This graph appeared in an exam question
They calculated the area by using $\frac{1}{2} \times (6+3.8) \times 8$.
I'm really confused because I only know how to find the area under the curve using 3 methods and it doesn't seem to use any of them:
- Trapezium Rule
- Simpson Rule
- Counting squares
What method are the answers using? I'm perplexed as to where 3.8 came from.
If you draw a trapezium with corners on the points (0,6), (8, 3.8), (8,0) and (0,0) over the figure, you will see that this trapezium has approximately the same area as the area under the curve that you are trying to approximate.
The formula is the formula for the area of this trapezium. See here for an explanation. The area of the trapezium is given by $\frac{1}{ 2}\times(\text{length of longest side}+ \text{length of shortest side}) \times \text{height}$. In our case, the length of the longest side is 6, the shortest side is 3.8 and the height is 6.
The reason 3.8 was chosen is because it is a fair approximation. If a lower number, like 3, was chosen, there would still be a lot of the curve outside of the trapezium and the trapezium would be too small. If a higher number, like 4 was chosen, the trapezium would be too big and your answer would be an overestimate. Try drawing the trapeziums with side lengths 3 and 4 if you are struggling to see this.
As far as I can see, there is no hard-and-fast rule that has been used to get 3.8. You need to be able to see that it is a good approximation. 3.7 or 3.9 would also do a decent job. I imagine that the exam mark scheme would allow for a range of values, say between 3.2 and 4.