Think about which functions differentiate to $2x$. Certainly $x^2$ does — but so do $x^2 + 1$, $x^2 - 7$ and $x^2 + 100$, because the derivative of a constant is zero. Differentiating throws the constant away, so integrating cannot recover it. All we can say is
Geometrically, $\displaystyle\int 2x\,dx$ is a whole family of curves, each a vertical translation of the others. At any given $x$ they all have the same gradient. One extra piece of information — a point on the curve — picks out a single member of the family.
To differentiate $x^n$ you multiply by the power and then reduce it by one. To integrate, do the opposite in the opposite order: increase the power by one, then divide by the new power.
$\displaystyle\int kx^n\,dx = \frac{k\,x^{n+1}}{n+1} + c$ and $\displaystyle\int k\,dx = kx + c$
The rule works for any rational power — positive, negative or fractional — except $n = -1$. Sums and differences are integrated term by term.
| Integrand | New power | Integral |
|---|---|---|
| $x^5$ | $6$ | $\tfrac16x^6 + c$ |
| $12x^3$ | $4$ | $3x^4 + c$ |
| $7$ | $1$ | $7x + c$ |
| $x^{-2}$ | $-1$ | $-x^{-1} + c = -\dfrac1x + c$ |
| $x^{1/2}$ | $\tfrac32$ | $\tfrac23x^{3/2} + c$ |
| $x^{-1/2}$ | $\tfrac12$ | $2x^{1/2} + c$ |
Find $\displaystyle\int \left(6x^2 - 4x + 5\right)dx$.
Find $\displaystyle\int \left(\frac{3}{x^2} + 4\sqrt{x}\right)dx$.
The power rule only applies to a sum of terms $kx^n$. There is no product rule or quotient rule for integration at this level, so brackets must be expanded and fractions split first.
- Expand any brackets.
- Split a fraction with a single-term denominator into separate terms.
- Write every root and reciprocal as a power of $x$.
- Integrate term by term and add $c$.
Find $\displaystyle\int (x+2)(x-3)\,dx$.
Find $\displaystyle\int \frac{2x^2 + 5}{\sqrt{x}}\,dx$.
A curve has gradient $\dfrac{dy}{dx} = 3x^2 - 6x + 1$ and passes through $(2, 5)$. Find its equation.
Given $\mathrm{f}'(x) = 6\sqrt{x} - \dfrac{2}{x^2}$ for $x > 0$ and $\mathrm{f}(4) = 20$, find $\mathrm{f}(x)$.
A curve has $\dfrac{dy}{dx} = kx^2 - 12$, where $k$ is a constant. The curve has a stationary point at $(2, -10)$. Find $k$ and the equation of the curve.
Any rate of change can be integrated to recover the quantity itself. The constant of integration is then fixed by an initial condition — the value at the start.
Water flows into a tank so that, $t$ minutes after a tap is opened, $\dfrac{dV}{dt} = 30 - 2t$ litres per minute, for $0 \leqslant t \leqslant 15$. Initially the tank holds $100$ litres. Find $V$ in terms of $t$ and the volume when the tap closes at $t = 15$.
Power rule
Add one to the power, divide by the new power.
Exception
$n = -1$ is not covered at AS.
Constant
$\int k\,dx = kx + c$.
$+\,c$
Every indefinite integral needs it.
Rewrite first
Expand brackets; split fractions; roots to powers.
Roots
$\sqrt{x} = x^{1/2}$, $\dfrac{1}{\sqrt x} = x^{-1/2}$.
Finding $c$
Integrate first, then substitute the known point.
Stationary point
Gives $\dfrac{dy}{dx} = 0$ and a point on the curve.
Check
Differentiate your answer.
Find $\displaystyle\int 4x^3\,dx$.
▶ Show solution
$\dfrac{4x^4}{4} + c = x^4 + c$
Find $\displaystyle\int \left(5x^4 - 6x^2 + 3\right)dx$.
▶ Show solution
$x^5 - 2x^3 + 3x + c$
Find $\displaystyle\int \frac{6}{x^3}\,dx$.
▶ Show solution
$\dfrac{6}{x^3} = 6x^{-3}$.
$\dfrac{6x^{-2}}{-2} + c = -3x^{-2} + c = -\dfrac{3}{x^2} + c$
Find $\displaystyle\int \left(3\sqrt{x} - \frac{1}{\sqrt{x}}\right)dx$.
▶ Show solution
Rewrite: $3x^{1/2} - x^{-1/2}$.
$3x^{1/2} \to \dfrac{3x^{3/2}}{3/2} = 2x^{3/2}$ and $-x^{-1/2} \to -\dfrac{x^{1/2}}{1/2} = -2x^{1/2}$
$= 2x^{3/2} - 2\sqrt{x} + c$
Find $\displaystyle\int (2x - 1)^2\,dx$.
▶ Show solution
Expand: $(2x-1)^2 = 4x^2 - 4x + 1$.
$\dfrac43x^3 - 2x^2 + x + c$
Find $\displaystyle\int \frac{x^4 + 3x}{x^3}\,dx$.
▶ Show solution
Split: $\dfrac{x^4}{x^3} + \dfrac{3x}{x^3} = x + 3x^{-2}$.
$\tfrac12x^2 + \dfrac{3x^{-1}}{-1} + c = \tfrac12x^2 - \dfrac{3}{x} + c$
A curve has $\dfrac{dy}{dx} = 4x - 3$ and passes through $(1, 6)$. Find its equation.
▶ Show solution
$y = 2x^2 - 3x + c$
At $(1, 6)$: $6 = 2 - 3 + c$, so $c = 7$.
$y = 2x^2 - 3x + 7$
$\mathrm{f}'(x) = 3x^2 + \dfrac{2}{x^2}$ and $\mathrm{f}(1) = 4$. Find $\mathrm{f}(x)$ and hence $\mathrm{f}(2)$.
▶ Show solution
$\mathrm{f}'(x) = 3x^2 + 2x^{-2}$, so $\mathrm{f}(x) = x^3 - 2x^{-1} + c = x^3 - \dfrac2x + c$.
$\mathrm{f}(1) = 1 - 2 + c = 4$, so $c = 5$.
$\mathrm{f}(x) = x^3 - \dfrac2x + 5$, and $\mathrm{f}(2) = 8 - 1 + 5 = 12$.
The gradient of a curve is $kx - 4$, where $k$ is a constant. The curve has a stationary point at $(2, -3)$. Find the equation of the curve.
▶ Show solution
Stationary at $x = 2$: $2k - 4 = 0$, so $k = 2$.
$\dfrac{dy}{dx} = 2x - 4$, so $y = x^2 - 4x + c$.
At $(2, -3)$: $-3 = 4 - 8 + c$, so $c = 1$.
$y = x^2 - 4x + 1$
The height $h$ cm of a plant, $t$ weeks after it is planted, is modelled by $\dfrac{dh}{dt} = 2 + 3\sqrt{t}$ for $t \geqslant 1$. When $t = 1$ the plant is $5$ cm tall.
(a) Find $h$ in terms of $t$. (b) Find the height when $t = 4$. (c) Show that the plant grows $20$ cm between $t = 1$ and $t = 4$. (d) Give one reason why the model is unlikely to be valid for large values of $t$.
▶ Show solution
(a) $\dfrac{dh}{dt} = 2 + 3t^{1/2}$, so $h = 2t + \dfrac{3t^{3/2}}{3/2} + c = 2t + 2t^{3/2} + c$.
When $t = 1$, $h = 5$: $5 = 2 + 2 + c$, so $c = 1$.
$h = 2t + 2t^{3/2} + 1$
(b) $h(4) = 8 + 2(8) + 1 = 25$ cm.
(c) $h(4) - h(1) = 25 - 5 = 20$ cm ✓
(d) The rate of growth $2 + 3\sqrt t$ keeps increasing, so the model predicts the plant grows faster and faster without limit. Real plants reach a maximum height, so the model must fail eventually.