Hey there!
With multiple applications of the trapezoidal rule on some function \(f(x)\) over some domain \(x \in [a, b]\) partitioned by x-values \(x_1, x_2 ... x_n\) such that \(a < x_1 < x_2 < ... < x_n < b\), we have that the area obtained by using the trapezoidal rule is \(\frac{h}{2}\left(f(a) + f(b) + 2(f(x_1) + f(x_2)+...+ f(x_n))\right)\), assuming, of course, that each of the partitions are equally spaced, represented by the value h \((\text{ie. }x_1 - a = x_2 - x_1 = x_3 - x_2 = ... = b - x_n = h\)) . Essentially what this means is that we have the space between each given value of x, divided by 2, multiplied by the total of the sum of the first and last terms of the function output and the sum of twice all the middle terms.
On this application, we have that \(h = 5\), and that \(f(a) = 0, f(b) = 0\). Thus, we have that the approx. area is roughly \(\frac{5}{2} (0+ 0 +2(10.2 +13.2 + 16.3 + 13.8+16.0+17.0+18.6+10.8 )) \\ = 5(115.9) = 579.5 \text{ square metres}\). Try this with any other trapezoidal rule questions you might have!
Hope this helps