Calculating Areas (Edexcel IGCSE Further Maths)

Definite Integrals

Spec point — 9A

What is a definite integral?

• A definite integral is defined by the following formula

  • $\int_{a}^{b} \mathrm{f}^{'} ( x ) \mathrm{d} x = [ \mathrm{f} ( x ) ]_{a}^{b} = \mathrm{f} ( b ) - \mathrm{f} ( a )$

    • i.e. integrate as usual to find $\mathrm{f} ( x )$
    • then substitute to find $\mathrm{f} ( b )$ and $\mathrm{f} ( a )$
    • and subtract $\mathrm{f} ( a )$ from $\mathrm{f} ( b )$ to find the value of the definite integral

  • $a$ and $b$ are numbers and are called the integration limits

    • $a$ is the lower limit
    • $b$ is the upper limit
    • the integral is 'from $a$ to $b$'

  • A constant of integration (“$+c$”) is not needed with definite integrals

• Note that the answer to a definite integral is a number

  • The answer to an indefinite integral is another function

Exam Hint

• Be careful when substituting in to find $\mathrm{f} ( b ) - \mathrm{f} ( a )$

  • It's quite easy to make mistakes here
  • Especially when fractions and negative numbers are involved

• Your calculator may be able to find the value of definite integrals

  • You can use this to check your work

• Look out for phrases in exam questions like "Use algebraic integration" or "Using calculus"

  • These mean that full working out of the integral 'by hand' is required
  • A calculator answer without working would not score marks

Worked Example Show that

$\int_{2}^{4} 3 x ( x^{2} - 2 ) \mathrm{d} x = 144$

Start by expanding the brackets inside the integral

$\int_{2}^{4} ( 3 x^{3} - 6 x ) \mathrm{d} x$

Integrate as usual (here it's a 'powers of $x$' integration)

Write the answer in square brackets with the integration limits outside

$\begin{gathered} \int_{2}^{4} ( 3 x^{3} - 6 x ) \mathrm{d} x & = [ 3 ( \frac{x^{3 + 1} }{3 + 1} ) - 6 ( \frac{x^{1 + 1} }{1 + 1} ) ]_{2}^{4} \\ = [ \frac{3}{4} x^{4} - 3 x^{2} ]_{2}^{4} \end{gathered}$

Now substitute 4 into that function And subtract from it the function with 2 substituted in

$\begin{gathered} [ \frac{3}{4} x^{4} - 3 x^{2} ]_{2}^{4} & = ( \frac{3}{4} ( 4 )^{4} - 3 ( 4 )^{2} ) - ( \frac{3}{4} ( 2 )^{4} - 3 ( 2^{2} ) ) \\ = ( 192 - 48 ) - ( 12 - 12 ) \\ = 144 - 0 \\ = 144 \end{gathered}$

And that's the answer we were asked to show!

$integral_{ } 2^{ } 4 3 x \text{stretchy} ( x ^{2} - 2 \text{stretchy} ) d x = 144$

Calculating Areas with Definite Integrals

Spec point — 9C

How can I calculate areas using definite integrals?

• Region $R$ in the following diagram is the 'area under a curve' between $x=a$ and $x=b$

  • It is the region bounded by

    • the curve $y = \mathrm{f} ( x )$
    • the $x$-axis
    • and the lines $x=a$ and and $x=b$

• The exact area of a region like region $R$ in the diagram can be found by evaluating the definite integral

  • $\text{Area} = \int_{a}^{b} \mathrm{f} ( x ) \mathrm{d} x$

    • i.e. definite integrals can be used as area calculators!

  • Note that for this to work, $a$ must be 'on the left' and $b$ must be 'on the right'

    • i.e. $a\leq b$

How do I form a definite integral to find the area under a curve?

• The curve $y=f(x)$ and the $x$-axis should be obvious boundaries

  • but the trick can be identifying $a$ and $b$

    • i.e. the lower and upper limits of the definite integral

• If a diagram is given in the question, this can help locate the limits

  • If a diagram is not given, then sketch one

• The 'left' and 'right' boundaries may be vertical lines

  • In that case their equations give the integral limits

    • $x=a$ and $x=b$

• The $y$-axis may be one of the (vertical) boundaries

  • in that case one of the limits will be $x=0$

• The 'left' and 'right' boundaries don't have to be vertical lines

  • One or both of them could be where $y = \mathrm{f} ( x )$ intercepts the$x$-axis

    • i.e., one of the roots of the equation $f(x)=0$

  • In this case solve the equation$f(x)=0$ to find the limit(s)

Exam Hint

• Look out for questions that ask you to find an indefinite integral in one part

  • where '$+c$' is needed in the answer

  • then in a later part use the same integral as a definite integral

    • where '$+c$' is not needed

• Add information to any diagram provided in the question

  • axes intercepts
  • values of limits
  • mark and shade the area you’re trying to find

• If no diagram is provided, sketch one!

Worked Example The following diagram shows a part of the graph of the curve with equation $y=3+2x-x^{2}$. The region labelled $R$ is bounded by the curve, the positive $y$-axis, and the positive $x$-axis.

Find the exact area of region $R$.

Start by finding the integration limits.

The $y$-axis is the left boundary, so $x=0$ will be the lower integration limit

The right boundary is where the curve intercepts the $x$-axis Solve the equation $y=0$ to find its $x$-coordinate

$\begin{gathered} 3 + 2 x - x^{2} & = 0 \\ x^{2} - 2 x - 3 & = 0 \\ ( x + 1 ) ( x - 3 ) & = 0 \end{gathered}$

$x=-1\mathrm{or}x=3$

We want the positive intercept, so the upper integral limit will be $x=3$

Put all that info into the definite integral

$\begin{array}{rcl} \text{Area} = \int_{0}^{3} ( 3 + 2 x - x^{2} ) \mathrm{d} x \\ = [ 3 x + 2 ( \frac{x^{1 + 1} }{1 + 1} ) - ( \frac{x^{2 + 1} }{2 + 1} ) ]_{0}^{3} \\ = [ 3 x + x^{2} - 1 \frac{1}{3} x^{3} ]_{0}^{3} \\ = ( 3 ( 3 ) + ( 3 )^{2} - 1 \frac{1}{3} ( 3 )^{3} ) - ( 3 ( 0 ) + ( 0 )^{2} - 1 \frac{1}{3} ( 0 )^{3} ) \\ = ( 9 + 9 - 9 ) - ( 0 + 0 - 0 ) \\ = 9 - 0 \\ = 9 \end{array}$

$9\mathrm{units}^{2}$

Negative Integrals

Spec point — 9A

What do we mean by a 'negative integral'

• The answer to a definite integral is a number

  • This number can be positive or negative (or zero!)

• The area between a curve and the$x$-axis may lie fully or partially below the $x$-axis

  • This occurs when the function$f(x)$ takes negative values within the boundaries of the area

• A definite integral used to find such an area

  • will be negative if the area is fully under the$x$-axis

  • possibly negative if the area is partially under the$x$-axis

    • even if positive, the integral will not calculate the correct area in this case
    • the 'negative areas' will subtract from the 'positive areas'

How do I find the area under a curve when the curve is fully under the x-axis?

• STEP 1 Write the definite integral to find the area as usual

  • This may involve finding the lower and upper integration limits

• STEP 2 The answer to the definite integral will be negative

  • But the answer will have the same 'size' as the area

  • So just remove the minus sign to get the area

    • e.g. if the value of the integral is $-36$
    • then the area will be $36$ (square units)

How do I find the area under a curve when the curve is partially under the x-axis?

• STEP 1 Split the area into parts

  • the area(s) that are above the $x$-axis
  • and the area(s) that are below the $x$-axis

• STEP 2 Write the definite integral for each part

  • This may involve finding the lower and upper integration limits for each part

    • e.g. solving $\mathrm{f} ( x ) = 0$ to find where $y = \mathrm{f} ( x )$ crosses the $x$-axis

• STEP 3 Find the value of each definite integral separately

• STEP 4 Change the negative values to positive

  • Then find the total area by summing the 'positive versions' of each integral

Exam Hint

• If no diagram is provided, sketch one

  • This lets you see where the curve is above and below the $x$-axis
  • Then you can split up your integrals accordingly

Worked Example The diagram below shows the graph of $y=f(x)$, where $f(x)=(x+4)(x-1)(x-5)$.

The region$R_{1}$ is bounded by the curve $y=f(x)$ and the positive $x$- and $y$-axes.

The region$R_{2}$ is bounded by the curve $y=f(x)$, the positive $x$-axis, and the line $x=3$.

(a) Determine the coordinates of the point labelled$P$.

That point is one of the $x$-axis intercepts of the function

Solve $\mathrm{f} ( x ) = 0$ to find what those area

$( x + 4 ) ( x - 1 ) ( x - 5 ) = 0$

$x=-4,1,5$

It has to be positive, because the point is to the right of the origin And it has to be less than 3, because it's to the left of the line $x=3$ So it must be the intercept at $x=1$

$(1,0)$

(b) Find the exact total area of the shaded regions $R_{1}$ and $R_{2}$.

$R_{1}$ is the region between $x=0$ and $x=1$

It lies totally above the $x$-axis, so the definite integral will calculate the area directly

| Area of R _ 1 | = integral _ 0 ^ 1 ( x + 4 ) ( x - 1 ) ( x - 5 ) d x | | _ = integral _ 0 ^ 1 ( x ^3 - 2 x ^2 - 19 x + 20 ) d x | | _ = [ 1 fourth x ^ 4 - 2 / 3 x ^3 - 19 / 2 x ^2 + 20 x ] _ 0 ^ 1 | | _ = ( 1 fourth ( 1 ) ^ 4 - 2 / 3 ( 1 ) ^3 - 19 / 2 ( 1 ) ^2 + 20 ( 1 ) ) - ( 1 fourth ( 0 ) ^ 4 - 2 / 3 ( 0 ) ^3 - 19 / 2 ( 0 ) ^2 + 20 ( 0 ) ) | | _ = ( 1 fourth - 2 / 3 - 19 / 2 + 20 ) - ( 0 - 0 - 0 + 0 ) | | _ = 121 / 12 - 0 | | _ = 121 / 12 | |

$R_{2}$ is the region between $x=1$ and $x=3$

It lies totally below the $x$-axis, so the integral will give us the negative version of the area

| - ( Area of R _ 2 ) | = integral _ 1 ^ 3 ( x ^3 - 2 x ^2 - 19 x + 20 ) d x | | _ = [ 1 fourth x ^ 4 - 2 / 3 x ^3 - 19 / 2 x ^2 + 20 x ] _ 1 ^ 3 | | _ = ( 1 fourth ( 3 ) ^ 4 - 2 / 3 ( 3 ) ^3 - 19 / 2 ( 3 ) ^2 + 20 ( 3 ) ) - ( 1 fourth ( 1 ) ^ 4 - 2 / 3 ( 1 ) ^3 - 19 / 2 ( 1 ) ^2 + 20 ( 1 ) ) - | | _ = ( 81 / 4 - 18 - 171 / 2 + 60 ) - ( 1 fourth - 2 / 3 - 19 / 2 + 20 ) | | _ = - 93 / 4 - 121 / 12 | | _ = - 400 / 12 | |

So the area of $R_{2}$ is $\frac{400}{12}$* *$( = \frac{100}{3} )$

Add the two areas together to get the total area

$\mathrm{total} \mathrm{area}=\frac{121}{12}+\frac{400}{12}=\frac{521}{12}$

Final answer: $\frac{521}{12}\mathrm{units}^{2}$