Area under curves
The trapezium rule
To find the area under a curve, we have to split the space into very thin strips and look at them individually. Each strip is approximately the same shape as a trapezium, and we use the trapezium formula to estimate the area under the curve.
If we looked at the total area of the trapeziums, you would get:
\({A} = \frac {1} {2}({y_{first}}~+~{y_1})h~+~\frac {1} {2}(y_1~+~y_2)h~+~\frac {1} {2}(y_2~+~y_3)h~+~\)
\(\frac {1} {2}(y_3~+~y_4)h~+~\frac {1} {2}(y_4~+~y_{last})h\)
If we factorise the common factors, the following is true:
\({A} = \frac {1} {2}\times{h}({y_{first}}~+~y_1~+~y_1~+~y_2~+~y_2~+~\)
\( y_3~+~y_3~+~y_4~+~y_4~+~y_{last})\)
You may notice that there are two of all the y values apart from the first and last y values. This means we can simplify the formula to:
\({A} = \frac {1} {2}\times{h}({y_{first}}~+~y_{last}~+~2(y_1~+~y_2~+~y_3~+~y_4))\)
This formula can be made more general for any amount of trapeziums.
This is called the trapezium rule:
\({A} = \frac {1} {2}\times{h}({y_{first}}~+~y_{last}~+~2(Sum~of~the~rest))\)
Example
Use the trapezium rule, with the values from the table to estimate the area under the curve between \(\text{x = 0}\) and \(\text{x = 8}\).
Write out the formula we are using.
\({A} = \frac {1} {2}\times{h}({y_{first}}~+~y_{last}~+~2(Sum~of~the~rest))\)
Substitute the correct values in your formula. Don’t forget the numbers inside the brackets are the ‘y’ values and the height is the gap between the x values, 2 in this case.
\({A} = \frac {1} {2}\times{2}({5}~+~{4.5}~+~2({6}~+~{6.2}~+~{4.8}))\)
At this point we could just use our calculators. However if we didn’t have a calculator, we work out the inner brackets first.
\({A} = \frac {1} {2}\times{2}({5}~+~{4.5}~+~2({17}))\)
Then multiply it out by 2.
\({A} = \frac {1} {2}\times{2}({5}~+~{4.5}~+~{34})\)
Then it’s the second bracket’s turn.
\({A} = \frac {1} {2}\times{2}\times({43.5})\)
Finally multiply all the values together.
\({A} = 43.5\) square units
Velocity-Time graph
If we find the area under velocity-time graphs, then we actually find the distance travelled for the section of graph we are using.
Question
The graph and table shows information about the velocity of a motor bike. Estimate the distance travelled in the first minute of the motorbike’s journey.
\({A} = \frac {1} {2}\times{h}({y_{first}}~+~y_{last}~+~2(Sum~of~the~rest))\)
\({A} = \frac {1} {2}\times{10}\times~({0}~+~{4}~+~2({0.5}~+~{1}~+~{1.5}~+~{2}~+~{3}))\)
\({A} = {100}\)
The total distance is 100 m.
You may not always be given a table, sometimes you will have to read the values from the graphs itself.
Example
The following graph shows a journey made by a motorcyclist on returning home from the nearest bike shop.
Use the trapezium rule with five strips to approximate the distance travelled between \(\text {t = 3}\) and \(\text {t = 18}\) seconds.
The distance needed is between \(\text {t = 3}\) and \(\text {t = 18}\), and the corresponding \(\text {y}\) coordinates would be:
\({t} = {3},~{y_{first}} = {5}\)
\({t} = {6},~{y_2} = {10}\)
\({t} = {9},~{y_3} = {20}\)
\({t} = {12},~{y_4} = {35}\)
\({t} = {15},~{y_5} = {55}\)
\({t} = {18},~{y_{last}} = {55}\)
\({A} = \frac {1} {2}\times{h}({y_{first}}~+~y_{last}~+~2(Sum~of~the~rest))\)
\({A} = \frac {1} {2}\times{3}\times({5}~+~55~+~2({10}~+~20~+~35~+~55))\)
\({A} = {450}~{m}\)
