Step-by-step worked examples and graded practice questions on integration — reversing the power rule, integrating negative and fractional powers, finding the constant of integration, evaluating definite integrals, and finding the area under a curve, including regions that dip below the \(x\)-axis. Written to the Edexcel Pure Year 1 specification and equally suitable for AQA and OCR A.
📚 Pure Year 1 (AS)✅ 15 Practice Questions🔍 6 Worked Examples⚠️ Common Mistakes
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Integration reverses differentiation: given a gradient function \(f'(x)\), integration finds the original function \(f(x)\). This topic completes Pure Year 1 by closing the loop between the two halves of calculus — differentiation finds a gradient from a curve, integration finds a curve from a gradient, and (via the definite integral) finds the exact area under that curve.
Five skills make up this topic:
Reversing the power rule — \(\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+c\) for \(n \neq -1\), applied term by term
Negative and fractional powers — rewriting roots and fractions as indices before integrating
The constant of integration — finding \(c\) using a given point on the curve
Definite integrals — evaluating \(\displaystyle\int_a^b f(x)\,dx\) between two limits
Area under a curve — using definite integration to find an exact area, including regions that cross the \(x\)-axis
Reversing the power rule
Since differentiating \(x^n\) gives \(nx^{n-1}\), integrating reverses this: increase the power by 1, then divide by the new power. This only works for \(n \neq -1\) (dividing by zero is undefined). Every indefinite integral needs a constant of integration, \(+c\), since any constant differentiates to zero — infinitely many functions share the same gradient function.
Integrate term by term: \(3x^2 \to \dfrac{3x^3}{3}=x^3\)
2
\(-4x \to \dfrac{-4x^2}{2}=-2x^2\); \(5 \to 5x\)
3
Add the constant of integration
Answer\(x^3-2x^2+5x+c\)
Negative and fractional powers
The power rule applies just as well to negative and fractional indices — but only once a root or fraction has been rewritten in index form: \(\sqrt{x}=x^{1/2}\), \(\dfrac{1}{x^n}=x^{-n}\). Never apply the power rule directly to a term still written as a root or a fraction.
Worked Example 2
Find \(\displaystyle\int \left(2\sqrt{x} + \frac{3}{x^2}\right) dx\), for \(x>0\).
Given a gradient function \(f'(x)\) and one point the curve passes through, integrate first, then substitute the point into the integrated function \(f(x)\) — never into \(f'(x)\) — to find the specific value of \(c\).
Worked Example 3
The curve \(C\) has gradient function \(f'(x)=6x^2-4x+1\) and passes through the point \((1,5)\). Find \(f(x)\).
1
Integrate: \(f(x)=2x^3-2x^2+x+c\)
2
Substitute \((1,5)\): \(2-2+1+c=5\)
3
\(1+c=5 \implies c=4\)
Answer\(f(x)=2x^3-2x^2+x+4\)
Definite integrals
A definite integral is evaluated between two limits: find the integrated function \(F(x)\) (no \(+c\) needed — it cancels between the limits), then compute \(F(b)-F(a)\).
\[\int_a^b f(x)\,dx = F(b)-F(a)\]
Worked Example 4
Evaluate \(\displaystyle\int_1^3 (2x+3)\,dx\).
1
Integrate: \(F(x)=x^2+3x\)
2
\(F(3)=9+9=18\); \(F(1)=1+3=4\)
3
\(F(3)-F(1)=18-4\)
Answer\(14\)
Area under a curve
When a curve stays entirely above the \(x\)-axis over \([a,b]\), the definite integral \(\displaystyle\int_a^b f(x)\,dx\) gives the exact area between the curve and the \(x\)-axis directly.
Worked Example 5
Find the area enclosed between the curve \(y=4x-x^2\) and the \(x\)-axis.
1
Find where the curve meets the \(x\)-axis: \(4x-x^2=0 \implies x(4-x)=0 \implies x=0\) or \(x=4\)
2
Area \(=\displaystyle\int_0^4 (4x-x^2)\,dx = \left[2x^2-\dfrac{x^3}{3}\right]_0^4\)
The shaded region between \(y=4x-x^2\) and the \(x\)-axis, bounded by the curve's two roots at \((0,0)\) and \((4,0)\) — area \(=\int_0^4(4x-x^2)\,dx=\dfrac{32}{3}\). Not to scale.
When a curve crosses the \(x\)-axis within the interval, a single definite integral no longer gives the true area — the parts below the axis contribute negatively, and can even cancel out parts above it. Split the integral at every root, and take the absolute value of any negative result before adding the parts together.
Worked Example 6
Find the total area of the region bounded by the curve \(y=x^2-4x\), the \(x\)-axis, and the line \(x=6\).
1
Roots: \(x^2-4x=0 \implies x=0\) or \(x=4\) — the curve is below the axis on \((0,4)\) and above it on \((4,6)\)
Total area \(=\left|-\dfrac{32}{3}\right|+\dfrac{32}{3}=\dfrac{64}{3}\) — note that integrating straight through from \(0\) to \(6\) gives \(0\), the wrong answer
Answer\(\dfrac{64}{3}\) square units
Exam tips
💡 Tip 1
Never forget "+c" on an indefinite integral
Every indefinite integral is only correct up to a constant — an otherwise perfect answer missing \(+c\) still loses a mark.
💡 Tip 2
Rewrite roots and fractions before integrating
\(\sqrt{x}=x^{1/2}\) and \(\dfrac{1}{x^n}=x^{-n}\) — do this conversion first; the power rule cannot be applied to a term still in root or fraction form.
💡 Tip 3
Substitute into \(f(x)\), never \(f'(x)\), to find \(c\)
The given point lies on the original curve — integrate first, then substitute the point into the result.
💡 Tip 4
Drop "+c" once limits are given
In a definite integral the constant cancels automatically between the two limits — including it just adds unnecessary working.
💡 Tip 5
Always check for roots inside the interval before finding "the area"
If the curve crosses the \(x\)-axis anywhere in \([a,b]\), a single definite integral gives a signed (and possibly misleading) result — split at every root and take absolute values first.
Common mistakes
Common Mistake 1
Omitting the constant of integration
Writing \(\int 2x\,dx = x^2\) instead of \(x^2+c\) — every indefinite integral needs it, since any constant term differentiates to zero.
Common Mistake 2
Increasing the power but forgetting to divide by it
Writing \(\int 3x^2\,dx = 3x^3\) instead of \(x^3\) — the new power in the denominator is not optional.
Common Mistake 3
Integrating a root or fraction without rewriting it as an index first
The power rule only applies once every term is in the form \(x^n\) — a term still written as \(\sqrt{x}\) or \(\dfrac{1}{x^2}\) cannot be integrated directly.
Common Mistake 4
Integrating straight through a root without splitting
Taking a single definite integral across a region that dips below the \(x\)-axis can give a small, or even zero, signed result — the true area needs the interval split at each root, with the negative part's absolute value added, not its signed value.
Common Mistake 5
Substituting the given point into the gradient function
Substituting into \(f'(x)\) instead of the integrated \(f(x)\) when finding \(c\) — the given point lies on the curve itself, not on its gradient function.
Practice questions
Work through each question before checking the worked solution.
Q15The curve \(y=f(x)\) is such that \(f'(x)=3x^2-12x+9\). The curve has a stationary point at \(x=1\) and passes through \((0,4)\). Find \(f(x)\) and state the coordinates of the other stationary point.A* Challenge
\(f'(x)=3x^2-12x+9=3(x-1)(x-3)=0 \implies x=1\) or \(x=3\)
\(f(x)=x^3-6x^2+9x+4\); the other stationary point is at \(x=3\), where \(f(3)=4\), so \((3,4)\)
Summary
\(\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+c\) for \(n \neq -1\); integrate a sum term by term.
Rewrite roots and fractions as fractional/negative indices before integrating.
Find the constant of integration by substituting a given point into \(f(x)\), not \(f'(x)\).
Definite integrals \(\displaystyle\int_a^b f(x)\,dx = F(b)-F(a)\) need no \(+c\) — it cancels.
A definite integral only equals the true area directly when the curve stays on one side of the \(x\)-axis; split at every root and take absolute values where it crosses.
This topic completes Pure Year 1, reversing the power rule from Differentiation and reusing the root-finding and curve-sketching skills from Graphs & Transformations and the index laws from GCSE Indices. Next in the Pure Maths calculus strand: Further Integration & Differential Equations (Year 2), which extends these methods to trigonometric, exponential and logarithmic functions, integration by substitution, and solving differential equations.
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