Integration Overview – A Level Maths Year 1

Integration is differentiation run backwards. If you know the gradient function of a curve, integration gets you back to the curve. And — remarkably — the same process measures the area under a curve. Those two facts, joined by the fundamental theorem of calculus, are the whole of this topic in Year 1.

Differentiation asks: "given $y$, what is $\dfrac{dy}{dx}$?" Integration asks the reverse question: "given $\dfrac{dy}{dx}$, what was $y$?" Because adding a constant does not change a gradient, the answer is only ever known up to a constant — the famous $+\,c$.

What is new compared with GCSE

Reversing differentiation · the constant of integration
Definite integrals with limits · exact areas under curves
Areas below the axis · areas between a curve and a line

DfE content area H · Integration. At AS you integrate $x^n$ (for any rational $n \neq -1$) and sums of such terms, evaluate definite integrals, and use them to find areas. Year 2 extends this to $e^x$, $\dfrac1x$, trigonometric functions, substitution and integration by parts. Calculators can evaluate definite integrals — but when a question says "show detailed reasoning" or asks for an exact answer, you must write the integration and the substitution of limits in full.

Word	Meaning	Example
Integrand	The function being integrated.	$3x^2$ in $\displaystyle\int 3x^2\,dx$
Indefinite integral	A family of functions whose derivative is the integrand.	$\displaystyle\int 3x^2\,dx = x^3 + c$
Constant of integration	The unknown constant $c$, lost when differentiating.	$x^3 + 5$ and $x^3 - 2$ both differentiate to $3x^2$
Definite integral	An integral with limits; its value is a number.	$\displaystyle\int_1^2 3x^2\,dx = 7$
Limits	The lower and upper values of $x$.	$1$ and $2$ above
$dx$	Tells you the variable you are integrating with respect to.	$\displaystyle\int t^2\,dt = \tfrac13 t^3 + c$
Square brackets	Hold the integrated expression before limits are substituted.	$\left[x^3\right]_1^2 = 8 - 1$

Indefinite means $+\,c$. An indefinite integral without a constant is marked wrong. A definite integral never needs one — the constant cancels when you subtract.

1 — Integrating a power of $x$

$\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + c, \qquad n \neq -1$

2 — Rewrite first

Expand brackets, split fractions and turn roots into indices before integrating

3 — Finding a curve from its gradient

Integrate $\mathrm{f}'(x)$, then substitute a known point to find $c$

4 — Evaluating a definite integral

$\displaystyle\int_a^b \mathrm{f}(x)\,dx = \Big[\mathrm{F}(x)\Big]_a^b = \mathrm{F}(b) - \mathrm{F}(a)$

5 — Areas

Area under a curve above the axis $= \displaystyle\int_a^b y\,dx$; below the axis the integral is negative
Area between curves $= \displaystyle\int_a^b (\text{top} - \text{bottom})\,dx$

Worked Example — Several techniques in one question

A curve has gradient $\dfrac{dy}{dx} = 3x^2 - 6x$ and passes through $(0, 4)$. Find its equation, and find the area of the region between the curve and the $x$-axis for $0 \leqslant x \leqslant 2$.

①Integrate: $y = x^3 - 3x^2 + c$.

②At $(0, 4)$: $4 = 0 - 0 + c$, so $c = 4$ and $y = x^3 - 3x^2 + 4$.

③Check the sign for $0 \leqslant x \leqslant 2$: $y = (x+1)(x-2)^2 \geqslant 0$ there, so the curve stays above the axis.

④$\displaystyle\int_0^2 \left(x^3 - 3x^2 + 4\right)dx = \left[\tfrac14x^4 - x^3 + 4x\right]_0^2 = (4 - 8 + 8) - 0 = 4$

⑤The area is $4$ square units.

Step ③ is the one people skip. Before using an integral as an area, always ask whether the curve dips below the axis anywhere in the interval.

Everything in DfE content area H that is examined at AS, and the page that covers it.

HIndefinite IntegrationIntegration as the reverse of differentiation, integrating $x^n$ and sums, the constant of integration, and finding a curve through a point.
HDefinite Integrals and AreasThe fundamental theorem of calculus, evaluating definite integrals, areas under curves, regions below the axis and areas between a curve and a line.

Integration leans on differentiating powers of $x$ and on indices and surds. If rewriting $\dfrac{3}{\sqrt{x}}$ as $3x^{-1/2}$ is not automatic yet, revise those first.

Forgetting $+\,c$. Every indefinite integral needs it, and "find the equation of the curve" questions depend on it.

Integrating a product or quotient term by term. $\displaystyle\int (x+1)(x+2)\,dx$ is not $\left(\tfrac12x^2 + x\right)\left(\tfrac12x^2 + 2x\right)$. Expand first, then integrate.

Dividing by the old power. $\displaystyle\int x^4\,dx = \tfrac15x^5$, not $\tfrac14x^5$. Raise the power first, then divide by the new power.

Negative powers. $\displaystyle\int x^{-3}\,dx = \dfrac{x^{-2}}{-2} = -\dfrac{1}{2x^2}$. The power goes from $-3$ to $-2$ (up, not down), and the divisor is negative.

Treating a negative integral as an area. If part of the region is below the $x$-axis, integrate each part separately and add the sizes. A single integral across a root can even give zero.

Bracket errors with limits. $\mathrm{F}(b) - \mathrm{F}(a)$ subtracts the whole of $\mathrm{F}(a)$. Put it in brackets and take care with negative lower limits.

Power rule

$\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + c$, $n \neq -1$.

Constant multiple

$\int kx^n\,dx = \dfrac{k\,x^{n+1}}{n+1} + c$.

Constant term

$\int k\,dx = kx + c$.

Sums

Integrate term by term.

Finding $c$

Substitute a known point after integrating.

Definite integral

$[\mathrm{F}(x)]_a^b = \mathrm{F}(b) - \mathrm{F}(a)$; no $c$.

Below the axis

Integral is negative; area is its modulus.

Between curves

$\int (\text{top} - \text{bottom})\,dx$ between intersections.

Check

Differentiate your answer: you should get the integrand back.

Ten questions drawing on the whole Integration topic. Work each one through before opening the solution.

Question 1

Find $\displaystyle\int \left(8x^3 + 2\right)dx$.

▶ Show solution

$\dfrac{8x^4}{4} + 2x + c = 2x^4 + 2x + c$

Question 2

Find $\displaystyle\int 5x^{3/2}\,dx$.

▶ Show solution

Add one to the power: $\tfrac32 + 1 = \tfrac52$.

$\dfrac{5x^{5/2}}{5/2} + c = 2x^{5/2} + c$

Question 3

Find $\displaystyle\int (x+1)(x+3)\,dx$.

▶ Show solution

Expand first: $(x+1)(x+3) = x^2 + 4x + 3$.

$\displaystyle\int \left(x^2 + 4x + 3\right)dx = \tfrac13x^3 + 2x^2 + 3x + c$

Question 4

A curve has $\dfrac{dy}{dx} = 6x^2 - 2$ and passes through $(1, 3)$. Find its equation.

▶ Show solution

$y = 2x^3 - 2x + c$

At $(1, 3)$: $3 = 2 - 2 + c$, so $c = 3$.

$y = 2x^3 - 2x + 3$

Question 5

Evaluate $\displaystyle\int_{-1}^{2} \left(3x^2 - 2x + 1\right)dx$.

▶ Show solution

$\left[x^3 - x^2 + x\right]_{-1}^{2}$

$= (8 - 4 + 2) - (-1 - 1 - 1) = 6 - (-3) = 9$

Question 6

Evaluate $\displaystyle\int_1^4 3\sqrt{x}\,dx$.

▶ Show solution

$3\sqrt{x} = 3x^{1/2}$, which integrates to $\dfrac{3x^{3/2}}{3/2} = 2x^{3/2}$.

$\left[2x^{3/2}\right]_1^4 = 2(8) - 2(1) = 14$

Question 7

Find the area of the region enclosed by the curve $y = 4 - x^2$ and the $x$-axis.

▶ Show solution

The curve meets the axis where $4 - x^2 = 0$, i.e. $x = \pm 2$, and lies above it in between.

$\displaystyle\int_{-2}^{2}\left(4 - x^2\right)dx = \left[4x - \tfrac13x^3\right]_{-2}^{2}$

$= \left(8 - \tfrac83\right) - \left(-8 + \tfrac83\right) = 16 - \tfrac{16}{3} = \dfrac{32}{3}$

Question 8

Find the area of the region between the curve $y = x^2 - 3x$ and the $x$-axis.

▶ Show solution

$x^2 - 3x = x(x - 3)$, which is zero at $x = 0$ and $x = 3$ and negative between them.

$\displaystyle\int_0^3 \left(x^2 - 3x\right)dx = \left[\tfrac13x^3 - \tfrac32x^2\right]_0^3 = 9 - \tfrac{27}{2} = -\tfrac92$

The negative sign shows the region is below the axis. Area $= \dfrac92$.

Question 9

Given that $\displaystyle\int_1^k 6x^2\,dx = 52$, find the value of $k$.

▶ Show solution

$\left[2x^3\right]_1^k = 2k^3 - 2 = 52$

$2k^3 = 54$, so $k^3 = 27$ and $k = 3$.

Question 10

A curve $y = \mathrm{f}(x)$ has $\mathrm{f}'(x) = 3x^2 - 6x$ and passes through the point $(0, 4)$.

(a) Find $\mathrm{f}(x)$. (b) Show that $\mathrm{f}(x) = (x+1)(x-2)^2$. (c) Sketch the curve, and find the exact area of the region bounded by the curve and the $x$-axis.

▶ Show solution

(a) $\mathrm{f}(x) = x^3 - 3x^2 + c$, and $\mathrm{f}(0) = 4$ gives $c = 4$. So $\mathrm{f}(x) = x^3 - 3x^2 + 4$.

(b) $(x-2)^2 = x^2 - 4x + 4$, and $(x+1)\left(x^2 - 4x + 4\right) = x^3 - 4x^2 + 4x + x^2 - 4x + 4 = x^3 - 3x^2 + 4$ ✓

(c) A positive cubic crossing the axis at $x = -1$ and touching it at $x = 2$ (the squared factor). Between these it lies above the axis, with a maximum at $(0, 4)$.

The bounded region is $-1 \leqslant x \leqslant 2$:

$\displaystyle\int_{-1}^{2}\left(x^3 - 3x^2 + 4\right)dx = \left[\tfrac14x^4 - x^3 + 4x\right]_{-1}^{2}$

$= (4 - 8 + 8) - \left(\tfrac14 + 1 - 4\right) = 4 - \left(-\tfrac{11}{4}\right) = \dfrac{27}{4}$

The area is $\dfrac{27}{4}$ square units.