Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 5 Calculus · 5.5 Anti-differentiation and area
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Anti-differentiate axⁿ, n ∈ ℤ, n ≠ −1, and sums of such terms | SL, HL | "Find ∫(8x³ − 6x² + 3) dx" (2 to 3 marks) |
| Rewrite before integrating: expand brackets, split fractions, negative powers | SL, HL | "Find ∫ (x² − 4)/x² dx" (3 marks) |
| Use a boundary condition to find the constant of integration | SL, HL | "Given f′(x) = … and f(1) = 6, find f(x)" (4 to 5 marks) |
| Explain the link between anti-derivatives, definite integrals and area | SL, HL | Short reasoning; underpins every area question |
| Evaluate a definite integral using technology | SL, HL | Paper 2: "Find the value of ∫ from 1 to 4 of …" (2 marks) |
| Find the area enclosed by a curve y = f(x) and the x-axis, where f(x) > 0 | SL, HL | "Write down an expression for the area of R. Hence find the area" (3 to 5 marks) |
| Apply this to velocity-time graphs | SL, HL | "Find the distance travelled in the first 5 seconds" (2 to 3 marks) |
Before you start
You need 5.3, because every integral here is checked by differentiating it back, and the rewriting skills are the same: expand, split, and write 1/xⁿ as x⁻ⁿ. You need to solve a quadratic for its x-intercepts (2.7), since those are often the limits of an area. The formula booklet gives both results on this page: ∫xⁿ dx = xⁿ⁺¹ ÷ (n + 1) + C for n ≠ −1, and the area under a curve A = ∫ y dx from a to b, for y > 0. Working out definite integrals by hand is 5.11; here the GDC does it.
1The idea in one paragraph
Differentiation takes a function and gives its gradient. Anti-differentiation, also called integration, goes the other way: given the gradient function, find a function it came from. For powers this means reversing the power rule: raise the power by one, then divide by the new power. There is always a family of answers, because adding a constant does not change a gradient, so every answer carries a + C, and one known point on the curve (a boundary condition) pins C down. The surprise is that this reversed process also measures area: the area under a curve between two x-values is a limit of sums of thin rectangles, called a definite integral, and it can be calculated from an anti-derivative. On Paper 2 your GDC evaluates definite integrals directly, so your job is to write the right one down.
2Reversing the power rule
To differentiate xⁿ you multiply by n and reduce the power by one. To undo that, do the opposite steps in the opposite order: increase the power by one, then divide by the new power.
∫ xⁿ dx = xⁿ⁺¹ ÷ (n + 1) + C, n ≠ −1. (Formula booklet.)
The symbol ∫ … dx means "the anti-derivative of … with respect to x", and the result is called an indefinite integral. The dx says which letter is the variable, just as it does in dy/dx.
Check any integral by differentiating it. d/dx (x⁴ ÷ 4) = 4x³ ÷ 4 = x³, so ∫ x³ dx = x⁴/4 + C is right. That check takes five seconds and catches nearly every slip.
Multiples and sums work exactly as for differentiation: a number in front stays in front, and a sum is done term by term.
Two entries catch people:
- A constant. ∫ 5 dx = 5x + C, because 5 = 5x⁰, and raising the power gives 5x¹ ÷ 1. The derivative of 5x is 5.
- Negative powers. ∫ 3/x² dx = ∫ 3x⁻² dx = 3 × x⁻¹ ÷ (−1) + C = −3x⁻¹ + C = −3/x + C. Raising −2 by one gives −1, not −3.
Why n ≠ −1. For n = −1 the rule would divide by n + 1 = 0, which is impossible. So ∫ 1/x dx cannot be found this way. It has an answer, a logarithm, which is 5.10; on this page every power will be an integer other than −1.
3Rewrite first, as you would to differentiate
The rule works only on sums of terms axⁿ. There is no product rule or quotient rule for integration at this level, so a bracket or a fraction must be rewritten before you start.
Expand brackets.
Integrating each bracket and multiplying the results is wrong, and scores nothing.
Split a fraction over a single term.
Check: d/dx (x + x⁻²) = 1 − 2x⁻³. ✓
4The constant of integration, and boundary conditions
Figure 1 shows why the + C is compulsory. The curves y = x² + C, for any C, all have derivative 2x: they are the same shape shifted up or down, with the same gradient at every x. Knowing dy/dx = 2x tells you the shape but not the height.
So ∫ 2x dx = x² + C describes the whole family. To pick one member you need one more fact, usually a point the curve passes through. That fact is called a boundary condition (or an initial condition when the variable is time).
Find y given dy/dx = 6x² − 2x and y = 5 when x = 1.
Figure 2 shows the family and the one curve through (1, 5).
With a negative power: given f′(x) = 1 − 4/x² and f(2) = 7, find f(x).
Two rules for this kind of question. Write the + C before you substitute, or there is nothing to find. And substitute into the integrated function, f, not into the derivative you were given: the boundary condition f(2) = 7 is about f.
In context the constant has a meaning. If v = ds/dt is a velocity, then s = ∫ v dt + C is a displacement, and C is fixed by where the object starts. Section 7 does one.
5Area under a curve, and the definite integral
Now a different-looking question: what is the area between the curve y = 6x − x² and the x-axis, from x = 0 to x = 6?
There is no formula for a curved shape, so approximate it with rectangles. Split the interval into strips, draw a rectangle on each whose height is the curve's height at the middle of the strip, and add up the areas. Figure 3 does it with 6 strips and then 24.
| number of rectangles | 6 | 24 |
|---|---|---|
| total area of rectangles | 36.5 | 36.03 |
More, thinner rectangles fit the curve better, and the totals settle on a limit, just as the chord gradients settled on a limit in 5.1. That limit is the exact area, and it is written
A = ∫_{a}^{b} y dx, the definite integral of y from x = a to x = b. For f(x) > 0 on a ≤ x ≤ b, it is the area between the curve and the x-axis. (Formula booklet.)
The numbers a and b are the limits of integration: a is the lower, b the upper. The ∫ sign is a stretched S, for "sum", and y dx is the area of one thin strip, a height times a tiny width.
The link with anti-derivatives. The guide asks you to be aware of this, and it is the central fact of calculus. If F is any anti-derivative of f, then
∫_{a}^{b} f(x) dx = F(b) − F(a).
For y = 6x − x²: an anti-derivative is F(x) = 3x² − x³/3, and F(6) − F(0) = (108 − 72) − 0 = 36, exactly where the rectangles were heading. That is why one symbol, ∫, is used for both ideas: the area is found by anti-differentiating. The + C does not matter here because it appears in F(b) and in F(a) and cancels. Doing this by hand is 5.11. On this page, the GDC does the evaluation and you do the setting up.
6Definite integrals and areas with technology
Every GDC evaluates a definite integral numerically: you enter the function, the variable and the two limits, and it returns a number. The mark scheme expects to see the integral you entered.
Write the integral down first, with its limits and dx, then give the value. An area with no expression scores only the final accuracy mark, if that.
The area under y = 8/x² + 1 from x = 1 to x = 4, shaded in Figure 4.
(If you want to see the link in action: F(x) = −8/x + x, and F(4) − F(1) = (−2 + 4) − (−8 + 1) = 2 + 7 = 9.)
A region enclosed by a curve and the x-axis. When a question says "the region enclosed by the curve and the x-axis" and gives no limits, the limits are where the curve meets the x-axis. Find them first.
For y = 8 − 2x², shaded as region R in Figure 5:
The guide limits this subtopic to regions where f(x) > 0, above the x-axis, so the definite integral is the area directly. Regions below the axis, where the integral comes out negative, and areas between two curves are 5.11. Always look at a sketch (or the GDC graph) to confirm the curve is above the axis between your limits.
A few GDC habits: set the mode to radians if the function has trigonometry in it; use the exact limits the question gives, or stored values if the limits came from an earlier part; and give the answer to 3 significant figures unless told otherwise.
7Velocity-time graphs
The guide names velocity-time graphs as the other context. Velocity is the rate of change of displacement, v = ds/dt, so displacement is an anti-derivative of velocity. And since the area under a v-t graph is (velocity) × (time) summed up over thin strips, it measures how far the object has gone.
When v(t) > 0 throughout, distance travelled from t = a to t = b is ∫_{a}^{b} v(t) dt, the area under the velocity-time graph.
A cyclist's velocity is v(t) = 2 + 1.5t − 0.25t² m s⁻¹ for 0 ≤ t ≤ 6. Figure 6 shows the graph, which stays above the axis.
The same question could ask for the displacement function. If the cyclist starts at s = 0:
Both routes give 21 m, which is the link from section 5 again: the area is the change in the anti-derivative.
8Where marks are lost
Forgetting + C. An indefinite integral without + C loses a mark, and in a boundary-condition question there is then nothing to find.
Dividing by the old power. ∫ x³ dx is x⁴/4, not x⁴/3. Raise the power first, then divide by the new one.
Getting negative powers wrong. ∫ x⁻³ dx = x⁻² ÷ (−2) = −1/(2x²). Raising −3 by one gives −2.
Integrating brackets or fractions piece by piece. There is no product rule for integrals. Expand or split first.
Substituting the boundary condition into the derivative. "f(2) = 7" is about f. Integrate first, then substitute.
No expression for an area. On Paper 2, "area = 21.3" alone risks losing the method marks. Write ∫_{−2}^{2} (8 − 2x²) dx, then the value.
Wrong limits. For a region enclosed by a curve and the x-axis, the limits are the x-intercepts, not the turning point and not the y-intercept.
Leaving out dx. ∫_{1}^{4} (8/x² + 1) without dx is incomplete notation, and some schemes penalise it.
9Work it right
- Rewrite as a sum of powers: expand, split, and change 1/xⁿ into x⁻ⁿ.
- Integrate term by term: raise each power by one, divide by the new power, and add + C once.
- Check by differentiating your answer back.
- For a boundary condition, substitute the given point into the integrated function, solve for C, and write the final function in full.
- For an area, sketch or graph first to see the region and check it is above the x-axis.
- Find the limits: given in the question, or the x-intercepts of the curve.
- Write the definite integral with limits and dx, then evaluate it on the GDC and give 3 s.f.
- For a velocity-time graph, distance travelled is the integral of v, and displacement needs a starting value to fix C.
10Try it
Marks in brackets. Q1 to Q3 are Paper 1 style, no calculator. Q4 and Q5 are Paper 2 style, with a GDC.
Q1. Find
(a) ∫ (8x³ − 6x² + 3) dx 2 marks
(b) ∫ (x² − 4) ÷ x² dx, for x ≠ 0. 3 marks
Q2. Let f′(x) = 3x² − 4/x², for x ≠ 0. Given that f(1) = 6, find f(x). 5 marks
Q3. The gradient of a curve at the point (x, y) is given by dy/dx = 2x − 6. The curve passes through the point (1, 0).
(a) Find the equation of the curve. 3 marks
(b) Find the coordinates of the point on the curve where the gradient is zero. 2 marks
Q4. Let f(x) = 3 + 2x − x². The region R is enclosed by the graph of f and the x-axis.
(a) Find the x-intercepts of the graph of f. 2 marks
(b) Write down an expression for the area of R. 2 marks
(c) Find the area of R. 1 mark
Q5. A particle moves in a straight line. Its velocity, v m s⁻¹, after t seconds is v(t) = 3 + 2t − 0.1t³, for 0 ≤ t ≤ 5. The velocity is positive throughout this time.
(a) Find the distance travelled by the particle in the first 5 seconds. 3 marks
(b) The particle's displacement from a fixed point O is s metres, and s = 4 when t = 0. Find an expression for s in terms of t. 3 marks
(c) Use your answer to (b) to verify your answer to (a). 1 mark
11In one breath
Anti-differentiation reverses differentiation: to integrate xⁿ, raise the power by one and divide by the new power, xⁿ⁺¹ ÷ (n + 1), for any integer n except −1, then add + C, because every vertical shift of a curve has the same derivative. Rewrite brackets and fractions as sums of powers first, since there is no product rule for integrals, and check each answer by differentiating it back. A boundary condition, one known point, fixes C: integrate, substitute the point into the integrated function, solve. The area under a curve above the x-axis is the limit of sums of thin rectangles, the definite integral ∫_{a}^{b} y dx, and it equals the change in any anti-derivative, F(b) − F(a), which is why the same symbol does both jobs. On Paper 2, find the limits (often the x-intercepts), write the integral with its limits and dx, then let the GDC evaluate it; for a velocity-time graph that area is the distance travelled.
Answers
Q1. (a) 2x⁴ − 2x³ + 3x + C. A1 for two terms correct, A1 for all three terms and + C.
(b) (x² − 4) ÷ x² = 1 − 4x⁻², so the integral is x − 4x⁻¹ ÷ (−1) + C = x + 4/x + C. M1 for splitting into 1 − 4x⁻², A1 for x, A1 for + 4x⁻¹ with + C.
Q2. f(x) = ∫ (3x² − 4x⁻²) dx = x³ − 4x⁻¹ ÷ (−1) + C = x³ + 4/x + C. Then f(1) = 1 + 4 + C = 6, so C = 1, and f(x) = x³ + 4/x + 1. M1 for attempting to integrate, A1 for x³, A1 for + 4/x, M1 for substituting (1, 6) into their f with + C, A1 for the final function.
Q3. (a) y = ∫ (2x − 6) dx = x² − 6x + C. At (1, 0): 0 = 1 − 6 + C, so C = 5, and y = x² − 6x + 5. A1 for x² − 6x, M1 for substituting (1, 0) with + C, A1 for the equation.
(b) 2x − 6 = 0 gives x = 3, and y = 9 − 18 + 5 = −4. The point is (3, −4). A1 for x = 3, A1 for (3, −4). Follow-through from their (a).
Q4. (a) 3 + 2x − x² = 0 ⇒ x² − 2x − 3 = 0 ⇒ (x − 3)(x + 1) = 0, so x = −1 and x = 3. M1 for setting f(x) = 0, A1 for both values.
(b) Area of R = ∫_{−1}^{3} (3 + 2x − x²) dx. A1 for the correct limits, A1 for the integrand with dx. Limits the wrong way round score A0.
(c) 10.7 (3 s.f.; exactly 32/3). A1.
Q5. (a) distance = ∫_{0}^{5} (3 + 2t − 0.1t³) dt = 24.4 m (24.375). M1 for an integral of v, A1 for the limits 0 and 5, A1 for 24.4.
(b) s = ∫ (3 + 2t − 0.1t³) dt = 3t + t² − 0.025t⁴ + C. When t = 0, s = 4, so C = 4, and s = 3t + t² − 0.025t⁴ + 4. A1 for 3t + t², A1 for −0.025t⁴, A1 for C = 4 in the final expression.
(c) s(5) − s(0) = (15 + 25 − 15.625 + 4) − 4 = 24.375 m, which matches (a). R1 for the change in s equal to the distance, valid because v > 0 throughout.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 5.5 Anti-differentiation and area. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.