Integrating Basic Functions (Edexcel IGCSE Further Maths)

Integrating Powers of x

Spec point — 9A

How do I integrate powers of x?

• Powers of$x$ are integrated according to the following formula:

  • \int x^{n \mathrm{d} x = \frac{x^{n} + 1} }{n + 1} + c (where$c$ is the constant of integration)

    • This is valid for any value of $n$ except $n=-1$
    • So you cannot integrate $\int x^{- 1} \mathrm{d} x = \int \frac{1}{x} \mathrm{d} x$ this way

• If $x^{n}$ is multiplied by a constant $a$ then

  • \int a x^{n \mathrm{d} x = a \int x^{n} \mathrm{d} x = \frac{a x^{n} + 1} }{n + 1} + c

    • This also is not valid for $n=-1$

• These formulae are not on the exam formula sheet, so you need to remember them

• Remember the special case:

  • $\int a \mathrm{d} x = a x + c$

    • e.g. $\int 4 \mathrm{d} x = 4 x + c$

  • This allows constant terms to be integrated

• Functions involving roots will need to be rewritten as fractional powers

  • e.g. $\int \sqrt[3]{x} \mathrm{d} x$

    • rewrite $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$
    • then integrate

• Fractions with $x$ in the denominator will need to be rewritten as negative powers

  • e.g. $\int \frac{1}{x^{2}} \mathrm{d} x$

    • rewrite $\frac{1}{x^{2}}$ as $x^{-2}$
    • then integrate

How do I integrate sums and differences of powers of x?

• The formulae can be used to integrate sums or differences of powers of$x$

  • Just integrate term by term

    • e.g. $\int ( 8 x^{3} - 2 x + 4 ) \mathrm{d} x$
    • $=\frac{8x^{3+1}}{3+1}-\frac{2x^{1+1}}{1+1}+4x+c$
    • $=2x^{4}-x^{2}+4x+c$

• Products and quotients cannot be integrated this way

  • You need to expand and/or simplify first

    • e.g. $\int 8 x^{2} ( 2 x - 3 ) \mathrm{d} x$
    • expand $8 x^{2} ( 2 x - 3 )$ as $16x^{3}-24x^{2}$
    • then integrate term by term
    • you cannot just multiply the integrals of $8x^{2}$ and $2x-3$ together

What might I be asked to do once I’ve integrated?

• You may be given the derivative of a function and asked to find the function

  • Integration and differentiation are inverse operations so

    • $\int \mathrm{f}^{'} ( x ) \mathrm{d} x = \mathrm{f} ( x ) + c$
    • $\int \frac{ \mathrm{d} y}{ \mathrm{d} x} \mathrm{d} x = y + c$

• With more information the constant of integration,$c$, can be found

• The area under a curve can also be found using integration

Exam Hint

• Remember the basic pattern of integrating powers of x

  • 'Raise the power by one and divide by the new power'
  • Lots of practice will improve your speed and accuracy

• It's easy to check your answer when integrating

  • Just differentiate your answer
  • It should turn back into the function you were integrating in the first place

Worked Example Given that $\frac{dy}{dx}=2x^{2}+3-\frac{1}{\sqrt{x}}$, find an expression for$y$ in terms of$x$.

Start by rewriting $\frac{dy}{dx}$entirely in powers of $x$

By laws of indices $\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$

$\frac{dy}{dx}=2x^{2}+3-x^{-\frac{1}{2}}$

Remember $\int \frac{ \mathrm{d} y}{ \mathrm{d} x} \mathrm{d} x = y + c$

We can integrate term by term using \int a x^{n \mathrm{d} x = \frac{a x^{n} + 1} }{n + 1} + c

$\begin{gathered} y = \int ( 2 x^{2} + 3 - x^{- 1 \frac{1}{2}} ) \mathrm{d} x \\ = 2 ( \frac{x^{2 + 1} }{2 + 1} ) + 3 x - \frac{x^{- 1 \frac{1}{2} + 1} }{- 1 \frac{1}{2} + 1} + c \\ = 2 ( \frac{x^{3}}{3} ) + 3 x - \frac{x^{1 \frac{1}{2}} }{1 \frac{1}{2}} + c \\ = \frac{2}{3} x^{3} + 3 x - 2 x^{1 \frac{1}{2}} + c \end{gathered}$

We can't find the value of $c$ without further info, so that's the answer to the question

It's 'nice' to turn $x^{\frac{1}{2}}$ back into$\sqrt{x}$ for the final answer, but you would also get the marks without doing that

$y=\frac{2}{3}x^{3}+3x-2\sqrt{x}+c$

Integrating Trig Functions

Spec point — 9A

How do I integrate sin and cos?

• You can integrate $\mathrm{sin}x$ and $\mathrm{cos}x$ by using the formulae

  • $\int \sin x d x = - \cos x + c$

  • $\int \cos x d x = \sin x + c$

    • $c$ is the constant of integration

• If $x$ is multiplied by a constant $a$ then

  • $\text{integral} \text{sin} a x d x = - \frac{1}{ } a \text{cos} a x + c$
  • $\text{integral} \text{cos} a x d x = \frac{1}{ } a \text{sin} a x + c$

• None of these formulae are on the exam formula sheet, so you need to remember them

• For calculus with trigonometric functions angles must be measured in radians

  • Make sure you know how to change the angle mode on your calculator

Exam Hint

• Remember to include 'c', the constant of integration, for any indefinite integrals

• As soon as you see a question involving integration and trigonometry

  • put your calculator into radians mode

Worked Example Given that $\mathrm{f}^{'} ( x ) = 4 \cos 3 x - 1 \frac{1}{2} \sin 2 x$, find an expression for $\mathrm{f} ( x )$ in terms of $x$.

Remember that $\int \mathrm{f}^{'} ( x ) \mathrm{d} x = \mathrm{f} ( x ) + c$

We can integrate term by term using the integration formulae for $\mathrm{sin}ax$ and $\mathrm{cos}ax$

$\begin{array}{rcl} \mathrm{f} ( x ) & = \int ( 4 \cos 3 x - 1 \frac{1}{2} \sin 2 x ) \mathrm{d} x \\ = 4 ( 1 \frac{1}{3} \sin 3 x ) - 1 \frac{1}{2} ( - 1 \frac{1}{2} \cos 2 x ) + c \\ = \frac{4}{3} \sin 3 x + 1 \frac{1}{4} \cos 2 x + c \end{array}$

We can't find the value of $c$ without further info, so that's the answer to the question

$f(x)=\frac{4}{3}\mathrm{sin}3x+\frac{1}{4}\mathrm{cos}2x+c$

Integrating e^x

Spec point — 9A

How do I integrate exponentials?

• $e^{x}$ can be integrated using the formula

  • $\text{integral} e^{ } x d x = e^{ } x + c$

    • $c$ is the constant of integration

• If $x$ is multiplied by a constant $a$ then

  • $\text{integral} e^{ } a x d x = \frac{1}{ } a e^{ } a x + c$

• These formulae are not on the exam formula sheet, so you need to remember them

Exam Hint

• Because '$e^{x}$ is its own integral' it is quite easy to integrate exponentials

  • Just be careful dealing with the constant in $e^{ax}$

Worked Example Given that $\mathrm{f}^{'} ( x ) = \frac{ \mathrm{e}^{2 x} - \mathrm{e}^{- 3 x} }{2}$, find an expression for $\mathrm{f} ( x )$ in terms of $x$.

Remember that $\int \mathrm{f}^{'} ( x ) \mathrm{d} x = \mathrm{f} ( x ) + c$

We can integrate term by term using the integration formula for $e^{ax}$

You might find it easier to split the fraction into two separate fractions first

$\begin{gathered} \mathrm{f} ( x ) & = \int ( 1 \frac{1}{2} \mathrm{e}^{2 x} - 1 \frac{1}{2} \mathrm{e}^{- 3 x} ) \mathrm{d} x \\ = 1 \frac{1}{2} ( 1 \frac{1}{2} \mathrm{e}^{2 x} ) - 1 \frac{1}{2} ( \frac{1}{- 3} \mathrm{e}^{- 3 x} ) + c \\ = 1 \frac{1}{4} \mathrm{e}^{2 x} + \frac{1}{6} \mathrm{e}^{- 3 x} + c \end{gathered}$

We can't find the value of $c$ without further info, so that's the answer to the question

$f(x)=\frac{1}{4}e^{2x}+\frac{1}{6}e^{-3x}+c$