx}{\sqrt{1-x^{2}}}

For the remaining integral put t = 1-x^{2}, dt = -2x\,dx:

\int\frac{x\,dx}{\sqrt{1-x^{2}}} = -\frac{1}{2}\int t^{-1/2}dt = -\sqrt{t} = -\sqrt{1-x^{2}}

Hence \int\sin^{-1}x\,dx = x\sin^{-1}x + \sqrt{1-x^{2}} + C.

Group C — Long answer (8 marks each)

1. (a) Derive the formula for integration by parts from the product rule. [3](b) Hence evaluate \int x^{2}\ln x\,dx. [5]
2. Evaluate \int\frac{2x+1}{(x-1)(x^{2}+1)}\,dx by partial fractions, stating the form of the decomposition and justifying it. [8]
3. (a) Evaluate \int_{1}^{3}(x^{2}+1)\,dx from first principles, as the limit of a sum. [5](b) Verify your answer using the fundamental theorem of calculus. [3]
4. Find the area of the region enclosed by the parabola y^{2}=4x and the line y = 2x-4. [8]
5. (a) Evaluate \int_{0}^{\pi/2}x\sin x\,dx. [4](b) Find the area of the region bounded by the curve y = 2x - x^{2} and the x-axis. [4]

Answers to Group C

1. (a) From the product rule, \frac{d}{dx}(uv) = u\frac{dv}{dx}+v\frac{du}{dx}. Integrating both sides with respect to x gives uv = \int u\frac{dv}{dx}dx + \int v\frac{du}{dx}dx, and rearranging,

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx

(b) By ILATE take u = \ln x and dv = x^{2}dx, so du = \frac{dx}{x} and v = \frac{x^{3}}{3}:

\int x^{2}\ln x\,dx = \frac{x^{3}}{3}\ln x - \int\frac{x^{3}}{3}\cdot\frac{dx}{x} = \frac{x^{3}}{3}\ln x - \frac{1}{3}\int x^{2}dx

= \frac{x^{3}}{3}\ln x - \frac{x^{3}}{9} + C

2. The denominator has one distinct linear factor and one non-repeated irreducible quadratic factor, so the decomposition is

\frac{2x+1}{(x-1)(x^{2}+1)} = \frac{A}{x-1} + \frac{Bx+D}{x^{2}+1}

(a quadratic factor needs a linear numerator, because the numerator must be allowed one degree less than the factor). Clearing denominators,

2x+1 = A(x^{2}+1) + (Bx+D)(x-1)

At x=1: 3 = 2A, so A = \frac32. Coefficient of x^{2}: A+B = 0, so B = -\frac32. Constant term: A - D = 1, so D = \frac12. Therefore

I = \frac{3}{2}\int\frac{dx}{x-1} - \frac{3}{2}\int\frac{x\,dx}{x^{2}+1} + \frac{1}{2}\int\frac{dx}{x^{2}+1}

= \frac{3}{2}\ln|x-1| - \frac{3}{4}\ln(x^{2}+1) + \frac{1}{2}\tan^{-1}x + C

3. (a) Here a=1, b=3, h = \frac{2}{n} and f(x) = x^{2}+1, so f(1+rh) = (1+rh)^{2}+1 = 2 + 2rh + r^{2}h^{2} and

\int_1^3 (x^2+1)dx = \lim_{n\to\infty} h\sum_{r=1}^{n}\left(2+2rh+r^{2}h^{2}\right)

= \lim_{n\to\infty}\left[2nh + 2h^{2}\cdot\frac{n(n+1)}{2} + h^{3}\cdot\frac{n(n+1)(2n+1)}{6}\right]

With h = 2/n this is

\lim_{n\to\infty}\left[4 + \frac{4(n+1)}{n} + \frac{8}{6}\cdot\frac{(n+1)(2n+1)}{n^{2}}\right] = 4 + 4 + \frac{8}{6}(1)(2) = \frac{32}{3}

(b) \int_1^3 (x^2+1)dx = \left[\frac{x^{3}}{3}+x\right]_1^3 = (9+3) - \left(\frac13+1\right) = 12 - \frac43 = \frac{32}{3}. The two agree.

4. Solve the two equations together. From the line, x = \frac{y+4}{2}; from the parabola, x = \frac{y^{2}}{4}. Equating,

\frac{y^{2}}{4} = \frac{y+4}{2} \;\Longrightarrow\; y^{2}-2y-8 = 0 \;\Longrightarrow\; (y-4)(y+2) = 0

so y = -2 and y = 4 (the points (1,-2) and (4,4)). Horizontal strips are easier here, because a vertical strip would change its lower boundary partway across. For -2 < y < 4 the line lies to the right of the parabola, so

A = \int_{-2}^{4}\left(\frac{y+4}{2} - \frac{y^{2}}{4}\right)dy = \left[\frac{y^{2}}{4} + 2y - \frac{y^{3}}{12}\right]_{-2}^{4}

= \left(4 + 8 - \frac{64}{12}\right) - \left(1 - 4 + \frac{8}{12}\right) = \frac{20}{3} - \left(-\frac{7}{3}\right) = 9 \ \text{square units}

5. (a) By ILATE take u = x and dv = \sin x\,dx, so v = -\cos x:

\int_{0}^{\pi/2}x\sin x\,dx = \left[-x\cos x\right]_{0}^{\pi/2} + \int_{0}^{\pi/2}\cos x\,dx

The bracket is zero at both limits, since \cos\frac{\pi}{2}=0 and x=0 at the lower limit. So the value is \left[\sin x\right]_{0}^{\pi/2} = 1.

(b) The curve meets the x-axis where x(2-x)=0, i.e. at x=0 and x=2, and it lies above the axis between them. Hence

A = \int_{0}^{2}(2x-x^{2})dx = \left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2} = 4 - \frac{8}{3} = \frac{4}{3} \ \text{square units}