Reference Sheet: Programme F.13 — Integration
Compact rules and procedures, in the order of the book's programme. Every rule links to the unit in the lesson that derives it — if a rule here looks arbitrary, the link is where it stops being arbitrary.
Logarithms and signs. This sheet writes \int\frac1x\,dx=\ln|x|+C, which is valid for x<0 as well. The book writes \ln x throughout F.13, assuming positive arguments; for positive arguments the two agree.
1. Integration
Derived in Unit 01.
Integration is the reverse of differentiation: \int f(x)\,dx is a function whose derivative is f(x). f(x) is the integrand; dx names the variable. Check any integral by differentiating it.
Example. \frac{d}{dx}x^5=5x^4, so \int5x^4\,dx=x^5+C.
2. The constant of integration
Derived in Unit 01.
Differentiation destroys constants, so an indefinite integral always carries +C. A condition ("I=\dots when x=\dots") fixes C: substitute it and solve.
Example. I=\int6x^2\,dx=2x^3+C; if I=5 at x=1, then C=3.
3. Standard integrals
Derived in Unit 02.
| f(x) | \int f(x)\,dx |
|---|---|
| x^n | \frac{x^{n+1}}{n+1}+C\quad(n\neq-1) |
| a (constant) | ax+C |
| \frac1x | \ln\lvert x\rvert+C |
| e^x | e^x+C |
| a^x | \frac{a^x}{\ln a}+C |
| \sin x | -\cos x+C |
| \cos x | \sin x+C |
| \sec^2x | \tan x+C |
A constant factor stays outside: \int a\,f(x)\,dx=a\int f(x)\,dx.
Examples. \int x^{-3}\,dx=-\frac{x^{-2}}{2}+C. \int\sqrt x\,dx=\frac23x^{3/2}+C. \int3^x\,dx=\frac{3^x}{\ln3}+C.
4. Integrating polynomials
Derived in Unit 02.
Integrate term by term; collect all the constants into one C. To evaluate at a point after fixing C, use nested form.
Example. \int(4x^3-3x^2+2)\,dx=x^4-x^3+2x+C.
5. Functions of a linear function of x
Derived in Unit 03.
Replace x in the standard result by ax+b, and divide by a:
| Integrand | Integral |
|---|---|
| (ax+b)^n | \frac{(ax+b)^{n+1}}{a(n+1)}+C |
| e^{ax+b} | \frac{e^{ax+b}}a+C |
| \cos(ax+b) | \frac{\sin(ax+b)}a+C |
| \sin(ax+b) | -\frac{\cos(ax+b)}a+C |
| \frac1{ax+b} | \frac{\ln\lvert ax+b\rvert}a+C |
Example. \int(1-4x)^3\,dx=\frac{(1-4x)^4}{4\times(-4)}+C=-\frac{(1-4x)^4}{16}+C.
6. Integration by partial fractions
Derived in Unit 04; the splitting rules are those of Programme F.8.
- If the numerator's degree is not lower than the denominator's, divide first.
- Factorise the denominator and split into partial fractions.
- Integrate each term: \int\frac{A}{ax+b}\,dx=\frac Aa\ln\lvert ax+b\rvert+C.
Example. \frac{3x+1}{(x-1)(x+1)}=\frac{2}{x-1}+\frac{1}{x+1}, so the integral is 2\ln\lvert x-1\rvert+\ln\lvert x+1\rvert+C.
7. Areas under curves: the definite integral
Derived in Unit 06.
The area under y=f(x) from x=a to x=b (curve above the axis) is
a and b are the limits. The constant of integration cancels, so it is omitted. The reason: the area function A(x) grows at rate \frac{dA}{dx}=f(x), so it is an antiderivative.
Example. \int_1^2(3x^2+1)\,dx=\big[x^3+x\big]_1^2=(8+2)-(1+1)=8.
8. Integration as a summation
Derived in Unit 05.
— the limit of the total of thin rectangles f(x)\,\delta x. \int is an elongated S for sum. For a rate, the integral is the accumulated amount: \int v\,dt is distance, \int i\,dt is charge, \int q\,dt is volume.
Example. With strips indexed x_r=a+r\,\delta x, the right-hand sum for y=x^2 on [0,3] is \frac{9(n+1)(2n+1)}{2n^2}\to9=\int_0^3x^2\,dx.
9. Negative values: areas above and below the axis
Derived in Unit 07.
Where f(x)<0, each strip f(x)\,\delta x is negative, so \int_a^bf\,dx is the net (signed) area. For the physical area: sketch, find the crossings, integrate each part separately, and add the magnitudes.
Example. \int_0^{2\pi}\sin x\,dx=0, but the area between y=\sin x and the axis over [0,2\pi] is 2+2=4.
10. The area between a curve and an intersecting line
Derived in Unit 08.
with a, b the x-coordinates of the intersections (solve y_{\text{top}}=y_{\text{bottom}}). The position of the axis does not matter.
Example. y=9-x^2 and y=5 meet at x=\pm2: A=\int_{-2}^2(4-x^2)\,dx=\left[4x-\frac{x^3}3\right]_{-2}^2=\frac{16}3-\left(-\frac{16}3\right)=\frac{32}3.
Traps
- Forgetting +C in an indefinite integral — or keeping it in a definite one, where it cancels.
- n=-1 in the power rule. \int x^{-1}\,dx is \ln\lvert x\rvert, not \frac{x^0}{0}.
- Not dividing by a. \int e^{3x}\,dx=\frac13e^{3x}+C; differentiate to check.
- \int\sin x\,dx=-\cos x, with the minus sign; \int\cos x\,dx=+\sin x.
- Integrating a quotient top and bottom. There is no such rule: use partial fractions, dividing first if the fraction is improper.
- Order of subtraction. F(\text{upper})-F(\text{lower}).
- Signed area. A definite integral across an axis crossing gives net area, not physical area. Sketch first.
Self-check
Q1. Find \int(3x^2-4x+5)\,dx.
Q2. I=\int(4x-1)\,dx and I=6 when x=2. Find I when x=3.
Q3. Integrate (a) (3x-2)^4, (b) \sin(2x+1), (c) 4e^{2x}, (d) \frac5{2x+3}.
Q4. Find \int\frac{x+7}{(x-1)(x+2)}\,dx.
Q5. Evaluate \int_0^2(3x^2+2x)\,dx.
Q6. Evaluate \int_0^\pi\sin x\,dx.
Q7. Find the area between y=x^2-2x and the x-axis from x=0 to x=3, and the value of \int_0^3(x^2-2x)\,dx.
Q8. Find the area enclosed between y=6-x^2 and y=2.
Q9. Water flows into a tank at q=0.2t-0.01t^2 m³/min. How much enters in the first 10 minutes?
Solutions
Q1. x^3-2x^2+5x+C. Check: \frac{d}{dx} gives 3x^2-4x+5.
Q2. I=2x^2-x+C. At x=2: 8-2+C=6, so C=0. At x=3: I=18-3=15.
Q3. (a) \frac{(3x-2)^5}{5\times3}+C=\frac{(3x-2)^5}{15}+C. (b) -\frac{\cos(2x+1)}2+C. (c) \frac{4e^{2x}}2+C=2e^{2x}+C. (d) \frac52\ln\lvert2x+3\rvert+C.
Q4. x+7=A(x+2)+B(x-1). At x=1: 8=3A, A=\frac83. At x=-2: 5=-3B, B=-\frac53. So
Check at x=0: \frac{8/3}{-1}-\frac{5/3}{2}=-\frac83-\frac56=-\frac72, and \frac{7}{(-1)(2)}=-\frac72.
Q5. \big[x^3+x^2\big]_0^2=(8+4)-0=12.
Q6. \big[-\cos x\big]_0^\pi=(-\cos\pi)-(-\cos0)=1+1=2.
Q7. The curve crosses the axis at x=0 and x=2. \int_0^2(x^2-2x)\,dx=\left[\frac{x^3}3-x^2\right]_0^2=\frac83-4=-\frac43 (below). \int_2^3(x^2-2x)\,dx=(9-9)-\left(\frac83-4\right)=\frac43 (above). Area =\frac43+\frac43=\frac83; the integral straight through is -\frac43+\frac43=0.
Q8. They meet where 6-x^2=2, x=\pm2, with the parabola on top: \int_{-2}^2(4-x^2)\,dx=\left[4x-\frac{x^3}3\right]_{-2}^2=\left(8-\frac83\right)-\left(-8+\frac83\right)=\frac{32}3=10.67.
Q9. V=\int_0^{10}(0.2t-0.01t^2)\,dt=\left[0.1t^2-\frac{0.01t^3}{3}\right]_0^{10}=10-3.333=6.67 m³.