Question

For the definite integral of f(x)=-0.1x^2+7 from x=1 to x=4.

Approximate the integral by finding T10, T20, and T50 using the trapezoidal rule. do these valuse overestimate the integral or underestimate it?

Answer 1

So that is what your function looks like.
Now to do trapezoidal approximation, you'll fill in different portions of these equations:
\[\Delta x = \frac{ b-a }{ n }\]
\[A = \frac{ 1 }{ 2 }\Delta x[f(a)+2f(a+\Delta x)+2f(a+2\Delta x)+...+2f(b-2\Delta x)+2f(b-\Delta x)+f(b)]\]
Delta X is your step size, meaning how far you move right before you take another chunk, so as you use more an more trapezoids, you'll see that step size get smaller and smaller.

A is the definite integral of that curve as approximated by the trapezoidal rule. And while that second equation looks ridiculous, it's actually pretty simple (when you aren't using more than 10 or so trapezoids haha).

Now if the problem is to do this by hand, I'm sorry but that'll take ages. I just Googled for a trapezoidal rule calculator and found the following values for T10, T20, and T50, where it plugs your a and b values into your function and allows you to set a custom amount of trapezoids for the computation:

(1/2) * (0.3) * [ f(1) + 2*f(1.3) + 2*f(1.6) + 2*f(1.9) + 2*f(2.2) + 2*f(2.5) + 2*f(2.8) + 2*f(3.1) + 2*f(3.4) + 2*f(3.7) + f(4) ]
T10 = 18.8955

(1/2) * (0.15) * [ f(1) + 2*f(1.15) + 2*f(1.3) + 2*f(1.45) + 2*f(1.6) + 2*f(1.75) + 2*f(1.9) + 2*f(2.05) + 2*f(2.2) + 2*f(2.35) + 2*f(2.5) + 2*f(2.65) + 2*f(2.8) + 2*f(2.95) + 2*f(3.1) + 2*f(3.25) + 2*f(3.4) + 2*f(3.55) + 2*f(3.7) + 2*f(3.85) + f(4) ]
T20 = 18.898875

(1/2) * (0.06) * [ f(1) + 2*f(1.06) + 2*f(1.12) + 2*f(1.18) + 2*f(1.24) + 2*f(1.3) + 2*f(1.36) + 2*f(1.42) + 2*f(1.48) + 2*f(1.54) + 2*f(1.6) + 2*f(1.66) + 2*f(1.72) + 2*f(1.78) + 2*f(1.84) + 2*f(1.9) + 2*f(1.96) + 2*f(2.02) + 2*f(2.08) + 2*f(2.14) + 2*f(2.2) + 2*f(2.26) + 2*f(2.32) + 2*f(2.38) + 2*f(2.44) + 2*f(2.5) + 2*f(2.56) + 2*f(2.62) + 2*f(2.68) + 2*f(2.74) + 2*f(2.8) + 2*f(2.86) + 2*f(2.92) + 2*f(2.98) + 2*f(3.04) + 2*f(3.1) + 2*f(3.16) + 2*f(3.22) + 2*f(3.28) + 2*f(3.34) + 2*f(3.4) + 2*f(3.46) + 2*f(3.52) + 2*f(3.58) + 2*f(3.64) + 2*f(3.7) + 2*f(3.76) + 2*f(3.82) + 2*f(3.88) + 2*f(3.94) + f(4) ]
T50 = 18.89982

So you can see that as the amount of trapezoids you use gets larger and larger, the area gets a little bit more precise and increases a little bit. That's important to note when you look at the second question there, asking about underestimating or overestimating, and it's pretty easy to see if you imagine drawing in those trapezoids like so:

I apologize for the super crappy drawing but you can see that with the trapezoids, there is a tiny bit of area that isn't accounted for, which makes sense when you see that as you increase the amount of trapezoids, that area gets larger, because the trapezoids' sides gets closer and closer to actually fitting the line. Think of it kinda like this:

With two trapezoids, you see a big area completely left out, but as you make more trapezoids, they'll come closer to that line.
All in all, it just means that this trapezoidal approximation is an underestimate of the true integral in this scenario.



Sorry for such a long answer :P