Chapter 13 · Calculus · 3 questions in the paper · 8 min read
Integration
What the syllabus asks for 3
- Standard integrals and rules
- Methods: substitution and parts
- Definite integrals and area
Integration questions are the easy half of calculus: a standard integral, a definite integral, an area or an integration by substitution or parts. Learn the table and the three rules, and be sure to add +C for indefinite integrals.
- Standard integrals and linearity.
- Substitution (including \int\frac{f'}{f}dx=\ln|f|) and parts.
- Definite integrals and their properties, area under a curve.
13.1 Standard integrals and rules
| f(x) | \int f\,dx | f(x) | \int f\,dx |
|---|---|---|---|
| x^n (n\ne-1) | \frac{x^{n+1}}{n+1}+C | \sin x | -\cos x+C |
| \frac{1}{x} | \ln\lvert x\rvert+C | \cos x | \sin x+C |
| e^x | e^x+C | \sec^2x | \tan x+C |
| a^x | \frac{a^x}{\ln a}+C | \frac{1}{1+x^2} | \tan^{-1}x+C |
| \sqrt{x} | \frac{2}{3}x^{3/2}+C | \frac{1}{\sqrt{1-x^2}} | \sin^{-1}x+C |
Integrate term by term; constants pull out. \int(ax+b)^n dx=\frac{(ax+b)^{n+1}}{a(n+1)}+C and \int\frac{dx}{ax+b}=\frac{1}{a}\ln|ax+b|+C.
13.2 Methods: substitution and integration by parts
- Substitution: put u=g(x). In particular \int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+C, so \int\frac{2x}{x^2+1}dx=\ln(x^2+1)+C.
- By parts: \int u\,dv=uv-\int v\,du, choosing u in the order Inverse trig, Log, Algebraic, Trig, Exponential (ILATE). Example: \int xe^x dx=(x-1)e^x+C; \int_1^e\ln x\,dx=1.
13.3 Definite integrals and area
- \int_a^bf\,dx=F(b)-F(a). Reversing limits changes the sign; \int_a^bf=\int_a^cf+\int_c^bf.
- Over [-a,a]: an odd function gives 0 (\int_{-1}^1x^3dx=0); an even function gives 2\int_0^a.
- \int_0^\pi\sin x\,dx=2, \int_0^{\pi/2}\cos x\,dx=1.
- Area between y=f(x), the x-axis and x=a,b is \int_a^b|f|dx; between two curves, \int_a^b(f-g)dx with f\ge g.



13.4 Eight traps in integration
Integration questions are checked by differentiating, so most errors are about a missing factor or a forgotten limit. These eight traps are the ones that cost marks in the model paper.
- An indefinite integral needs +C; a definite integral does not. \int(3x^2+2x+1)\,dx=x^3+x^2+x+C, but \int_0^2(3x^2+2x+1)\,dx=8+4+2=14 with no constant. Examiners often put an option with and without C. Always check by differentiating your answer.
- The power rule fails at n=-1. \int x^n\,dx=\frac{x^{n+1}}{n+1} for n\ne-1, and \int\frac1x\,dx=\ln|x|+C (note the absolute value). The integral of a constant k is kx. Do not forget to add 1 to the power before dividing.
- Divide by the inner derivative. \int\cos2x\,dx=\frac12\sin2x+C and \int e^{ax}dx=\frac{e^{ax}}{a}+C. If you leave out the factor \frac1a your derivative check gives 2\cos2x, not \cos2x. This is the reverse of the chain rule.
- In substitution, change everything, including the limits. Put u=g(x), replace du=g'(x)\,dx and rewrite the limits in u, as for \int2x\cos(x^2)\,dx=\sin(x^2)+C (Fig 13.2). If you keep the old limits with the new variable the value is wrong. The integrand must contain g'(x) as a factor.
- By parts: \int u\,dv=uv-\int v\,du, with u chosen by LIATE. For \int xe^x\,dx take u=x and get (x-1)e^x+C. For \int\ln x\,dx take u=\ln x, dv=dx to get x\ln x-x+C. Choosing u=e^x makes the integral harder.
- A definite integral is F(b)-F(a). Swapping the limits changes the sign. Split at a point where the function is undefined or changes rule. For an odd function \int_{-a}^{a}f\,dx=0, and \int_{-1}^{1}x^2dx=\frac23 for an even one.
- Area is not the same as the signed integral. \int_0^{2\pi}\sin x\,dx=0 but the area between the curve and the axis is 4 (Fig 13.3). Split at the zeros and add the absolute values. The area between two curves is \int(\text{upper}-\text{lower})\,dx.
- Memorise the standard forms and verify by differentiation. \int\sec^2x\,dx=\tan x, \int\frac{dx}{1+x^2}=\tan^{-1}x, \int\tan x\,dx=\ln|\sec x|, \int a^xdx=\frac{a^x}{\ln a} and \int e^xdx=e^x. The Riemann sum picture (Fig 13.1) is why the integral of x^2 on [0,2] is \frac83.
13.4.1 Worked checks
Q. Evaluate \int(3x^2+2x+1)\,dx and \int_0^2(3x^2+2x+1)\,dx.
Answer. The indefinite integral is x^3+x^2+x+C. The definite integral is [x^3+x^2+x]_0^2=8+4+2-0=14. The constant cancels in the definite case.
Q. Find \int xe^x\,dx and \int_0^1xe^x\,dx.
Answer. By parts, \int xe^xdx=xe^x-e^x+C=(x-1)e^x+C. Then \int_0^1=[(x-1)e^x]_0^1=0-(-1)=1. Differentiating (x-1)e^x gives xe^x, which confirms the result.
Q. Find the area enclosed by y=x and y=x^2.
Answer. They meet where x=x^2, at x=0 and x=1, and y=x is above. Area =\int_0^1(x-x^2)\,dx=\frac12-\frac13=\frac16. This is the shaded region in Fig 13.3.
Formula and fact sheet
| Item | Result |
|---|---|
| Power rule | \int x^ndx=\frac{x^{n+1}}{n+1}+C |
| Log rule | \int\frac{f'}{f}dx=\ln\lvert f\rvert+C |
| Parts | uv-\int v\,du |
| Definite | F(b)-F(a) |
Chapter summary
- Add +C to every indefinite integral and omit it in definite ones.
- \int\frac{1}{x}dx=\ln|x|, not x^0.
- An odd function over a symmetric interval integrates to zero.
- Use parts when a product has a log, inverse-trig or x\times e^x.
- Area under y=x^2 from 0 to 3 is 9.
Practice questions
Set A — Recall (20 questions)
- \int x^2\,dx equalsax^3+Cb\frac{x^3}{3}+Cc2x+Cd\frac{x^2}{2}+C
- \int\frac{1}{x}\,dx equalsax^0+Cb-\frac{1}{x^2}+Cc\frac{1}{x^2}+Cd\ln|x|+C
- \int\cos x\,dx equalsa\sin x+Cb-\sin x+Cc-\cos x+Cd\cos x+C
- \int e^x\,dx equalsaxe^x+Cbe^{x+1}+Cce^x+Cd\ln x+C
- \int\sin x\,dx equalsa\cos x+Cb-\cos x+Cc\sin x+Cd-\sin x+C
- \int_0^1x\,dx equalsa1b0c2d\frac{1}{2}
- \int\sec^2x\,dx equalsa\sec x\tan x+Cb\cot x+Cc\tan x+Cd-\tan x+C
- The integral of zero isaa constantb0cxd1
- \int\frac{1}{1+x^2}\,dx equalsa\ln(1+x^2)+Cb\sin^{-1}x+Cc\frac{1}{x}+Cd\tan^{-1}x+C
- \int_0^23\,dx equalsa3b6c9d2
- \int(2x+3)\,dx equalsa2x^2+3x+Cbx^2+3+Ccx^2+3x+Cd2+C
- \int_0^\pi\sin x\,dx equalsa2b0c1d-2
- \int x^{-2}\,dx equalsa\frac{1}{x}+Cb-\frac{1}{x}+Cc-2x^{-3}+Cd\ln x+C
- \int a^x\,dx equalsaa^x+Cba^x\ln a+Ccxa^{x-1}+Cd\frac{a^x}{\ln a}+C
- \int_{-1}^{1}x^3\,dx equalsa0b\frac{1}{2}c2d\frac{1}{4}
- \int\sqrt{x}\,dx equalsa\frac{1}{2}x^{-1/2}+Cbx^{3/2}+Cc\frac{2}{3}x^{3/2}+Cd2\sqrt{x}+C
- \int_1^e\frac{1}{x}\,dx equalsaeb1c0de-1
- \int\frac{f'(x)}{f(x)}\,dx equalsaf(x)^2+Cbf'(x)+Cc\frac{1}{f(x)}+Cd\ln|f(x)|+C
- \int_0^{\pi/2}\cos x\,dx equalsa1b0c2d-1
- The area under y=x from x=0 to x=4 isa4b16c8d2
Answers and explanations — Set A
1. (b) — raise the power and divide by it.
2. (d) — the exception to the power rule.
3. (a) — reverse of differentiating sin.
4. (c) — eˣ is its own integral.
5. (b) — mind the minus sign.
6. (d) — [x²/2] from 0 to 1.
7. (c) — \dfrac{d}{dx} tan x = sec² x.
8. (a) — an arbitrary constant.
9. (d) — standard result.
10. (b) — 3 × 2.
11. (c) — x² + 3x + C.
12. (a) — [−cos x] = 1 + 1 = 2.
13. (b) — x⁻¹/(−1).
14. (d) — divide by ln a.
15. (a) — odd function over [−1, 1].
16. (c) — x^\dfrac{3/2}{3/2}.
17. (b) — ln e − ln 1.
18. (d) — substitution u = f(x).
19. (a) — sin(π/2) − sin 0.
20. (c) — triangle: ½ × 4 × 4 = 8.
Set B — Understanding (12 questions)
- \int(3x^2+2x+1)\,dx equalsa6x+2+Cbx^3+x^2+Cc3x^3+x^2+x+Cdx^3+x^2+x+C
- \int_1^22x\,dx equalsa3b4c2d1
- \int\frac{2x}{x^2+1}\,dx equalsa2\ln x+Cbx^2+1+Cc\ln(x^2+1)+Cd\tan^{-1}x+C
- \int_0^1e^x\,dx equalsaebe-1c1de+1
- \int xe^x\,dx equalsa(x-1)e^x+Cb(x+1)e^x+Ccxe^x+Cd\frac{e^x}{2}+C
- The area bounded by y=x^2, the x-axis and x=3 isa27b3c6d9
- \int_0^2x^2\,dx equalsa4b\frac{8}{3}c\frac{2}{3}d8
- \int\cos2x\,dx equalsa\sin2x+Cb2\sin2x+Cc\frac{\sin2x}{2}+Cd-\frac{\sin2x}{2}+C
- \int_{-2}^{2}x^2\,dx equalsa\frac{16}{3}b0c\frac{8}{3}d8
- \int\frac{dx}{2x+3} equalsa\ln|2x+3|+Cb2\ln|2x+3|+Cc-\frac{1}{(2x+3)^2}+Cd\frac{1}{2}\ln|2x+3|+C
- \int_0^{\pi/4}\sec^2x\,dx equalsa0b1c\sqrt{2}d\frac{\pi}{4}
- The area between y=x and y=x^2 from 0 to 1 isa\frac{1}{2}b\frac{1}{3}c\frac{1}{6}d\frac{1}{12}
Answers and explanations — Set B
1. (d) — integrate each term; (a) is the derivative.
2. (a) — [x²] from 1 to 2 = 4 − 1.
3. (c) — numerator is the derivative of the denominator.
4. (b) — e − 1.
5. (a) — by parts: xeˣ − eˣ.
6. (d) — 27/3.
7. (b) — 8/3.
8. (c) — divide by the inner coefficient 2.
9. (a) — 2 × (8/3).
10. (d) — remember the 1/2.
11. (b) — tan(π/4) − tan 0.
12. (c) — ½ − ⅓ = ⅙.
Set C — Application (8 questions)
- \int_0^1(x+1)^2\,dx equalsa3b\frac{7}{3}c\frac{4}{3}d\frac{1}{3}
- \int_0^{\pi/2}\sin x\cos x\,dx equalsa1b\frac{1}{4}c0d\frac{1}{2}
- \int_1^e\ln x\,dx equalsa1bece-1d0
- \int_0^1\frac{dx}{\sqrt{1-x^2}} equalsa1b\pic\frac{\pi}{2}d\frac{\pi}{4}
- The area under y=2x from x=0 to x=3 isa6b9c3d18
- \int\frac{x^2+1}{x}\,dx equalsax^2+\ln|x|+Cb2x-\frac{1}{x^2}+Cc\frac{x^2}{2}+Cd\frac{x^2}{2}+\ln|x|+C
- If \int_0^ax\,dx=8, then a isa2b8c4d16
- The area enclosed by y=4-x^2 and the x-axis isa\frac{32}{3}b\frac{16}{3}c8d16
Answers and explanations — Set C
1. (b) — [(x + 1)³/3] = (8 − 1)/3.
2. (d) — ½ ∫ sin 2x = ½ [−cos 2x / 2] = ½.
3. (a) — [x ln x − x] from 1 to e = 0 − (−1).
4. (c) — sin⁻¹ 1 − sin⁻¹ 0.
5. (b) — triangle: ½ × 3 × 6.
6. (d) — x + 1/x integrates to x²/2 + ln|x|.
7. (c) — a²/2 = 8.
8. (a) — [4x − x³/3] from −2 to 2 = 32/3.