P1.7 Integration
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Pure Mathematics 1 (AS), specification sections 5.1 and 5.2: indefinite integration of powers of \(x\). Definite integrals and areas follow in P2.
What this page covers
7 points
- Integration as the reverse process of differentiation
- Why every indefinite integral needs a constant of integration, and what the family of curves looks like
- Integrating \(x^n\) for any rational \(n \ne -1\), including a constant
- Constant multiples, sums and differences, and why products must be expanded first
- Rewriting roots and fractions as powers of \(x\) before integrating
- Finding the equation \(y = f(x)\) of a curve from \(f'(x)\) and one point on the curve
- Longer questions that combine integration with tangents and points of zero gradient
In one minute
6 points
- Integration undoes differentiation: if \(\frac{dy}{dx} = f'(x)\), then \(y = \int f'(x)\,dx\), and the answer always carries a constant of integration \(+\,c\).
- Power rule: raise the power by one, then divide by the new power. \(\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + c\) for \(n \ne -1\).
- Constants vanish when you differentiate, so an indefinite integral is a whole family of curves, one for each value of \(c\).
- Powers first: expand brackets and split fractions; write \(\sqrt{x} = x^{\frac12}\) and \(\frac{1}{x^2} = x^{-2}\). Never integrate a product or a quotient piece by piece.
- A point fixes \(c\): integrate, substitute the coordinates of a known point, solve for \(c\), and write the equation in full.
- Definite integrals and areas are in P2; \(\int x^{-1}\,dx\), exponentials and trigonometric functions are in P3.
Contents
12 sections
- What does integration undo?
- How do you integrate a power of \(x\)?
- How do you integrate sums, multiples and brackets?
- How do you integrate roots and fractions?
- How do you find a curve from its gradient function?
- How do longer integration questions fit together?
- Exam tips for P1.7
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What does integration undo?
In P1.6 you learned to start with a curve and find its gradient function. Integration runs the same machine backwards: you are given the gradient function and asked which curve it came from. That is useful whenever a rate of change is easier to know than the quantity itself, which is why integration appears later in kinematics, areas, volumes and differential equations.
Why the \(+\,c\)? Differentiate \(x^3 + 7\) and you get \(3x^2\). Differentiate \(x^3 - 2\) and you also get \(3x^2\). Any constant disappears when you differentiate, because a constant has zero gradient. So when you reverse the process, there is no way to know which constant was there. You record that uncertainty by writing \(+\,c\), the constant of integration.
Geometrically, adding a constant moves a graph up or down without changing its shape. Every member of the family has exactly the same gradient at each value of \(x\), so the gradient function cannot tell them apart. The figure shows four members of the family \(\int (2x - 2)\,dx = x^2 - 2x + c\).
Leaving out \(+\,c\). Without it you have found only one curve of the family, and the next part of the question, which usually asks for the curve through a given point, becomes impossible.
How do you integrate a power of \(x\)?
Start from what you know about differentiation. Differentiating \(x^{n+1}\) gives \((n+1)x^n\). That is almost \(x^n\); it is just \(n+1\) times too big. So differentiate \(\frac{x^{n+1}}{n+1}\) instead and you get exactly \(x^n\). Reversing this gives the rule.
- \(n\) can be any rational number (positive, negative or a fraction) except \(-1\); \(c\) is the constant of integration.
- For a constant: \(\int a\,dx = ax + c\). This is the case \(n = 0\), since \(a = ax^0\).
- In the formulae list? No — learn it: you will use it in almost every question.
In words: add one to the power, then divide by the new power. The order matters, because the division uses the new power, not the old one. For \(n = -1\) the new power would be \(0\) and you would divide by zero, so the rule breaks down; \(\int x^{-1}\,dx\) needs the natural logarithm, which is in P3.
| Integrand | As a power | New power | Integral |
|---|---|---|---|
| \(x^3\) | \(x^3\) | \(4\) | \(\frac14 x^4 + c\) |
| \(\sqrt{x}\) | \(x^{\frac12}\) | \(\frac32\) | \(x^{\frac32} \div \frac32 = \frac23 x^{\frac32} + c\) |
| \(\frac{1}{x^2}\) | \(x^{-2}\) | \(-1\) | \(x^{-1} \div (-1) = -\frac{1}{x} + c\) |
| \(\frac{1}{\sqrt{x}}\) | \(x^{-\frac12}\) | \(\frac12\) | \(x^{\frac12} \div \frac12 = 2\sqrt{x} + c\) |
| \(7\) | \(7x^0\) | \(1\) | \(7x + c\) |
Notice how negative powers behave. For \(x^{-2}\), "add one" means \(-2 + 1 = -1\), so the power moves towards zero, and the division by \(-1\) changes the sign. Students often write \(x^{-3}\) here because they subtract out of habit from differentiation. A quick derivative check catches it.
Find \(\int (8x^3 - 6x^2 + 5)\,dx\).
Integrate term by term, raising each power by one and dividing by the new power:
\(8 \times \dfrac{x^4}{4} - 6 \times \dfrac{x^3}{3} + 5x\)
Simplify and add the constant: \(2x^4 - 2x^3 + 5x + c\)
Check by differentiating: \(8x^3 - 6x^2 + 5\). ✓
It helps to see a gradient function and one of its integrals side by side. The first graph below is \(f'(x) = 3x^2 - 6x\). Integrating gives \(f(x) = x^3 - 3x^2 + c\); the second graph shows the member with \(c = 4\). Read them together: where \(f'(x)\) is positive the curve rises, where \(f'(x)\) is negative it falls, and where \(f'(x) = 0\) the curve has a stationary point.
Differentiate your answer in your head before moving on. It takes seconds and catches the two classic slips: multiplying by the new power instead of dividing, and lowering the power instead of raising it.
How do you integrate sums, multiples and brackets?
Differentiation is linear: a constant multiple stays in front, and a sum is differentiated term by term. Integration inherits exactly the same two properties, because it is the reverse process.
- Constant multiples come outside: \(\int k\,f(x)\,dx = k\int f(x)\,dx\).
- Sums and differences are integrated term by term: \(\int \big(f(x) \pm g(x)\big)\,dx = \int f(x)\,dx \pm \int g(x)\,dx\). One \(+\,c\) at the end covers all the terms, because several unknown constants added together are still just one unknown constant.
- Products are different. There is no rule that \(\int fg\,dx = \int f\,dx \times \int g\,dx\). Multiply out the brackets first, so that you have a sum of powers.
Find \(\int (x - 3)(2x + 1)\,dx\), giving each coefficient as an exact fraction or integer.
Step 1. Multiply out the brackets: \((x-3)(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3\).
Step 2. Integrate term by term: \(2 \times \dfrac{x^3}{3} - 5 \times \dfrac{x^2}{2} - 3x + c\).
Step 3. Simplify: \(\frac23 x^3 - \frac52 x^2 - 3x + c\)
Integrating each bracket separately and multiplying, \(\left(\frac12 x^2 - 3x\right)(x^2 + x)\), gives a quartic, and differentiating it does not give back the integrand.
The graph below shows the integrand from Worked example 2 and the integral with \(c = 0\). The integrand is the gradient of the integral, so the integral has its turning points exactly where the integrand crosses the \(x\)-axis, at \(x = -\frac12\) and \(x = 3\).
A constant on its own is a common source of slips. When you differentiate, a constant disappears; when you integrate, it does the opposite and gains an \(x\). The gradient function \(y = 5\) is a horizontal line, and the curves with that gradient everywhere are the straight lines \(y = 5x + c\).
Treating a constant as if it vanishes when integrating, as it does when differentiating. \(\int 5\,dx = 5x + c\), not \(0\) and not \(5\).
How do you integrate roots and fractions?
The power rule only accepts terms of the form \(ax^n\). So every root and every fraction must be rewritten using the laws of indices from P1.1 before any integrating happens. This rewriting line is where most of the real work in this topic lies.
- \(\sqrt{x} = x^{\frac12}\), \(\sqrt[3]{x} = x^{\frac13}\), \(x\sqrt{x} = x^{\frac32}\).
- \(\dfrac{1}{x^n} = x^{-n}\), \(\dfrac{1}{\sqrt{x}} = x^{-\frac12}\), \(\dfrac{4}{x^3} = 4x^{-3}\) (the 4 stays in front: it is not part of the denominator's power).
- A fraction with a sum on top and a single term underneath splits: \(\dfrac{x^2 - 5}{2\sqrt{x}} = \dfrac{x^2}{2x^{\frac12}} - \dfrac{5}{2x^{\frac12}}\).
- Dividing by a fractional power is multiplying by its reciprocal: \(x^{\frac32} \div \frac32 = \frac23 x^{\frac32}\). Dividing by \(\frac12\) doubles.
A fraction with a single term on top and a sum underneath, such as \(\frac{1}{x+1}\), cannot be split this way. Such integrals are not part of P1.
Find \(\int \left(6\sqrt{x} + \dfrac{4}{x^3}\right)dx\), \(x > 0\), simplifying your answer.
Step 1. Write each term as a power of \(x\): \(6x^{\frac12} + 4x^{-3}\).
Step 2. Raise each power by one and divide by the new power: \(6 \times x^{\frac32} \div \frac32 + 4 \times x^{-2} \div (-2)\).
Step 3. Simplify (\(6 \div \frac32 = 4\) and \(4 \div (-2) = -2\)): \(4x^{\frac32} - 2x^{-2} + c\), which can also be written \(4x\sqrt{x} - \dfrac{2}{x^2} + c\).
(a) Write \(\dfrac{x^2 - 5}{2\sqrt{x}}\) in the form \(Ax^p + Bx^q\), where \(A\), \(B\), \(p\) and \(q\) are constants. (b) Hence find \(\int \dfrac{x^2 - 5}{2\sqrt{x}}\,dx\).
(a) Split the fraction and use the index laws, \(x^2 \div x^{\frac12} = x^{\frac32}\):
\(\dfrac{x^2}{2x^{\frac12}} - \dfrac{5}{2x^{\frac12}} = \) \(\frac12 x^{\frac32} - \frac52 x^{-\frac12}\), so \(A = \frac12\), \(p = \frac32\), \(B = -\frac52\), \(q = -\frac12\).
(b) Integrate each term: \(\frac12 \times x^{\frac52} \div \frac52 - \frac52 \times x^{\frac12} \div \frac12\).
Simplify: \(\frac15 x^{\frac52} - 5x^{\frac12} + c\)
"Hence" means use part (a): you are expected to integrate the rewritten form, not the original fraction.
Integrating the top and the bottom of a fraction separately. \(\dfrac{x^2 - 5}{2\sqrt{x}}\) does not integrate to \(\left(\frac{x^3}{3} - 5x\right) \div \left(\frac43 x^{\frac32}\right)\). Split the fraction into separate powers of \(x\) first.
How do you find a curve from its gradient function?
Integrating \(f'(x)\) gives the whole family \(f(x) + c\). To pick out one particular curve you need one more piece of information, and in P1 that is almost always a point the curve passes through. Substituting its coordinates turns the equation into a single equation in \(c\).
- Integrate \(f'(x)\), including \(+\,c\).
- Substitute the coordinates \((x_1, y_1)\) of the given point and solve for \(c\).
- Write the final answer in full, \(y = \dots\) or \(f(x) = \dots\), with the value of \(c\) in place.
A curve has equation \(y = f(x)\), where \(f'(x) = 6x^2 - 2x - 3\). The curve passes through the point \((2, 7)\). Find \(f(x)\).
Step 1. Integrate: \(f(x) = 2x^3 - x^2 - 3x + c\).
Step 2. Use the point: when \(x = 2\), \(y = 7\), so \(7 = 2(8) - 4 - 6 + c = 6 + c\), giving \(c = 1\).
Step 3. State the function: \(f(x) = 2x^3 - x^2 - 3x + 1\)
Look at how the point does its job. All five curves have the same shape and the same turning points' \(x\)-coordinates; they differ only in height. The point \((2, 7)\) fixes the height, and with it the curve.
The constant is not always the \(y\)-intercept. It equals \(f(0)\) only when every other term is zero at \(x = 0\); with terms such as \(\frac{12}{x}\) the curve may not even meet the \(y\)-axis. Always find \(c\) by substituting the point you are given.
How do longer integration questions fit together?
Exam-length questions join this topic to differentiation. A typical structure is: find the curve by integrating, then use the gradient function again for a tangent, a normal or a point where the gradient is zero. Two habits keep these questions clean.
- Keep \(f'(x)\). It is given in the question, so you do not need to differentiate your answer to find a gradient. Use the given expression directly.
- Keep exact values (fractions, surds) until the last line. The form usually asked for, such as \(ax + by + d = 0\) with integers, needs them.
The curve \(C\) has equation \(y = f(x)\), \(x > 0\), and \(\dfrac{dy}{dx} = 3\sqrt{x} - \dfrac{12}{x^2}\). The point \(P(4, 9)\) lies on \(C\).
(a) Find the equation of \(C\). (b) Find the equation of the tangent to \(C\) at \(P\), giving your answer in the form \(ax + by + d = 0\), where \(a\), \(b\) and \(d\) are integers.
(a) Powers first: \(\dfrac{dy}{dx} = 3x^{\frac12} - 12x^{-2}\).
Integrate: \(y = 3 \times x^{\frac32} \div \frac32 - 12 \times x^{-1} \div (-1) + c = 2x^{\frac32} + \dfrac{12}{x} + c\).
Use \(P(4, 9)\), with \(4^{\frac32} = 8\): \(9 = 16 + 3 + c\), so \(c = -10\). \(y = 2x^{\frac32} + \dfrac{12}{x} - 10\)
(b) Gradient at \(P\) from the given \(\frac{dy}{dx}\): \(3\sqrt{4} - \dfrac{12}{16} = 6 - \dfrac34 = \dfrac{21}{4}\).
Tangent through \((4, 9)\): \(y - 9 = \frac{21}{4}(x - 4)\), so \(4y - 36 = 21x - 84\), giving \(21x - 4y - 48 = 0\).
The curve \(y = f(x)\) has \(f'(x) = 3x^2 + kx - 9\), where \(k\) is a constant. The gradient of the curve is zero where \(x = 3\), and the curve passes through the point \((2, -18)\).
(a) Find the value of \(k\). (b) Find \(f(x)\). (c) Find the coordinates of the other point on the curve where the gradient is zero.
(a) Zero gradient at \(x = 3\) means \(f'(3) = 0\): \(27 + 3k - 9 = 0\), so \(k = -6\).
(b) Integrate \(f'(x) = 3x^2 - 6x - 9\): \(f(x) = x^3 - 3x^2 - 9x + c\).
Use \((2, -18)\): \(-18 = 8 - 12 - 18 + c = -22 + c\), so \(c = 4\). \(f(x) = x^3 - 3x^2 - 9x + 4\)
(c) Solve \(f'(x) = 0\): \(3(x^2 - 2x - 3) = 3(x - 3)(x + 1) = 0\), so the other value is \(x = -1\). Then \(f(-1) = -1 - 3 + 9 + 4 = 9\), giving \((-1, 9)\).
These are the curve's stationary points. Deciding whether each is a maximum or a minimum is covered in P2.
Using the given point in \(f'(x)\) instead of \(f(x)\). The coordinates \((2, -18)\) lie on the curve, so they satisfy \(y = f(x)\); \(f'(2)\) is a gradient and has nothing to do with \(-18\).
Exam tips for P1.7
Before you integrate anything, rewrite the whole integrand as a sum of terms \(ax^n\). Write this line down: it shows your method, and most errors in this topic come from skipping it.
When the new power is a fraction, write the division out: \(x^{-\frac12}\) becomes \(x^{\frac12} \div \frac12 = 2x^{\frac12}\). Dividing by \(\frac12\) doubles; dividing by \(\frac32\) multiplies by \(\frac23\).
Every indefinite integral ends with \(+\,c\). If the question gives a point, finding \(c\) is expected, and the final answer is the equation with the number in place, not just "\(c = 1\)".
"Simplify" means combine coefficients (\(6 \div \frac32 = 4\)). "In the form \(Ax^p + Bx^q\)" means state the constants clearly. "\(ax + by + d = 0\) with integers" means clear every fraction.
"Hence" tells you to use the result you have just found, usually a rewritten integrand. "Show that" gives you the target: every step must be visible, and the final line should match it exactly.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The constant of integration is optional. | Infinitely many functions have the same derivative. | Always write \(+\,c\) on an indefinite integral. |
| \(\int x^3\,dx = 3x^2\). | That is differentiation: the power went down. | Raise the power and divide: \(\frac{x^4}{4} + c\). |
| A product can be integrated bracket by bracket. | Differentiating the product of the integrals does not give back the product. | Expand first, then integrate term by term. |
| \(\int \frac{1}{x^2}\,dx = \frac{1}{x^3/3}\). | The denominator has been integrated as if it were on its own. | Rewrite as \(x^{-2}\): the integral is \(-x^{-1} + c\). |
| The power rule works for \(x^{-1}\) too. | The new power would be 0, and dividing by 0 is undefined. | \(\int x^{-1}\,dx\) involves the natural logarithm (P3). |
| The constant \(c\) is the \(y\)-intercept. | Only true when every other term is zero at \(x = 0\). | Find \(c\) by substituting the given point. |
How this topic connects
P1.6 Differentiation and P2 stationary points
Integration reverses differentiation, and longer questions use both: integrate to find the curve, then use the gradient function for tangents, normals and points of zero gradient. Stationary points and their nature follow in P2.
P2 Definite integrals and areas
Putting limits on the integral removes the constant and gives the area between a curve and the \(x\)-axis; the trapezium rule estimates it when the function cannot be integrated exactly.
P3 and P4 further integration; M2 variable acceleration
\(\int x^{-1}\,dx\), exponentials and trigonometric functions come in P3; substitution, integration by parts, volumes of revolution and differential equations in P4. In M2, velocity is the integral of acceleration and displacement the integral of velocity, each with a constant fixed by a known starting value: exactly the "curve through a point" method of this page.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + c\), \(n \ne -1\) | integrating any power of \(x\) except \(x^{-1}\) | No — learn it |
| \(\int a\,dx = ax + c\) | integrating a constant | No — derive it (the case n = 0) |
| \(\int k\,f(x)\,dx = k\int f(x)\,dx\); \(\int (f \pm g)\,dx = \int f\,dx \pm \int g\,dx\) | constant multiples, sums and differences | No — learn it |
| \(\sqrt{x} = x^{\frac12}\), \(\dfrac{1}{x^n} = x^{-n}\), \(\dfrac{1}{\sqrt{x}} = x^{-\frac12}\) | rewriting before integrating | No — learn it (laws of indices) |
| \(y = \int f'(x)\,dx\), then substitute \((x_1, y_1)\) | equation of a curve from its gradient and a point | No — derive it |
Check your understanding
1. Find \(\int (5x^4 + 2)\,dx\).
Raise each power and divide: \(5 \times \frac{x^5}{5} + 2x\). \(x^5 + 2x + c\). Differentiate to check: \(5x^4 + 2\).
2. Find \(\int \dfrac{3}{\sqrt{x}}\,dx\), \(x > 0\).
\(3x^{-\frac12} \to 3 \times x^{\frac12} \div \frac12\). \(6\sqrt{x} + c\).
3. Explain why the rule \(\int x^n\,dx = \frac{x^{n+1}}{n+1} + c\) cannot be used for \(\frac{1}{x}\).
\(\frac1x = x^{-1}\), so \(n + 1 = 0\) and the rule would divide by zero. This integral needs the natural logarithm (P3).
4. A curve has gradient function \(2x + 4\) and passes through \((1, 0)\). Find its equation.
\(y = x^2 + 4x + c\); \(0 = 1 + 4 + c\), so \(c = -5\). \(y = x^2 + 4x - 5\).
5. Find \(\int \dfrac{x^2 + 1}{x^2}\,dx\), \(x \ne 0\).
Split first: \(1 + x^{-2}\). Then integrate: \(x + x^{-1} \div (-1)\). \(x - \dfrac{1}{x} + c\).
6. The curve \(y = f(x)\), \(x > 0\), has \(f'(x) = 4x - \dfrac{3}{x^2}\) and passes through \((1, 2)\). Find \(f(x)\).
\(f'(x) = 4x - 3x^{-2}\), so \(f(x) = 2x^2 - 3x^{-1} \div (-1) + c = 2x^2 + \frac{3}{x} + c\). At \((1, 2)\): \(2 = 2 + 3 + c\), so \(c = -3\). \(f(x) = 2x^2 + \dfrac{3}{x} - 3\).
Specification checklist
- (P1 5.1) I understand integration as the reverse process of differentiation, and I include a constant of integration in every indefinite integral.
- (P1 5.1) I can explain why an indefinite integral describes a family of curves that are vertical translations of one another.
- (P1 5.2) I can integrate \(ax^n\) for any rational \(n \ne -1\), together with sums and differences of such terms.
- (P1 5.2) I can rewrite products, roots and fractions as sums of powers of \(x\) before integrating, and give answers in a requested form.
- (P1 5.2) I can find the equation \(y = f(x)\) of a curve given \(f'(x)\) and a point on the curve, and use the result with tangents, normals and points where the gradient is zero.