How can the trapezium rule be used to estimate a definite integral?

Are we comfortable with an integral being the area under a curve on a graph? Well we'll draw a curve and think about how we could estimate this area. Would you agree that we could chop up the area under the curve with a number rectangles, or trapezia and add up the area of these shapes to estimate the integral? This is the trapezium rule- and we write it like :\int_{a}^{b} f(x) dx \approx h/2 [(y₀ + y_n) + 2(y₁ + y₂ + ... + y_n-1)]. Where we split the interval [a, b] into n evenly spaced x₀, x₁, ..., x_n with width h, and y_n=f(x_n).Derive where the formula comes from (has to be done by diagram really, but shows the construction of trapezia and sums the areas of each trapezia using A_i = h/2 (y_i-1 + y_i) and adding up these).Let's use an example to unpack how we're going to answer questions on this. Let's say we have a curve f(x) = x² and we are trying to estimate \int_{1}^{2} x² dx. Let's split our interval into 4 trapezia. We know that h is the width of each trapezium. The total width is 1 (2-1) so h = 0.25. x₀ = a so here x₀ = 1 then we keep adding our h until we hit b, in this case 2.x₁= 1.25x₂= 1.50x₃= 1.75x₄ = 2. We then create a table with our y_n s in. x_i y_i 1 1 1.25 1.5625 1.5 2.25 1.75 3.0625 2 4
Now we can but this into our formula: \int_{1}^{2} x² dx \approx 4/2 [ (1+4) + 2*(1.5625 + 2.25+3.6025)]= 2*(5 + 14.83) = 39.66.
Can we compare this to the exact answer?