This note covers integration as the reverse of differentiation, definite integrals as areas, and the trapezoidal rule for estimating an area from a formula or a table of values. Integration lets you recover totals, such as distance or volume, from rates of change.
Key ideas
Integration undoes differentiation. If \(F'(x)=f(x)\), then \(F\) is an antiderivative of \(f\), and we write \(\int f(x)\,dx = F(x)+c\). The constant \(c\) appears because every constant disappears when you differentiate, so a whole family of curves has the same gradient function.
To integrate a term \(ax^n\), raise the power by one and divide by the new power. This works for every \(n\) except \(n=-1\), with integer \(n\) at SL and fractional \(n\) as well at HL. Terms are integrated one at a time, and a constant multiple stays in front. Rewrite reciprocals as powers first, for example \(\dfrac{1}{x^2}=x^{-2}\). At HL, do the same for roots, such as \(\sqrt{x}=x^{1/2}\).
A boundary condition, such as a known point on the curve, lets you find \(c\). Substitute the point into the integrated function and solve.
A definite integral \(\int_a^b f(x)\,dx\) gives a number. When \(f(x)\ge 0\) on the interval, this number is the area between the curve, the \(x\)-axis and the lines \(x=a\) and \(x=b\). If the curve lies below the axis, the integral is negative, so the area is the positive size of that value. If the curve crosses the axis, split the interval at each crossing and add the positive areas.
You can evaluate a definite integral by hand using \([F(x)]_a^b=F(b)-F(a)\), or directly on your GDC. Always check the calculator value against the hand method when the integrand is simple.
The trapezoidal rule estimates an area when you cannot integrate exactly, or when you only have data. Split the interval into \(n\) strips of equal width \(h\), treat each strip as a trapezium, and add the areas. If the curve bends upwards (concave up), the tops of the trapezia lie above the curve and the rule overestimates. If it bends downwards, the rule underestimates.
Why it matters. Integration adds up many tiny pieces. The area under a speed–time graph is distance, the area under a power–time graph is energy, and the area under a rate-of-flow graph is volume. Whenever a question gives you a rate and asks for a total, think integration.
Reading a trapezoidal table. Check that the \(x\)-values are equally spaced before you use the rule. If they are not, find the area of each trapezium separately using \(\tfrac12(y_1+y_2)\times\text{width}\) and add them. Using more strips makes the estimate more accurate, because the straight tops follow the curve more closely.
Antiderivatives and checking. You can always check an integration by differentiating your answer. The result must match the original function exactly. This takes seconds and catches most slips with powers and fractions.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\int ax^n\,dx=\dfrac{ax^{n+1}}{n+1}+c,\ n\ne -1\) | Integrate each power term. Add \(c\) for an indefinite integral. |
| \(\int_a^b f(x)\,dx=F(b)-F(a)\) | Definite integral. No constant is needed because it cancels. |
| \(A=\int_a^b f(x)\,dx\) | Area under a curve that stays above the \(x\)-axis between \(x=a\) and \(x=b\). |
| \(h=\dfrac{b-a}{n}\) | Width of each strip in the trapezoidal rule with \(n\) strips. |
| \(A\approx \dfrac{h}{2}\left[y_0+2(y_1+y_2+\dots+y_{n-1})+y_n\right]\) | Trapezoidal rule. End values once, middle values twice. |
Worked examples
Example 1 A curve has gradient function \(f'(x)=6x^2-4x+5\) and passes through the point \((2,19)\). Find \(f(x)\).
Integrate each term: \(f(x)=\dfrac{6x^3}{3}-\dfrac{4x^2}{2}+5x+c=2x^3-2x^2+5x+c\).
Use \(f(2)=19\): \(2(8)-2(4)+5(2)+c=19\), so \(16-8+10+c=19\).
This gives \(18+c=19\), so \(c=1\).
Answer: \(f(x)=2x^3-2x^2+5x+1\)
Example 2 (HL, because of the fractional power) Find the area under the curve \(y=2x+3\sqrt{x}\) between \(x=1\) and \(x=4\).
Rewrite the root as a power: \(y=2x+3x^{1/2}\). The curve is above the \(x\)-axis for \(x\gt 0\).
Integrate: \(\int (2x+3x^{1/2})\,dx = x^2+3\cdot\dfrac{x^{3/2}}{3/2}=x^2+2x^{3/2}\).
At \(x=4\): \(16+2(8)=32\). At \(x=1\): \(1+2(1)=3\).
Subtract: \(32-3=29\).
Check on the GDC: enter the definite integral of \(2x+3\sqrt{x}\) from 1 to 4. It also gives 29.
Answer: area = 29 square units
Example 3 A cyclist's speed is measured every 2 seconds.
| \(t\) (s) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| \(v\) (m s\(^{-1}\)) | 0 | 3.1 | 5.2 | 6.4 | 6.9 |
Use the trapezoidal rule to estimate the distance travelled in the 8 seconds.
Distance is the area under the speed–time graph. There are 4 strips, each of width \(h=2\).
\(A\approx \dfrac{2}{2}\left[0+2(3.1+5.2+6.4)+6.9\right]\).
The middle sum is \(3.1+5.2+6.4=14.7\), so \(2\times 14.7=29.4\).
Then \(A\approx 1\times(0+29.4+6.9)=36.3\).
Answer: about 36.3 m
Example 4 Use the trapezoidal rule with 4 strips to estimate \(\int_1^3 \dfrac{1}{x}\,dx\). Say whether your estimate is too big or too small.
\(h=\dfrac{3-1}{4}=0.5\). The \(x\)-values are 1, 1.5, 2, 2.5, 3.
The \(y\)-values are \(1,\ 0.6667,\ 0.5,\ 0.4,\ 0.3333\) (to 4 d.p.).
\(A\approx \dfrac{0.5}{2}\left[1+2(0.6667+0.5+0.4)+0.3333\right]=0.25\times(1+3.1333+0.3333)\).
This is \(0.25\times 4.4667=1.1167\).
The graph of \(y=\dfrac1x\) curves upwards, so the trapezia sit above it. The estimate is too big. (The exact value, found with logarithms, is about 1.10.)
Answer: about 1.12, an overestimate
Common errors
- Forgetting \(+c\) in an indefinite integral, or putting \(+c\) in a definite integral.
- Using the trapezoidal rule with \(n\) values instead of \(n\) strips. There is always one more \(y\)-value than there are strips.
- Doubling the end values or leaving the middle values single.
- Integrating \(\sqrt{x}\) or \(\dfrac{1}{x^2}\) without first writing them as powers.
- Reading a negative definite integral as a negative area. Area is always positive, so use the size of the value or split at the axis crossing.
Exam tip: Write the integral you are evaluating before you press the calculator keys, and quote the answer to 3 s.f. unless it is exact. For the trapezoidal rule, list \(h\) and every \(y\)-value first so any slip is easy to spot.
Check yourself
- Find \(\int (4x^3-6x)\,dx\).
Answer
Raise each power and divide: \(\dfrac{4x^4}{4}-\dfrac{6x^2}{2}+c=x^4-3x^2+c\). - Evaluate \(\int_0^2 (3x^2+1)\,dx\).
Answer
\([x^3+x]_0^2=(8+2)-0=10\). - (HL) Find the area under \(y=\sqrt{x}\) from \(x=1\) to \(x=9\).
Answer
\(\int_1^9 x^{1/2}dx=\left[\dfrac{2}{3}x^{3/2}\right]_1^9=\dfrac23(27-1)=\dfrac{52}{3}\approx 17.3\) square units. - A graph passes through \((0,1),(1,4),(2,5),(3,3)\). Estimate the area under it from \(x=0\) to \(x=3\) with the trapezoidal rule.
Answer
\(h=1\), so \(A\approx\dfrac12[1+2(4+5)+3]=\dfrac12(22)=11\) square units.