Calculating definite integrals

Some of the Edfinity problems assume that you have a "calculator" to figure out the value of a definite integral. Here's an example from the textbook:

So, for the homework go ahead and use one of the several "calculators" we have available. Below I'll show you WolframAlpha and CoCalc for that integral, $$\int_0^2 2^t\,dt.$$

CoCalc

Here is a longer example of how to use CoCalc's "Integrate" command to calculate a definite integral:

integrate.pdf

It also has a clarification of one of the Edfinity problems.

WolframAlpha.com

On a test

But on a test you won't have any such calculator available. So I'll expect you to use an approximation method. Two possibilities (among others) To approximate $$\int_0^2 2^t\,dt.$$

...use geometry

Split up the area to be approximated into rectangles (area = $w\times h$) and triangles (area=$w\times h / 2$). E.g. Adding the areas of the triangle and the rectangle together I get $3+2=5$ so... $$\int_0^2 2^t\,dt\approx 5.$$ Visually, it looks like the area must be somewhat less than 5.

Left or right sums

I could calculate $f(t)=2^t$ for $t=0,1,2$: $f(0)=1$, $f(1)=2$, $f(2)=4$. A left sum (illustrated in the diagram) approximation of the area, with rectangles th at are 1 unit wide, would be: $$\text{left}=(f(0)*1+f(1)*1)=(1*1+2*1)=3.$$ A right sum approximation would be: $$\text{right}=(f(1)*1+f(2)*1)=(2*1+4*1)=6.$$ The average of these two values should be an even better approximation than either one individually: $$\int_0^2 2^t\,dt\approx \frac{\text{left}+\text{right}}2=9/2=4.5.$$ And 4.5 is within 4% of the exact answer, 4.328.