Integrating standard functionsEdexcel International A Level Maths: Revision notes

Section 1

Exponentials

Integration reverses differentiation. For a constant $k\neq0$ : $\int e^{kx}\,dx=\frac1ke^{kx}+c,\qquad \int a^x\,dx=\frac{a^x}{\ln a}+c.$ Example: $\int6e^{2x}\,dx=3e^{2x}+c$ . For $\int e^{-0.5t}\,dt$ you get $-2e^{-0.5t}+c$ , so $\int3e^{-0.5t}dt=-6e^{-0.5t}+c$ . Reverse of $\frac{d}{dx}a^x=a^x\ln a$ : divide by $\ln a$ , so $\int2^x\,dx=\frac{2^x}{\ln2}+c$ .

Key termsintegrationconstant of integration

Common mistake

Forgetting to divide by $k$ : $\int e^{2x}dx$ is $\frac12e^{2x}$ , not $e^{2x}$ or $2e^{2x}$ .

Section 2

The reciprocal function

$\int\frac1x\,dx=\ln|x|+c,\qquad x\neq0.$ The modulus is needed because $\ln$ is only defined for positive numbers. If $x>0$ is given, $\ln x$ is fine. A constant factor comes outside: $\int\frac{3}{x}dx=3\ln|x|+c$ , and $\int\frac{1}{2x}dx=\frac12\ln|x|+c$ . Likewise $\int\frac{5}{2x}dx=\frac52\ln|x|+c$ . Do not use the power rule on $x^{-1}$ ; it would need division by zero.

Key termsnatural logarithm

Common mistake

Writing $\int\frac{1}{2x}dx=\ln|2x|$ with no factor $\frac12$ . Write it as $\frac12\int\frac1x\,dx$ first.

Section 3

Sine and cosine

$\int\cos kx\,dx=\frac1k\sin kx+c,\qquad \int\sin kx\,dx=-\frac1k\cos kx+c.$ The minus sign for $\sin$ comes from $\frac{d}{dx}\cos kx=-k\sin kx$ . The angle must be in radians. Example: $\int4\sin3x\,dx=-\frac43\cos3x+c$ and $\int2\cos\frac x2\,dx=4\sin\frac x2+c$ , because $\frac1k=2$ when $k=\frac12$ .

Key termsradians

Exam tip

Check by differentiating your answer: you should get the original function back.

Section 4

Sums, differences and definite integrals

Integrate term by term, taking constant multiples outside. For a definite integral, integrate, substitute the upper limit, then subtract the value at the lower limit. No constant is needed. Example: $\int_0^{\pi/3}\left(4\sin3x+2\cos\frac x2\right)dx=\left[-\frac43\cos3x+4\sin\frac x2\right]_0^{\pi/3}=\left(\frac43+2\right)-\left(-\frac43\right)=\frac{14}{3}$ . Exact values: $\cos\pi=-1$ , $\sin\frac\pi6=\frac12$ , $e^{\ln3}=3$ , $e^{-\ln3}=\frac13$ .

Key termsdefinite integral

Common mistake

Subtracting in the wrong order, or forgetting to subtract the lower-limit value.

Section 5

Finding constants and applications

Given a point on the curve, substitute it into the indefinite integral to find $c$ . Example: $\frac{dy}{dx}=6e^{2x}-\frac3x$ through $(1,3e^2+5)$ gives $y=3e^{2x}-3\ln x+5$ . Velocity to displacement: $s=\int v\,dt$ . For $v=3e^{-0.5t}+2$ , $s=-6e^{-0.5t}+2t+c$ , and $s=5$ at $t=0$ gives $c=11$ . A definite integral of a gradient gives the change in $y$ : $\int_a^b\frac{dy}{dx}\,dx=y(b)-y(a)$ .

Key termsdisplacement

Exam tip

Find the constant of integration straight away using the given condition.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Integrating standard functions

The function $f$ is defined by $f(x)=6e^{2x}-\frac{3}{x}$ for $x>0$ .
The curve $y=F(x)$ has gradient $f(x)$ and passes through the point $(1,3e^2+5)$ . Find $F(x)$ .2 marks
The function $g$ is defined by $g(x)=4\sin 3x+2\cos\frac{x}{2}$ , where $x$ is in radians.
Given that $\frac{dF}{dx}=g(x)$ and $F(0)=0$ , find the exact value of $F(\pi)$ .2 marks
A particle moves along a straight line. Its velocity is $v=3e^{-0.5t}+2$ m s $^{-1}$ at time $t$ seconds, for $t\geq0$ , and its position relative to a fixed point $O$ is $s$ metres.
Find the displacement of the particle between $t=0$ and $t=4$ , giving your answer to 3 significant figures.3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).